Contents / Real Analysis / Sequences & Series

Chapter 2

Sequences & Series

Limits of sequences by epsilon-N, the algebra of limits, squeeze and the standard limits; monotone convergence and Bolzano-Weierstrass; the Cauchy criterion and its relation to completeness; series, the harmonic divergence, and the full battery of convergence tests; absolute against conditional convergence with Riemann's rearrangement theorem; and power series with Cauchy-Hadamard and term-by-term operations.

Introduction

Everything in analysis is a limit of something simpler. A derivative is a limit of difference quotients, an integral a limit of finite sums, a continuous function one whose values respect limits. Before any of that can be made precise, the limit of a sequence has to be defined and its properties proved, because every other limiting process in the subject is reduced to this one.

This chapter does that in three movements. The first builds the theory of convergent sequences from the $ε$ – $N$ definition: how limits interact with arithmetic and order, and the two existence theorems — monotone convergence and Bolzano–Weierstrass — that produce a limit out of nothing but boundedness. Those two are where the completeness of $R$ , taken as an axiom in The Real Number System, does its work; without it they are false, as the rationals demonstrate.

The second movement asks when convergence can be recognised from inside the sequence, with no candidate limit to compare against. That is the Cauchy condition, and its equivalence with convergence is the most useful reformulation of completeness — and what makes series possible, since a series is almost never given together with its sum.

The third movement is series and power series. A series is not a new object; it is the sequence of its partial sums. But the questions change: one rarely wants the value of $\sum a_{n}$ , only whether it exists, and the convergence tests answer that from the shape of $a_{n}$ alone. Absolute convergence then separates the series whose sum is a genuine number from those whose sum is an artefact of the order of addition, and power series close the chapter by turning infinite sums back into functions that can be differentiated and integrated term by term.

Notation throughout: $(a_{n})$ denotes a sequence indexed by $n \in N$ , $\sum a_{n}$ a series, and $s_{N} = \sum_{n = 1}^{N} a_{n}$ its $N$ -th partial sum. The supremum and infimum, the triangle inequality and the completeness axiom are all taken from The Real Number System and used without reproof; every result cited from that chapter is named.

2.1Sequences as functions on the integers

Definition 2.1 (Sequence). A sequence in $R$ is a function $a : N \to R$ . The value at $n$ is written $a_{n}$ rather than $a (n)$ , and the sequence itself as $(a_{n})_{n = 1}^{\infty}$ , or simply $(a_{n})$ . The range of the sequence is the set ${a_{n} : n \in N}$ of values it takes.

Defining a sequence as a function rather than as "an infinite list" is what keeps order and repetition in the picture. The set ${1, 0, 1, 0, \dots}$ is just ${0, 1}$ , a two-element set; the sequence $a_{n} = \frac{1}{2} (1 + (- 1)^{n + 1})$ is an infinite object that visits those two values in a specific alternating order, and that order is the whole content of its divergence. Whenever an argument below says "the sequence", it is the function that is meant, never its range.

A sequence may be given by a closed formula, as $a_{n} = n / (n + 1)$ ; by a recursion, as $a_{1} = 1$ , $a_{n + 1} = \frac{1}{2} (a_{n} + 2/ a_{n})$ ; or by a description that admits neither, as " $a_{n}$ is the $n$ -th decimal digit of $π$ ". The recursive form matters most, because it is how sequences arise in practice — Newton's method, bisection, the partial sums of a series — and it is exactly the form for which no closed expression for $a_{n}$ can be inspected.

Intuition. A sequence is a film strip: frame $n$ shows the number $a_{n}$ . Rearranging the frames gives a different film even when the pictures are the same, and this is why the sequence, not its range, is the object of study.

The constant sequence $a_{n} = 5$ is a perfectly good, perfectly boring, sequence. So is one that lists the digits of $π$ , which has no pattern anyone can describe — but it is still a well-defined function on $N$ , and the theory applies to it verbatim.

Exercises for §2.1 →

2.2Monotonicity, boundedness, and subsequences

Two familiar families supply the standard examples. An arithmetic sequence has $a_{n} = a + (n - 1) d$ , a constant difference; a geometric sequence has $a_{n} = a r^{n - 1}$ , a constant ratio $r$ . The arithmetic ones are uninteresting for convergence — unless $d = 0$ they run off to $\pm \infty$ — while the geometric ones are the most important family in the chapter, since the ratio and root tests both work by comparing a series against a geometric one.

Three adjectives carry real weight in the theorems ahead, so they are stated once and precisely.

Definition 2.2 (Monotone, bounded, subsequence). A sequence $(a_{n})$ is increasing if $a_{n} \leq a_{n + 1}$ for every $n$ , decreasing if $a_{n} \geq a_{n + 1}$ for every $n$ , and monotone if it is one or the other. It is bounded above if there is $M$ with $a_{n} \leq M$ for all $n$ , bounded below if there is $m$ with $a_{n} \geq m$ for all $n$ , and bounded if there is $M$ with $∣ a_{n} ∣ \leq M$ for all $n$ .

A subsequence of $(a_{n})$ is a sequence $(a_{n_{k}})_{k = 1}^{\infty}$ where $n_{1} < n_{2} < n_{3} < \dots$ is a strictly increasing sequence of indices.

Two remarks on the definition. "Increasing" here permits equality, so a constant sequence is both increasing and decreasing; when strictness is needed it is said explicitly. And a subsequence must keep the original order — $(a_{2}, a_{1}, a_{4}, a_{3}, \dots)$ is a rearrangement, not a subsequence, and the two behave completely differently, a difference the rearrangement theorem at the end of the chapter turns into a theorem. Since $n_{k} \geq k$ always (a strictly increasing sequence of positive integers grows at least as fast as the index), a subsequence reaches as far out into the tail as the original does, which is why subsequences see the limit and not the early terms.

Example 2.3 (Classifying three sequences). Decide for each of $a_{n} = 3 (1/2)^{n - 1}$ , $b_{n} = (- 1)^{n}$ , and $c_{n} = n / (n + 1)$ whether it is monotone and whether it is bounded, and exhibit one convergent subsequence of $(b_{n})$ .

Solution.

$(a_{n})$ : the ratio $a_{n + 1} / a_{n} = 1/2 < 1$ with all terms positive, so it is decreasing, hence monotone. It is bounded: $0 < a_{n} \leq 3$ .
$(b_{n})$ : the terms are $- 1, 1, - 1, 1, \dots$ , so $b_{1} < b_{2}$ but $b_{2} > b_{3}$ — neither increasing nor decreasing, hence not monotone. It is bounded, with $∣ b_{n} ∣ = 1$ .
$(c_{n})$ : from $c_{n + 1} - c_{n} = \frac{n + 1}{n + 2} - \frac{n}{n + 1} = \frac{1}{( n + 1 ) ( n + 2 )} > 0$ it is increasing, and $0 < c_{n} < 1$ , so it is bounded.
A convergent subsequence of $(b_{n})$ : take $n_{k} = 2 k$ , giving the constant subsequence $b_{2 k} = 1$ for all $k$ , which converges to $1$ . Taking $n_{k} = 2 k - 1$ instead gives the constant $- 1$ . A bounded sequence can therefore have subsequences with different limits, which is precisely why boundedness alone cannot force convergence.

□

Exercises for §2.2 →

2.3Convergence and divergence, informally

The sequence $a_{n} = 1/ n$ has terms $1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \dots$ and one wants to say they "approach $0$ ". The sequence $b_{n} = (- 1)^{n}$ has terms $- 1, 1, - 1, \dots$ and one wants to say they approach nothing. Both intuitions are correct, and the whole difficulty is that neither phrasing survives contact with a harder example.

Three candidate descriptions of convergence show where the difficulty lies. "The terms get closer and closer to $L$ " is satisfied by $a_{n} = 1 - 1/ n$ with $L = 5$ : each term is strictly closer to $5$ than the last, and getting closer is not getting close. "The terms eventually equal $L$ " is far too strong: no term of $1/ n$ equals $0$ . "The terms get arbitrarily close to $L$ " is ambiguous: $(- 1)^{n}$ has infinitely many terms exactly equal to $1$ and still does not converge to $1$ , because the terms do not stay there.

The correct statement needs both halves: arbitrarily close, and from some point onward permanently. Divergence correspondingly has distinct flavours. A sequence can be unbounded and run to infinity ( $a_{n} = n$ ), bounded and oscillate ( $(- 1)^{n}$ ), or both ( $(- 1)^{n} n$ ). Only the first is written $a_{n} \to + \infty$ , an abbreviation for "for every $M$ there is $N$ such that $n > N$ implies $a_{n} > M$ ". There is no real number $\infty$ , so such a sequence has no limit in $R$ ; it diverges, with a description of how.

Intuition. Convergence means the film settles: after some frame, every single picture is inside a target zone of your choosing, no matter how narrow the zone.

The oscillating film keeps flipping between two pictures forever. Some frames are exactly on target, but "some frames" is not "all frames from here on", and the definition insists on the latter.

Exercises for §2.3 →

2.4The $ε$ – $N$ definition

Definition 2.4 (Limit of a sequence). The sequence $(a_{n})$ converges to $L \in R$ , written $lim_{n \to \infty} a_{n} = L$ or $a_{n} \to L$ , if for every $ε > 0$ there exists $N \in N$ such that

n > N ⟹ ∣ a_{n} - L ∣ < ε .

A sequence is convergent if such an $L$ exists, and divergent otherwise.

Read the quantifiers in the order written, because they do not commute. The $ε$ is a challenge issued by an adversary; $N$ is the response, and it may depend on $ε$ — it must, for any sequence not eventually constant — but on nothing else. In particular $N$ may not depend on $n$ : it is chosen first, and then every index beyond it has to comply. Swapping the two quantifiers produces "there is $ε > 0$ such that for all $N$ …", which is a statement about a single tolerance and says essentially nothing.

The inequality $∣ a_{n} - L ∣ < ε$ is the same as $L - ε < a_{n} < L + ε$ , so the definition says: every horizontal band around $L$ , however thin, contains all but finitely many terms. That reformulation also makes obvious that convergence is a property of the tail alone — changing $a_{1}, \dots, a_{1000}$ to arbitrary values changes no limit, since $N$ can always be enlarged past $1000$ .

Figure 2.5. The sequence

a_{n} = 2 + 1/ n

inside the band

∣ y - 2∣ < ε

for

ε = 0.2

. The band is entered at

n = 6

and never left, so

N = 5

answers this challenge; a narrower band is entered later, and the definition asks only that some entry point exist for every width.

Proving a limit from the definition is always the same two steps, the first mathematics and the second bookkeeping. Bound $∣ a_{n} - L ∣$ by something simple — an estimate of the form $C / n$ or $C r^{n}$ , crude but easy to invert — and then choose $N$ to make that estimate smaller than $ε$ .

Example 2.6 (A limit from the definition). Prove that $a_{n} = \frac{3 n + 1}{n + 2} \to 3$ .

Solution.

Estimate the distance to the proposed limit, combining over a common denominator:

\frac{3 n + 1}{n + 2} - 3 = \frac{3 n + 1 - 3 n - 6}{n + 2} = \frac{5}{n + 2} .

Bound it crudely: $\frac{5}{n + 2} < \frac{5}{n}$ for every $n \geq 1$ . Working with $5/ n$ rather than $5/ (n + 2)$ costs nothing and inverts more cleanly.
Given $ε > 0$ , choose $N$ to be any integer with $N \geq 5/ ε$ ; such an $N$ exists by the Archimedean property of $R$ , itself a consequence of the definition Upper bound, supremum, infimum applied to $N$ .
Verify: if $n > N$ then $n > 5/ ε$ , so $5/ n < ε$ , and therefore $∣ a_{n} - 3∣ = \frac{5}{n + 2} < \frac{5}{n} < ε$ . Since $ε$ was arbitrary, $a_{n} \to 3$ .
Concrete check at $ε = 0.01$ : the rule gives $N = 500$ , and $a_{501} = 1504/503 \approx 2.99006$ , at distance $0.0099 < 0.01$ . The choice is not the sharpest — the inequality already holds at $n = 498$ — and it does not have to be: the definition asks for some $N$ , not the least.

□

Pitfall. A proof that opens "let $ε = 1/ n$ " has already failed. The quantifier order forbids $ε$ from depending on $n$ : $ε$ is fixed before $N$ is chosen, and $n$ does not exist yet when $ε$ is named. The same error in disguise is choosing $N$ in terms of $n$ , for instance "take $N = n /2$ ", which makes the implication vacuous.

Example 2.7 (Proving divergence from the definition). Show that $b_{n} = (- 1)^{n}$ does not converge.

Solution.

Negate the definition. To say $(b_{n})$ does not converge to $L$ is to say: there exists $ε > 0$ such that for every $N$ there is some $n > N$ with $∣ b_{n} - L ∣ \geq ε$ .
Fix any candidate $L$ and take $ε = 1$ . Beyond any $N$ there are both an even index $m$ with $b_{m} = 1$ and an odd index $k$ with $b_{k} = - 1$ .
If both $∣1 - L ∣ < 1$ and $∣ - 1 - L ∣ < 1$ held, the triangle inequality from The Real Number System would give

2 = ∣1 - (- 1) ∣ \leq ∣1 - L ∣ + ∣ L - (- 1) ∣ < 1 + 1 = 2,

a contradiction. So at least one of the two indices violates the tolerance. 4. Hence for every $N$ some $n > N$ has $∣ b_{n} - L ∣ \geq 1$ , and since $L$ was arbitrary, $(b_{n})$ converges to no real number. 5. Note what was not needed: only that two values a fixed distance apart both recur forever. This is the model argument for every oscillatory divergence.□

Exercises for §2.4 →

2.5Uniqueness of the limit and the algebra of limits

The definition permits a sequence to converge to $L$ ; it does not, on its face, forbid the same sequence from also converging to some other $M$ . That it cannot is the first theorem, and it is what licenses the notation $lim a_{n}$ , a symbol for the limit.

Theorem 2.8 (Uniqueness of limits). A sequence of real numbers has at most one limit: if $a_{n} \to L$ and $a_{n} \to M$ , then $L = M$ .

Proof. Suppose $a_{n} \to L$ and $a_{n} \to M$ with $L \neq = M$ , and set $ε = ∣ L - M ∣/2 > 0$ . By the definition Limit of a sequence applied with this $ε$ , there is $N_{1}$ with $∣ a_{n} - L ∣ < ε$ for $n > N_{1}$ , and $N_{2}$ with $∣ a_{n} - M ∣ < ε$ for $n > N_{2}$ . Choose any $n > max (N_{1}, N_{2})$ ; both inequalities hold at that single index. The triangle inequality from The Real Number System then gives

∣ L - M ∣ \leq ∣ L - a_{n} ∣ + ∣ a_{n} - M ∣ < ε + ε = ∣ L - M ∣,

so $∣ L - M ∣ < ∣ L - M ∣$ , which is absurd. Hence $L = M$ .∎

The proof is a template worth absorbing: two $ε$ -statements are made to hold simultaneously by taking the larger of the two thresholds, and then a single index is used to derive a contradiction. Every later proof that juggles two convergent sequences does exactly this, and the step "let $N = max (N_{1}, N_{2})$ " will be performed without further comment.

Theorem 2.9 (Algebra of limits). Suppose $a_{n} \to L$ and $b_{n} \to M$ , and let $c \in R$ . Then

a_{n} + b_{n} \to L + M, c a_{n} \to c L, a_{n} b_{n} \to L M,

and, provided $M \neq = 0$ , the terms $b_{n}$ are nonzero for all large $n$ and $\frac{a _{n}}{b _{n}} \to \frac{L}{M}$ .

Proof. Sum. Let $ε > 0$ . Choose $N_{1}$ with $∣ a_{n} - L ∣ < ε /2$ for $n > N_{1}$ and $N_{2}$ with $∣ b_{n} - M ∣ < ε /2$ for $n > N_{2}$ . For $n > max (N_{1}, N_{2})$ ,

∣ (a_{n} + b_{n}) - (L + M) ∣ \leq ∣ a_{n} - L ∣ + ∣ b_{n} - M ∣ < \frac{ε}{2} + \frac{ε}{2} = ε .

Scalar multiple. If $c = 0$ the sequence is constantly $0$ and there is nothing to prove. Otherwise, given $ε > 0$ choose $N$ with $∣ a_{n} - L ∣ < ε /∣ c ∣$ for $n > N$ ; then $∣ c a_{n} - c L ∣ = ∣ c ∣ ∣ a_{n} - L ∣ < ε$ .

Product. First note that $(a_{n})$ is bounded: applying the definition with $ε = 1$ gives $N_{0}$ with $∣ a_{n} ∣ \leq ∣ L ∣ + 1$ for $n > N_{0}$ , and the finitely many earlier terms are bounded by their maximum, so $∣ a_{n} ∣ \leq K$ for all $n$ , where $K = max (∣ a_{1} ∣, \dots, ∣ a_{N_{0}} ∣, ∣ L ∣ + 1)$ . Now insert and remove the cross term:

∣ a_{n} b_{n} - L M ∣ = ∣ a_{n} b_{n} - a_{n} M + a_{n} M - L M ∣ \leq ∣ a_{n} ∣ ∣ b_{n} - M ∣ + ∣ M ∣ ∣ a_{n} - L ∣ \leq K ∣ b_{n} - M ∣ + ∣ M ∣ ∣ a_{n} - L ∣.

Given $ε > 0$ , make the first summand less than $ε /2$ by taking $∣ b_{n} - M ∣ < ε / (2 K)$ and the second less than $ε /2$ by taking $∣ a_{n} - L ∣ < ε / (2 (∣ M ∣ + 1))$ ; both hold beyond the maximum of the two thresholds.

Quotient. It suffices to prove $1/ b_{n} \to 1/ M$ and then apply the product rule. Since $M \neq = 0$ , apply the definition to $(b_{n})$ with $ε = ∣ M ∣/2$ : there is $N_{1}$ such that $∣ b_{n} - M ∣ < ∣ M ∣/2$ for $n > N_{1}$ , whence $∣ b_{n} ∣ > ∣ M ∣/2 > 0$ — in particular $b_{n} \neq = 0$ there, which is the claim that the terms are eventually nonzero. For such $n$ ,

\frac{1}{b _{n}} - \frac{1}{M} = \frac{∣ M - b _{n} ∣}{∣ b _{n} ∣ ∣ M ∣} < \frac{2}{∣ M ∣ ^{2}} ∣ b_{n} - M ∣,

and given $ε > 0$ the right-hand side is below $ε$ once $∣ b_{n} - M ∣ < ε ∣ M ∣^{2} /2$ .∎

Three things in that proof deserve notice. The product rule needed $(a_{n})$ bounded, and boundedness came free from convergence — the third paragraph proves in passing that every convergent sequence is bounded, a fact used repeatedly below. The quotient rule needed $(b_{n})$ bounded away from zero, which convergence to a nonzero limit supplies but convergence alone does not. And every hypothesis is about limits already known to exist: writing $lim (a_{n} b_{n}) = lim a_{n} \cdot lim b_{n}$ for $a_{n} = n$ , $b_{n} = 1/ n$ produces $1 = \infty \cdot 0$ , which is what that hypothesis exists to prevent.

Remark (Limits and order). The same techniques give the order statement: if $a_{n} \to L$ , $b_{n} \to M$ and $a_{n} \leq b_{n}$ for all large $n$ , then $L \leq M$ . For if $L > M$ , taking $ε = (L - M) /2$ eventually puts $a_{n}$ above $(L + M) /2$ and $b_{n}$ below it, contradicting $a_{n} \leq b_{n}$ . Note that strict inequality is not preserved: $0 < 1/ n$ for every $n$ , yet the limits satisfy $0 \leq 0$ , not $0 < 0$ . Limits turn $<$ into $\leq$ , always.

Example 2.10 (Rational expressions in $n$ ). Compute $n \to \infty lim \frac{4 n ^{2} - 3 n + 7}{2 n ^{2} + n}$ , justifying each step by a named rule.

Solution.

Divide numerator and denominator by the highest power $n^{2}$ present in the denominator:

\frac{4 n ^{2} - 3 n + 7}{2 n ^{2} + n} = \frac{4 - 3/ n + 7/ n ^{2}}{2 + 1/ n} .

The sequence $1/ n \to 0$ , and $1/ n^{2} = (1/ n) (1/ n) \to 0$ by the product rule in the theorem Algebra of limits.
By the scalar and sum rules the numerator tends to $4 - 0 + 0 = 4$ and the denominator to $2 + 0 = 2$ .
The denominator's limit $2$ is nonzero, so the quotient rule applies and the limit is $4/2 = 2$ .
Sanity check at $n = 1000$ : the expression is $\approx 1.9985$ . Dividing by $n^{2}$ was not a trick but the only route to a form where the algebra of limits applies — as written, numerator and denominator both diverge, and the theorem says nothing about that.

□

Exercises for §2.5 →

2.6The squeeze theorem, monotone convergence, and Bolzano–Weierstrass

The algebra of limits computes new limits from old. The results in this section do something harder: they prove that a limit exists without producing it, which is the only way to handle a sequence given by a recursion or by a formula too complicated to invert.

Theorem 2.11 (Squeeze theorem). Suppose $g_{n} \leq a_{n} \leq h_{n}$ for all $n$ beyond some index, and that $g_{n} \to L$ and $h_{n} \to L$ . Then $a_{n} \to L$ .

Proof. Let $ε > 0$ and let $N_{0}$ be an index beyond which the inequalities $g_{n} \leq a_{n} \leq h_{n}$ hold. Choose $N_{1}$ with $∣ g_{n} - L ∣ < ε$ for $n > N_{1}$ and $N_{2}$ with $∣ h_{n} - L ∣ < ε$ for $n > N_{2}$ . For $n > max (N_{0}, N_{1}, N_{2})$ we have both $L - ε < g_{n}$ and $h_{n} < L + ε$ , so

L - ε < g_{n} \leq a_{n} \leq h_{n} < L + ε,

which is exactly $∣ a_{n} - L ∣ < ε$ .∎

Observe what the hypotheses do not include: nothing about $(a_{n})$ beyond the two inequalities. The theorem manufactures the existence of $lim a_{n}$ , which is why it is the tool of choice whenever the terms involve something uncontrollable like $sin n$ .

Proposition 2.12 (Four standard limits).

(i) \frac{1}{n ^{p}} \to 0 (p > 0), (ii) r^{n} \to 0 (∣ r ∣ < 1), (iii) n^{1/ n} \to 1, (iv) \frac{n ^{k}}{c ^{n}} \to 0 (c > 1, k \in N) .

Proof. (i) Given $ε > 0$ , take any integer $N > ε^{- 1/ p}$ , available by the Archimedean property. For $n > N$ , $n^{p} > ε^{- 1}$ and so $∣1/ n^{p} ∣ < ε$ .

(ii) If $r = 0$ the claim is trivial. Otherwise write $1/∣ r ∣ = 1 + h$ with $h > 0$ . By Bernoulli's inequality — or directly from the binomial theorem, all of whose terms are positive — $(1 + h)^{n} \geq 1 + nh > nh$ , so

∣ r^{n} ∣ = \frac{1}{( 1 + h ) ^{n}} < \frac{1}{nh} \to 0,

and the squeeze theorem with $g_{n} = 0$ finishes it.

(iii) Write $n^{1/ n} = 1 + d_{n}$ with $d_{n} \geq 0$ for $n \geq 1$ . The binomial theorem gives, for $n \geq 2$ ,

n = (1 + d_{n})^{n} \geq (2 n) d_{n}^{2} = \frac{n ( n - 1 )}{2} d_{n}^{2},

since every binomial term is non-negative. Rearranging, $d_{n}^{2} \leq 2/ (n - 1)$ , so $0 \leq d_{n} \leq 2/ (n - 1) \to 0$ by part (i), and the squeeze theorem gives $d_{n} \to 0$ , that is $n^{1/ n} \to 1$ .

(iv) Write $c = 1 + h$ with $h > 0$ and take $n > 2 (k + 1)$ . Keeping only the $(k + 1)$ -st binomial term,

c^{n} = (1 + h)^{n} \geq (k + 1 n) h^{k + 1} \geq \frac{( n /2 ) ^{k + 1}}{( k + 1 )!} h^{k + 1},

using $(k + 1 n) \geq (\frac{n - k}{k + 1})^{k + 1} \geq \frac{( n /2 ) ^{k + 1}}{( k + 1 )!}$ for $n$ that large. Hence

0 \leq \frac{n ^{k}}{c ^{n}} \leq \frac{2 ^{k + 1} ( k + 1 )!}{h ^{k + 1}} \cdot \frac{n ^{k}}{n ^{k + 1}} = \frac{C}{n},

with $C$ a constant independent of $n$ , and the squeeze theorem applies.∎

Part (iv) is the statement that exponentials beat polynomials, and it is used constantly in the series tests: it is why $\sum n^{10} / 2^{n}$ converges despite the enormous numerator.

Example 2.13 (Squeezing an uncontrollable numerator). Find $n \to \infty lim \frac{sin ( n ^{2} ) + cos n}{n}$ and $n \to \infty lim (3^{n} + 4^{n})^{1/ n}$ .

Solution.

For the first, no formula for $sin (n^{2})$ is available and none is needed. Since $∣ sin (n^{2}) ∣ \leq 1$ and $∣ cos n ∣ \leq 1$ , the numerator lies in $[- 2, 2]$ , so

- \frac{2}{n} \leq \frac{sin ( n ^{2} ) + cos n}{n} \leq \frac{2}{n} .

Both bounds tend to $0$ by the proposition Four standard limits (i) and the scalar rule, so the limit is $0$ by the theorem Squeeze theorem.
For the second, bound the inside between its largest term and twice that: $4^{n} \leq 3^{n} + 4^{n} \leq 2 \cdot 4^{n}$ . Taking $n$ -th roots, which preserves order for positive numbers,

4 \leq (3^{n} + 4^{n})^{1/ n} \leq 4 \cdot 2^{1/ n} .

Now $2^{1/ n} \to 1$ : indeed $1 \leq 2^{1/ n} \leq n^{1/ n}$ for $n \geq 2$ , and $n^{1/ n} \to 1$ by part (iii), so a squeeze applies. Hence the upper bound tends to $4$ and the limit is $4$ .
Sanity check at $n = 20$ : $(3^{20} + 4^{20})^{1/20} \approx 4.0064$ . The largest base dominates, and $(\sum_{j} c_{j}^{n})^{1/ n} \to max_{j} c_{j}$ by exactly this argument.

□

Both existence theorems now follow. The first converts monotonicity plus boundedness into convergence, and its proof is where the completeness of $R$ enters the chapter for the first time.

Theorem 2.14 (Monotone convergence theorem). Every increasing sequence that is bounded above converges, and its limit is $sup {a_{n} : n \in N}$ . Symmetrically, every decreasing sequence bounded below converges, to the infimum of its range.

Proof. Let $(a_{n})$ be increasing and bounded above, and let $S = {a_{n} : n \in N}$ , a non-empty set bounded above. By the completeness axiom stated in the definition The reals, axiomatically, $L = sup S$ exists in $R$ .

Let $ε > 0$ . By the proposition The supremum is unique, and characterised by an epsilon condition, $L - ε$ is not an upper bound of $S$ , so there is an index $N$ with $a_{N} > L - ε$ . For any $n > N$ monotonicity gives $a_{n} \geq a_{N} > L - ε$ , while $a_{n} \leq L$ because $L$ bounds $S$ . Hence

L - ε < a_{n} \leq L < L + ε, so ∣ a_{n} - L ∣ < ε

for every $n > N$ . Since $ε$ was arbitrary, $a_{n} \to L = sup S$ .

For the decreasing case apply what has just been proved to $(- a_{n})$ , which is increasing and bounded above, and use $sup (- S) = - in f S$ together with the scalar rule of the theorem Algebra of limits.∎

Example 2.15 (A recursion with no closed form). Let $a_{1} = 1$ and $a_{n + 1} = 2 + a_{n}$ . Show that $(a_{n})$ converges and find its limit.

Solution.

Bounded above by $2$ , by induction: $a_{1} = 1 < 2$ , and if $a_{n} < 2$ then $a_{n + 1} = 2 + a_{n} < 4 = 2$ .
Increasing, also by induction. First $a_{2} = 3 \approx 1.732 > 1 = a_{1}$ . If $a_{n} \geq a_{n - 1}$ then $2 + a_{n} \geq 2 + a_{n - 1}$ , and the square root is increasing, so $a_{n + 1} \geq a_{n}$ .
By the theorem Monotone convergence theorem the limit $L$ exists. This is the essential move: $L$ is now known to be a real number before any attempt to compute it.
Compute it. The subsequence $(a_{n + 1})$ has the same limit $L$ , and $x \mapsto 2 + x$ is continuous, so letting $n \to \infty$ in the recursion gives $L = 2 + L$ , hence $L^{2} - L - 2 = 0$ , that is $(L - 2) (L + 1) = 0$ .
Since every term is positive, $L \geq 0$ by the remark on limits and order, so $L = 2$ .
Sanity check: $a_{1} = 1$ , $a_{2} \approx 1.732$ , $a_{3} \approx 1.932$ , $a_{4} \approx 1.983$ , $a_{5} \approx 1.996$ — climbing to $2$ and never passing it. Step 4 is worthless without step 3: the same manipulation applied to $b_{n + 1} = 2 b_{n}$ , $b_{1} = 1$ , "gives" $L = 2 L$ , hence $L = 0$ , nonsense precisely because no limit exists.

□

Theorem 2.16 (Bolzano–Weierstrass). Every bounded sequence of real numbers has a convergent subsequence.

Proof. Let $(a_{n})$ satisfy $∣ a_{n} ∣ \leq M$ for all $n$ . Call an index $n$ a peak if $a_{n} \geq a_{m}$ for every $m > n$ — that is, no later term ever exceeds it.

Case 1: there are infinitely many peaks. List them as $n_{1} < n_{2} < n_{3} < \dots$ . By the definition of a peak, $a_{n_{1}} \geq a_{n_{2}} \geq a_{n_{3}} \geq \dots$ , so the subsequence $(a_{n_{k}})$ is decreasing, and it is bounded below by $- M$ . By the theorem Monotone convergence theorem it converges.

Case 2: there are finitely many peaks. Let $N$ exceed every peak index and set $n_{1} = N + 1$ . Since $n_{1}$ is not a peak, there is $n_{2} > n_{1}$ with $a_{n_{2}} > a_{n_{1}}$ ; since $n_{2}$ is not a peak either, there is $n_{3} > n_{2}$ with $a_{n_{3}} > a_{n_{2}}$ , and recursively one obtains a strictly increasing subsequence $(a_{n_{k}})$ . It is bounded above by $M$ , so again the theorem Monotone convergence theorem delivers a limit.

In both cases a convergent subsequence exists.∎

Intuition. Bolzano–Weierstrass says that infinitely many points confined to a bounded interval must pile up somewhere. Cut the interval in half: one half contains infinitely many terms, so cut that half in half, and so on. The nested intervals shrink to a point, and the terms chosen from them converge to it.

The peak proof above is the same idea organised so that monotone convergence does all the work. Either some terms dominate everything after them infinitely often — and those form a decreasing subsequence — or after a while every term is beaten by a later one, and those beatings chain into an increasing subsequence.

Pitfall. Bolzano–Weierstrass produces a convergent subsequence, not convergence. The sequence $(- 1)^{n}$ is bounded and has subsequences converging to $1$ and to $- 1$ ; it converges to neither. A bounded sequence converges exactly when all of its convergent subsequences share one limit, and that extra condition is not supplied by boundedness.

Exercises for §2.6 →

2.7Cauchy sequences

Every criterion so far requires a candidate limit. That is a genuine obstacle: the recursion in the last example was tractable only because the fixed-point equation could be solved, and the partial sums of $\sum 1/ n^{2}$ converge to $π^{2} /6$ , a number nobody could have guessed in advance. What is wanted is a condition on the sequence alone, mentioning no limit, that is equivalent to convergence.

Definition 2.17 (Cauchy sequence). A sequence $(a_{n})$ is Cauchy if for every $ε > 0$ there exists $N$ such that

m, n > N ⟹ ∣ a_{m} - a_{n} ∣ < ε .

The quantifiers deserve the same scrutiny as before. $N$ depends on $ε$ alone; after it is chosen, both indices range freely over everything beyond it, independently of each other. That independence is the entire content of the definition, and the commonest misreading weakens it to consecutive terms only.

Pitfall. " $∣ a_{n + 1} - a_{n} ∣ \to 0$ " is not the Cauchy condition and does not imply convergence. The partial sums of the harmonic series, $s_{n} = \sum_{k \leq n} 1/ k$ , satisfy $∣ s_{n + 1} - s_{n} ∣ = 1/ (n + 1) \to 0$ , and yet $∣ s_{2 n} - s_{n} ∣ \geq 1/2$ for every $n$ , so they are not Cauchy and, as proved below, they diverge. Consecutive steps shrinking says nothing about how far the sequence can travel over many steps.

Intuition. Cauchy says: after a certain frame of the film, any two frames you pick — adjacent or a million apart — are nearly identical. You never have to know what the final picture looks like; you observe only that the frames have stopped disagreeing with each other.

That is exactly the situation of a numerical algorithm. One can watch successive iterates agree to more and more decimal places without having any idea what number they are agreeing on.

Example 2.18 (Checking the condition directly). Show that $a_{n} = 1/ n$ is Cauchy, and that $b_{n} = (- 1)^{n}$ is not.

Solution.

For $(a_{n})$ , let $ε > 0$ and choose $N > 2/ ε$ . For $m, n > N$ , the triangle inequality gives

\frac{1}{m} - \frac{1}{n} \leq \frac{1}{m} + \frac{1}{n} < \frac{1}{N} + \frac{1}{N} = \frac{2}{N} < ε .

For $(b_{n})$ , negate the definition: it suffices to exhibit one $ε$ that no $N$ can serve. Take $ε = 1$ . Given any $N$ , pick $n = 2 N$ and $m = 2 N + 1$ , both beyond $N$ ; then $∣ b_{m} - b_{n} ∣ = ∣ - 1 - 1∣ = 2 \geq 1$ .
So $(b_{n})$ fails the condition, and by the theorem Cauchy criterion below it cannot converge — recovering the earlier divergence proof with no reference to a candidate limit at all.

□

Exercises for §2.7 →

2.8The Cauchy criterion

Theorem 2.19 (Cauchy criterion). A sequence of real numbers converges if and only if it is Cauchy.

Proof. ( $\Rightarrow$ ) Convergent implies Cauchy. Suppose $a_{n} \to L$ and let $ε > 0$ . Choose $N$ with $∣ a_{n} - L ∣ < ε /2$ for all $n > N$ . Then for $m, n > N$ ,

∣ a_{m} - a_{n} ∣ \leq ∣ a_{m} - L ∣ + ∣ L - a_{n} ∣ < \frac{ε}{2} + \frac{ε}{2} = ε .

This direction uses nothing about $R$ beyond the triangle inequality, and so holds in any metric space.

( $\Leftarrow$ ) Cauchy implies convergent. Let $(a_{n})$ be Cauchy. The argument has three steps.

Step 1: $(a_{n})$ is bounded. Apply the definition with $ε = 1$ to get $N_{0}$ such that $∣ a_{m} - a_{n} ∣ < 1$ for $m, n > N_{0}$ . Fixing $n = N_{0} + 1$ gives $∣ a_{m} ∣ \leq ∣ a_{N_{0} + 1} ∣ + 1$ for every $m > N_{0}$ , and the finitely many terms $a_{1}, \dots, a_{N_{0}}$ are bounded by their maximum absolute value. Taking the larger of the two bounds bounds the whole sequence.

Step 2: some subsequence converges. By Step 1 and the theorem Bolzano–Weierstrass, there is a subsequence $(a_{n_{k}})$ converging to some $L \in R$ . This is the step that uses completeness, inherited by Bolzano–Weierstrass from monotone convergence and hence from the supremum axiom.

Step 3: the whole sequence converges to that same $L$ . Let $ε > 0$ . Choose $N$ from the Cauchy condition so that $∣ a_{m} - a_{n} ∣ < ε /2$ for $m, n > N$ , and choose $k$ large enough that both $n_{k} > N$ and $∣ a_{n_{k}} - L ∣ < ε /2$ ; this is possible because $n_{k} \geq k \to \infty$ and $a_{n_{k}} \to L$ . Then for every $n > N$ ,

∣ a_{n} - L ∣ \leq ∣ a_{n} - a_{n_{k}} ∣ + ∣ a_{n_{k}} - L ∣ < \frac{ε}{2} + \frac{ε}{2} = ε .

∎

Step 3 is the pivot, and is worth stating in words: a Cauchy sequence with a convergent subsequence converges. Being Cauchy means the terms cannot drift apart, so once a subsequence settles near $L$ , the rest is trapped near it too. The same lemma reappears wherever completeness is proved — in Metric Spaces, where the definition Complete metric space takes this conclusion as its defining property.

Example 2.20 (Convergence with an unknown limit). Let $a_{1} = 1$ and $a_{n + 1} = 1 + \frac{1}{1 + a _{n}}$ . Assuming $a_{n} \in [1, 2]$ for all $n$ , show that $(a_{n})$ is Cauchy, and identify its limit.

Solution.

Estimate how the recursion moves two consecutive gaps. Subtracting the recursion from itself at consecutive indices,

∣ a_{n + 1} - a_{n} ∣ = \frac{1}{1 + a _{n}} - \frac{1}{1 + a _{n - 1}} = \frac{∣ a _{n} - a _{n - 1} ∣}{( 1 + a _{n} ) ( 1 + a _{n - 1} )} \leq \frac{∣ a _{n} - a _{n - 1} ∣}{4},

because each factor in the denominator is at least $2$ . 2. Iterating, $∣ a_{n + 1} - a_{n} ∣ \leq 4^{- (n - 1)} ∣ a_{2} - a_{1} ∣ = \frac{1}{2} \cdot 4^{- (n - 1)}$ . 3. For $m > n$ , sum the steps with the triangle inequality and bound the finite geometric sum by the infinite one:

∣ a_{m} - a_{n} ∣ \leq j = n \sum m - 1 ∣ a_{j + 1} - a_{j} ∣ \leq \frac{1}{2} j = n \sum \infty 4^{- (j - 1)} = \frac{1}{2} \cdot \frac{4 ^{- (n - 2)}}{1 - 1/4} = \frac{2}{3} 4^{- (n - 2)} .

Given $ε > 0$ , the right-hand side is below $ε$ for all $n$ beyond some $N$ , since $4^{- n} \to 0$ by the proposition Four standard limits (ii). As the bound depends only on the smaller index $n$ , the same $N$ works for every $m > n$ , so $(a_{n})$ is Cauchy and converges by the theorem Cauchy criterion.
The limit satisfies $L = 1 + 1/ (1 + L)$ , that is $L^{2} = 2$ , and $L \in [1, 2]$ forces $L = 2 \approx 1.41421$ .
Sanity check: $1, 1.5, 1.4, 1.41 \overline{6}, 1.41379, \dots$ — the continued-fraction expansion of $2$ , each gap about a quarter of the last.

□

Exercises for §2.8 →

2.9Cauchy sequences and completeness

The theorem Cauchy criterion is usually summarised as " $R$ is complete", and it is worth being precise about what that abbreviation asserts. The forward direction is free in any setting where distances obey the triangle inequality. The converse is not: it is equivalent to the supremum axiom, and it fails in $Q$ .

The failure is concrete. Newton's method for $x^{2} = 2$ , namely $q_{1} = 1$ and $q_{n + 1} = \frac{1}{2} (q_{n} + 2/ q_{n})$ , produces only rational numbers and is Cauchy, but its only possible limit is $2$ , which is not rational by the proposition Irrationality of the square root of two. Inside $Q$ that sequence is Cauchy and divergent. Nothing about the sequence changed; what changed is the ambient supply of limits, and completeness is exactly the statement that $R$ has no such gaps. That is why the definition The Cauchy construction builds $R$ by adjoining a limit to every Cauchy sequence of rationals.

Several statements now in hand are logically interchangeable, each derivable from any other over the ordered-field axioms with the Archimedean property:

every non-empty set bounded above has a supremum (the completeness axiom);
every bounded monotone sequence converges (the theorem Monotone convergence theorem);
every bounded sequence has a convergent subsequence (the theorem Bolzano–Weierstrass);
every Cauchy sequence converges (the theorem Cauchy criterion);
every nested sequence of closed bounded intervals has a common point.

The chapter proved $1 \Rightarrow 2 \Rightarrow 3 \Rightarrow 4$ . Statement 5, the form used in bisection arguments, follows from 2 in the same way: the left endpoints increase and are bounded above by every right endpoint, so they converge to a point lying in every interval. Its hypotheses are easy to get wrong — the nested open intervals $(0, 1/ n)$ have empty intersection, and so do the nested closed but unbounded $[n, \infty)$ , so both "closed" and "bounded" are load-bearing. That pair of words is exactly what the theorem Heine-Borel of Metric Spaces identifies with compactness on the line.

Exercises for §2.9 →

2.10Series as sequences of partial sums

Definition 2.21 (Series and its sum). Given a sequence $(a_{n})$ , its $N$ -th partial sum is $s_{N} = a_{1} + a_{2} + \dots + a_{N}$ . The series $\sum_{n = 1}^{\infty} a_{n}$ is said to converge to $S$ if the sequence $(s_{N})$ of partial sums converges to $S$ , and in that case one writes $\sum_{n = 1}^{\infty} a_{n} = S$ . If $(s_{N})$ diverges, the series diverges.

The definition gives meaning to a symbol that otherwise has none. An infinite sum is not a sum — addition is binary, and no finite number of applications produces $\sum_{n = 1}^{\infty} a_{n}$ . The symbol abbreviates a limit, so every theorem about sequences applies verbatim with $(s_{N})$ as the sequence. The Cauchy criterion in particular reads: $\sum a_{n}$ converges if and only if for every $ε > 0$ there is $N$ with $\sum_{k = n + 1}^{m} a_{k} < ε$ whenever $m > n > N$ .

Two elementary consequences of the sequence theory, both used without comment below: linearity, that $\sum (α a_{n} + β b_{n}) = α \sum a_{n} + β \sum b_{n}$ when both series converge, which is the sum and scalar rules of the theorem Algebra of limits applied to partial sums; and tail-independence, that changing finitely many terms changes whether the series converges not at all, though it does change the sum.

Theorem 2.22 (Geometric series). For $r \in R$ with $∣ r ∣ < 1$ ,

n = 0 \sum \infty r^{n} = \frac{1}{1 - r},

and for $∣ r ∣ \geq 1$ the series diverges.

Proof. For $r \neq = 1$ the partial sums telescope: multiplying $s_{N} = 1 + r + \dots + r^{N}$ by $r$ and subtracting gives $s_{N} - r s_{N} = 1 - r^{N + 1}$ , hence

s_{N} = \frac{1 - r ^{N + 1}}{1 - r} .

If $∣ r ∣ < 1$ then $r^{N + 1} \to 0$ by the proposition Four standard limits (ii), so by the theorem Algebra of limits $s_{N} \to \frac{1}{1 - r}$ .

If $∣ r ∣ \geq 1$ then $∣ r^{n} ∣ \geq 1$ for every $n$ , so the terms do not tend to $0$ and the series diverges by the theorem Term test below. (For $r = 1$ directly: $s_{N} = N + 1 \to \infty$ .)∎

Example 2.23 (Two series whose partial sums can be computed). Evaluate $n = 1 \sum \infty \frac{3}{4 ^{n}}$ and $n = 1 \sum \infty \frac{1}{n ( n + 1 )}$ .

Solution.

The first is geometric after factoring out the first term: $\sum_{n = 1}^{\infty} 3 \cdot 4^{- n} = \frac{3}{4} \sum_{k = 0}^{\infty} 4^{- k}$ . By the theorem Geometric series with $r = 1/4$ , the inner sum is $\frac{1}{1 - 1/4} = \frac{4}{3}$ , so the total is $\frac{3}{4} \cdot \frac{4}{3} = 1$ .
The second is a telescoping series. Partial fractions give $\frac{1}{n ( n + 1 )} = \frac{1}{n} - \frac{1}{n + 1}$ , so

s_{N} = (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + \dots + (\frac{1}{N} - \frac{1}{N + 1}) = 1 - \frac{1}{N + 1} .

Hence $s_{N} \to 1$ and the sum is $1$ .
Sanity check on both: $3/4 + 3/16 + 3/64 = 0.984 \dots$ and $s_{3} = 1 - 1/4 = 0.75$ , each creeping up on $1$ from below as the positive terms require.
Geometric and telescoping are essentially the only families whose sums can be read off from the partial sums. For everything else the question is convergence, not value, and the tests below answer it without ever forming $s_{N}$ .

□

Exercises for §2.10 →

2.11The term test and the harmonic series

Theorem 2.24 (Term test). If $\sum a_{n}$ converges, then $a_{n} \to 0$ . Equivalently, if $a_{n} \neq \to 0$ then $\sum a_{n}$ diverges.

Proof. Suppose $\sum a_{n}$ converges to $S$ , so $s_{N} \to S$ . The shifted sequence $(s_{N - 1})$ also converges to $S$ , since for any $ε$ the threshold $N$ that works for $(s_{N})$ needs only be increased by one. Then $a_{N} = s_{N} - s_{N - 1} \to S - S = 0$ by the sum rule of the theorem Algebra of limits.∎

The test is one-directional and is useful only for proving divergence. Its converse is false, and the standard witness is the harmonic series — a series whose terms shrink to zero and whose sum is nevertheless infinite.

Theorem 2.25 (Divergence of the harmonic series). The series $n = 1 \sum \infty \frac{1}{n}$ diverges to $+ \infty$ .

Proof. Group the terms into blocks of doubling length. For each $k \geq 1$ , the block from index $2^{k - 1} + 1$ to $2^{k}$ contains $2^{k - 1}$ terms, each at least $2^{- k}$ since the largest index in the block is $2^{k}$ . Hence

n = 2^{k - 1} + 1 \sum 2^{k} \frac{1}{n} \geq 2^{k - 1} \cdot \frac{1}{2 ^{k}} = \frac{1}{2} .

Summing over $k = 1, \dots, K$ and adding the first term gives

s_{2^{K}} \geq 1 + \frac{K}{2} .

The partial sums are increasing, and this subsequence of them is unbounded, so $(s_{N})$ is unbounded and cannot converge — a convergent sequence is bounded, as established in the proof of the theorem Algebra of limits. Since $(s_{N})$ is increasing and unbounded, $s_{N} \to + \infty$ .∎

The same grouping as an upper estimate gives $s_{2^{K}} \leq 1 + K$ , so the growth is logarithmic: $s_{N} \approx ln N + γ$ with $γ \approx 0.5772$ . Divergence here is therefore invisible to computation — reaching $s_{N} > 100$ takes about $e^{100} \approx 1 0^{43}$ terms — which is precisely why a proof is needed. The theorem Term test, by contrast, is one-directional: it can only ever report "diverges" or "no information", and $\sum 1/ n$ , whose terms tend to $0$ and whose sum is infinite, shows how little "no information" is worth.

Exercises for §2.11 →

2.12The convergence tests

For a series of non-negative terms, the partial sums are increasing, so by the theorem Monotone convergence theorem exactly one of two things happens: the partial sums are bounded and the series converges, or they are unbounded and it diverges to $+ \infty$ . Every test below for positive series is therefore a device for bounding partial sums, and this single observation explains why they all look alike.

Theorem 2.26 (Comparison and limit comparison). Let $0 \leq a_{n}$ and $0 \leq b_{n}$ for all large $n$ .

(Comparison.) If $a_{n} \leq b_{n}$ for all large $n$ and $\sum b_{n}$ converges, then $\sum a_{n}$ converges. Contrapositively, if $\sum a_{n}$ diverges then so does $\sum b_{n}$ .
(Limit comparison.) If $a_{n}, b_{n} > 0$ and $a_{n} / b_{n} \to c$ with $0 < c < \infty$ , then $\sum a_{n}$ and $\sum b_{n}$ either both converge or both diverge.

Proof. 1. By tail-independence assume the inequalities hold from $n = 1$ . Write $s_{N} = \sum_{n \leq N} a_{n}$ and $t_{N} = \sum_{n \leq N} b_{n}$ . Then $s_{N} \leq t_{N}$ for every $N$ , and $(t_{N})$ converges, hence is bounded, say by $B$ . So $(s_{N})$ is increasing and bounded above by $B$ , and the theorem Monotone convergence theorem gives convergence.

2. Apply the definition of the limit with $ε = c /2$ : there is $N$ beyond which $\frac{a _{n}}{b _{n}} - c < \frac{c}{2}$ , that is

\frac{c}{2} b_{n} < a_{n} < \frac{3 c}{2} b_{n} .

If $\sum b_{n}$ converges, the right-hand inequality and part 1 give convergence of $\sum a_{n}$ . If $\sum b_{n}$ diverges, the left-hand inequality and part 1 (in contrapositive form, comparing $\frac{c}{2} b_{n} \leq a_{n}$ ) give divergence of $\sum a_{n}$ . Reversing the roles, using $b_{n} / a_{n} \to 1/ c \in (0, \infty)$ , gives the converse implications.∎

Limit comparison is the practical form, because it needs only the asymptotic size of $a_{n}$ , which is usually visible at a glance: $\frac{3 n ^{2} - n}{n ^{4} + 5}$ behaves like $3/ n^{2}$ , so it converges; $\frac{1}{n ^{2} + 1}$ behaves like $1/ n$ , so it diverges. To use it one needs a stock of reference series, and the integral test supplies the whole $p$ -family at once.

Theorem 2.27 (Integral test and $p$ -series). Let $f$ be positive, decreasing and continuous on $[1, \infty)$ , and set $a_{n} = f (n)$ . Then $\sum_{n = 1}^{\infty} a_{n}$ converges if and only if $\int_{1}^{\infty} f (x) d x$ converges. Consequently $\sum_{n = 1}^{\infty} n^{- p}$ converges precisely when $p > 1$ .

Proof. Because $f$ decreases, on the interval $[n, n + 1]$ it satisfies $f (n + 1) \leq f (x) \leq f (n)$ , so integrating over that interval of length $1$ gives

a_{n + 1} \leq \int_{n}^{n + 1} f (x) d x \leq a_{n} .

Summing the left inequality for $n = 1, \dots, N - 1$ gives $s_{N} - a_{1} \leq \int_{1}^{N} f$ , and summing the right one gives $\int_{1}^{N} f \leq s_{N - 1}$ . Hence

\int_{1}^{N} f (x) d x \leq s_{N - 1} \leq s_{N} \leq a_{1} + \int_{1}^{N} f (x) d x .

If the improper integral converges, the right inequality bounds the increasing sequence $(s_{N})$ , which therefore converges by the theorem Monotone convergence theorem. If the integral diverges, then since $f > 0$ the quantity $\int_{1}^{N} f$ increases without bound, and the left inequality forces $s_{N} \to \infty$ .

For the $p$ -series take $f (x) = x^{- p}$ , which is positive, continuous and decreasing on $[1, \infty)$ for $p > 0$ . For $p \neq = 1$ , $\int_{1}^{N} x^{- p} d x = \frac{N ^{1 - p} - 1}{1 - p}$ , which converges as $N \to \infty$ exactly when $1 - p < 0$ , that is $p > 1$ ; for $p = 1$ , $\int_{1}^{N} x^{- 1} d x = ln N \to \infty$ . For $p \leq 0$ the terms do not tend to $0$ and the theorem Term test gives divergence.∎

Figure 2.28. The integral test as a picture, for

f (x) = 1/ x

. Each bar has width

1

and height

f (n)

, so the bars total

s_{N}

, and each overhangs the curve on its own interval. The bars therefore out-total the area under the curve,

ln N

, which is unbounded — so the harmonic series diverges.

The next two tests compare against a geometric series rather than an integral, and in exchange for a stronger conclusion — they give absolute convergence — they are inconclusive in a whole family of borderline cases.

Theorem 2.29 (Ratio and root tests). Let $(a_{n})$ have nonzero terms and set

L = n \to \infty lim \frac{a _{n + 1}}{a _{n}} (when it exists), ρ = n \to \infty lim sup ∣ a_{n} ∣^{1/ n} .

If $L < 1$ or $ρ < 1$ , then $\sum a_{n}$ converges absolutely. If $L > 1$ or $ρ > 1$ , the terms are unbounded and $\sum a_{n}$ diverges. If $L = 1$ or $ρ = 1$ , no conclusion follows.

Proof. Root test, convergent case. Suppose $ρ < 1$ and pick $r$ with $ρ < r < 1$ . By the definition of $lim sup$ , only finitely many $n$ have $∣ a_{n} ∣^{1/ n} > r$ , so there is $N$ with $∣ a_{n} ∣ \leq r^{n}$ for all $n > N$ . Since $\sum r^{n}$ converges by the theorem Geometric series, the comparison part of the theorem Comparison and limit comparison applied to $\sum ∣ a_{n} ∣$ gives absolute convergence.

Root test, divergent case. If $ρ > 1$ then $∣ a_{n} ∣^{1/ n} > 1$ , hence $∣ a_{n} ∣ > 1$ , for infinitely many $n$ . So $a_{n} \neq \to 0$ and the theorem Term test gives divergence.

Ratio test. Suppose $L < 1$ and pick $r$ with $L < r < 1$ . There is $N$ with $∣ a_{n + 1} / a_{n} ∣ < r$ for all $n > N$ , so by induction $∣ a_{N + k} ∣ < r^{k} ∣ a_{N} ∣$ for every $k \geq 1$ . Comparison with the convergent geometric series $∣ a_{N} ∣ \sum_{k} r^{k}$ gives absolute convergence. If $L > 1$ , choose $r$ with $1 < r < L$ ; then eventually $∣ a_{n + 1} ∣ > r ∣ a_{n} ∣$ , so $∣ a_{n} ∣$ grows at least geometrically, $a_{n} \neq \to 0$ , and the series diverges.

Inconclusiveness at $1$ . For $a_{n} = 1/ n$ both $L$ and $ρ$ equal $1$ — for $ρ$ , because $n^{1/ n} \to 1$ by the proposition Four standard limits (iii) — and the series diverges. For $a_{n} = 1/ n^{2}$ both again equal $1$ , and the series converges. So no verdict is possible from the value $1$ alone.∎

The $lim sup$ in the root test is not fussiness: it makes the test apply to every series, whereas the ratio test needs a limit that may not exist. For $a_{n} = 2^{- n}$ at even $n$ and $3^{- n}$ at odd $n$ the ratios oscillate and $L$ does not exist, while $ρ = 1/2 < 1$ settles it at once. The root test is always at least as strong; the ratio test is usually the easier computation, which is why it is tried first.

Example 2.30 (Choosing a test). Determine convergence of (a) $n = 1 \sum \infty \frac{n ^{2} + 3}{n ^{4} - n}$ , (b) $n = 1 \sum \infty \frac{n !}{n ^{n}}$ , (c) $n = 2 \sum \infty \frac{1}{n ln n}$ .

Solution.

(a) The terms behave like $n^{2} / n^{4} = 1/ n^{2}$ . Formally, with $b_{n} = 1/ n^{2}$ ,

\frac{a _{n}}{b _{n}} = \frac{n ^{2} ( n ^{2} + 3 )}{n ^{4} - n} = \frac{1 + 3/ n ^{2}}{1 - 1/ n ^{3}} \to 1 \in (0, \infty),

so by limit comparison and the $p$ -series with $p = 2 > 1$ , the series converges. 2. (b) Factorials call for the ratio test:

\frac{a _{n + 1}}{a _{n}} = \frac{( n + 1 )!}{( n + 1 ) ^{n + 1}} \cdot \frac{n ^{n}}{n !} = \frac{n ^{n}}{( n + 1 ) ^{n}} = \frac{1}{( 1 + 1/ n ) ^{n}} \to \frac{1}{e} \approx 0.368 < 1,

so the series converges absolutely. 3. (c) Ratio and root both give $1$ here, so use the integral test with $f (x) = 1/ (x ln x)$ , positive and decreasing on $[2, \infty)$ . The substitution $u = ln x$ gives $\int_{2}^{N} \frac{d x}{x l n x} = ln (ln N) - ln (ln 2) \to \infty$ , so the series diverges. 4. Sanity check on (c) against intuition: $1/ (n ln n)$ is smaller than $1/ n$ yet still diverges, and $\sum 1/ (n (ln n)^{2})$ — same integral, now $\int u^{- 2} d u$ — converges. The boundary between convergence and divergence is not located at any single power of $n$ ; it is approached but never reached by these logarithmic refinements.□

The last test handles signs, and it is the only standard test that can certify a series whose absolute version diverges.

Theorem 2.31 (Alternating series test). Let $b_{n} \geq 0$ be decreasing with $b_{n} \to 0$ . Then $\sum_{n = 1}^{\infty} (- 1)^{n + 1} b_{n}$ converges to a sum $S$ , and the truncation error obeys $∣ S - s_{N} ∣ \leq b_{N + 1}$ .

Proof. Consider the even partial sums. Grouping in pairs,

s_{2 k + 2} - s_{2 k} = b_{2 k + 1} - b_{2 k + 2} \geq 0,

so $(s_{2 k})$ is increasing; and writing $s_{2 k} = b_{1} - (b_{2} - b_{3}) - \dots - (b_{2 k - 2} - b_{2 k - 1}) - b_{2 k}$ exhibits $s_{2 k} \leq b_{1}$ , since every bracket is non-negative. By the theorem Monotone convergence theorem, $s_{2 k} \to S$ for some $S$ .

The odd partial sums satisfy $s_{2 k + 1} = s_{2 k} + b_{2 k + 1} \to S + 0 = S$ , using $b_{n} \to 0$ and the theorem Algebra of limits. Since both the even-indexed and the odd-indexed partial sums converge to $S$ , so does the whole sequence: given $ε > 0$ , take the larger of the two thresholds, and every index, even or odd, lies beyond it.

For the error bound, note the same argument shows $(s_{2 k + 1})$ is decreasing, so for every $k$ the sum $S$ lies between $s_{2 k}$ and $s_{2 k + 1}$ , consecutive partial sums. Hence $∣ S - s_{N} ∣ \leq ∣ s_{N + 1} - s_{N} ∣ = b_{N + 1}$ .∎

Pitfall. Monotonicity of $(b_{n})$ is a genuine hypothesis, not decoration. Take $b_{n} = 1/ n$ for even $n$ and $b_{n} = 1/ n^{2}$ for odd $n$ ; then $b_{n} \to 0$ but $(b_{n})$ does not decrease, and the alternating series splits into $\sum 1/ n^{2}$ over odd indices, which converges, minus $\sum 1/ n$ over even indices, which diverges — so the series diverges. Terms tending to zero is not enough even with alternating signs.

Exercises for §2.12 →

2.13Absolute convergence

Definition 2.32 (Absolute and conditional convergence). The series $\sum a_{n}$ converges absolutely if $\sum ∣ a_{n} ∣$ converges. It converges conditionally if $\sum a_{n}$ converges but $\sum ∣ a_{n} ∣$ does not.

Theorem 2.33 (Absolute convergence implies convergence). If $\sum ∣ a_{n} ∣$ converges then $\sum a_{n}$ converges, and $∣ \sum a_{n} ∣ \leq \sum ∣ a_{n} ∣$ .

Proof. For each $n$ set $c_{n} = a_{n} + ∣ a_{n} ∣$ . Since $- ∣ a_{n} ∣ \leq a_{n} \leq ∣ a_{n} ∣$ , we have $0 \leq c_{n} \leq 2∣ a_{n} ∣$ , so $\sum c_{n}$ converges by comparison with the convergent series $2 \sum ∣ a_{n} ∣$ , using the theorem Comparison and limit comparison. Then

\sum a_{n} = \sum (c_{n} - ∣ a_{n} ∣)

converges as the difference of two convergent series, by linearity.

For the inequality, apply the triangle inequality to each partial sum: $∣ s_{N} ∣ = \sum_{n \leq N} a_{n} \leq \sum_{n \leq N} ∣ a_{n} ∣$ . Both sides converge, and the remark on limits and order preserves $\leq$ in the limit.∎

An alternative proof runs through the Cauchy criterion and is worth knowing, because it is the one that generalises to any complete space: for $m > n$ , $\sum_{k = n + 1}^{m} a_{k} \leq \sum_{k = n + 1}^{m} ∣ a_{k} ∣$ , so if the right side can be made small the left side can too, and the criterion applied to $(s_{N})$ does the rest.

Example 2.34 (The alternating harmonic series). Classify $n = 1 \sum \infty \frac{( - 1 ) ^{n + 1}}{n}$ and $n = 1 \sum \infty \frac{( - 1 ) ^{n + 1}}{n ^{2}}$ as absolutely convergent, conditionally convergent, or divergent.

Solution.

For the first, $b_{n} = 1/ n$ decreases to $0$ , so the theorem Alternating series test gives convergence; its sum is $ln 2 \approx 0.693$ .
Its absolute version is $\sum 1/ n$ , which diverges by the theorem Divergence of the harmonic series. So the convergence is conditional.
For the second, the absolute version is $\sum 1/ n^{2}$ , a $p$ -series with $p = 2 > 1$ , which converges. So the series converges absolutely, and by the theorem Absolute convergence implies convergence no separate argument for the signed series is needed.
Sanity check on the difference in quality: for the first, the error bound $b_{N + 1} = 1/ (N + 1)$ means $1000$ terms buy two decimal places at best; for the second, $\sum_{n > N} 1/ n^{2} < 1/ N$ bounds the absolute tail, needing no cancellation. Absolute convergence is convergence that does not depend on the signs, and the next section prices that dependence.

□

Exercises for §2.13 →

2.14Rearrangements and Riemann's theorem

A rearrangement of $\sum a_{n}$ is a series $\sum a_{σ (n)}$ where $σ : N \to N$ is a bijection. For a finite sum, rearranging changes nothing; commutativity of addition guarantees it. For an infinite series the guarantee evaporates, because the sum is a limit of partial sums and a rearrangement builds completely different partial sums.

Theorem 2.35 (Rearrangement: the absolute and the conditional case).

If $\sum a_{n}$ converges absolutely to $S$ , then every rearrangement converges, absolutely, to the same $S$ .
(Riemann.) If $\sum a_{n}$ converges conditionally, then for every $M \in R$ there is a rearrangement converging to $M$ , and there are rearrangements diverging to $+ \infty$ , to $- \infty$ , and oscillating without limit.

Proof. 1. Let $ε > 0$ . Since $\sum ∣ a_{n} ∣$ converges, its partial sums are Cauchy, so there is $N$ with $\sum_{n > N} ∣ a_{n} ∣ < ε$ . Choose $K$ large enough that the rearranged indices $σ (1), \dots, σ (K)$ include all of $1, \dots, N$ ; this is possible because $σ$ is onto, so each of the finitely many indices $1, \dots, N$ is hit at some finite position. For $k \geq K$ , the difference between $\sum_{j \leq k} a_{σ (j)}$ and $s_{N}$ consists only of terms with index above $N$ , each appearing at most once, so

j \leq k \sum a_{σ (j)} - s_{N} \leq n > N \sum ∣ a_{n} ∣ < ε,

and likewise $∣ S - s_{N} ∣ \leq \sum_{n > N} ∣ a_{n} ∣ < ε$ . Hence $\sum_{j \leq k} a_{σ (j)} - S < 2 ε$ for all $k \geq K$ , so the rearrangement converges to $S$ . Applying the same argument to $\sum ∣ a_{n} ∣$ shows the rearrangement is absolutely convergent.

2. Split the terms by sign: let $p_{n} = max (a_{n}, 0)$ and $q_{n} = max (- a_{n}, 0)$ , so that $a_{n} = p_{n} - q_{n}$ and $∣ a_{n} ∣ = p_{n} + q_{n}$ , with $p_{n}, q_{n} \geq 0$ .

Both $\sum p_{n}$ and $\sum q_{n}$ diverge. If both converged, then $\sum ∣ a_{n} ∣ = \sum (p_{n} + q_{n})$ would converge, contradicting conditionality. If exactly one converged — say $\sum p_{n} < \infty$ and $\sum q_{n} = \infty$ — then $\sum a_{n} = \sum p_{n} - \sum q_{n}$ would diverge to $- \infty$ , contradicting convergence of $\sum a_{n}$ .

Construction. Let $P_{1}, P_{2}, \dots$ be the positive terms of the series in their original order and $Q_{1}, Q_{2}, \dots$ the strictly negative ones. Fix a target $M$ . Take positive terms $P_{1}, \dots, P_{k_{1}}$ until their sum first exceeds $M$ , which is possible because $\sum P_{j}$ diverges; then append negative terms $Q_{1}, \dots, Q_{m_{1}}$ until the running total first drops below $M$ , possible because $\sum Q_{j}$ diverges to $- \infty$ ; then more positive terms until it exceeds $M$ again, and so on forever. Every term is used exactly once — at each stage the next unused term of the appropriate sign is taken — so this is a genuine rearrangement.

It converges to $M$ . At each switch the running total overshoots $M$ by at most the size of the last term used, because before that term was added the total was on the other side of $M$ . Since $\sum a_{n}$ converges, $a_{n} \to 0$ by the theorem Term test, so those last-term sizes tend to $0$ . Any partial sum lies between $M$ minus and $M$ plus the largest term size used since the previous switch, and that bound tends to $0$ ; hence the partial sums converge to $M$ .

Divergent rearrangements. Run the same construction against a moving target: use positive terms until the total exceeds $1$ , one negative term, positives until the total exceeds $2$ , one negative term, and so on; the partial sums exceed every integer, so the rearrangement diverges to $+ \infty$ . Reversing the roles of the signs gives $- \infty$ , and alternating between the targets $0$ and $1$ gives a rearrangement whose partial sums oscillate between them without converging.∎

Intuition. A conditionally convergent series has an infinite supply of positive material and an infinite supply of negative material, and its sum is whatever these two infinities happen to cancel to in the given order. Change the order and you change the cancellation.

Absolute convergence removes the leverage: the total mass $\sum ∣ a_{n} ∣$ is finite, so no matter how the terms are shuffled, only a small amount of material sits beyond any late stage, and every arrangement is forced to the same total.

Example 2.36 (Steering the alternating harmonic series). The alternating harmonic series sums to $ln 2 \approx 0.6931$ . Carry out the first few steps of a rearrangement aimed at $M = 1.5$ , and explain why it must succeed.

Solution.

The positive terms are $1, \frac{1}{3}, \frac{1}{5}, \dots$ with divergent sum, and the negative ones $- \frac{1}{2}, - \frac{1}{4}, - \frac{1}{6}, \dots$ with divergent sum — exactly the situation the proof of the theorem Rearrangement: the absolute and the conditional case requires.
Positives until above $1.5$ : $1 + \frac{1}{3} = 1.3 \overline{3}$ , still below; $+ \frac{1}{5} = 1.5 \overline{3}$ , above. Stop.
Negatives until below $1.5$ : $1.5 \overline{3} - \frac{1}{2} = 1.0 \overline{3}$ . Stop.
Positives again: $1.0 \overline{3} + \frac{1}{7} + \frac{1}{9} + \frac{1}{11} + \frac{1}{13} \approx 1.4526$ , still below; $+ \frac{1}{15} \approx 1.5193$ , above. Stop, and switch back.
Each overshoot is at most the size of the term that caused it — here $0.2$ , then $0.5$ , then about $0.067$ — and those sizes tend to $0$ because the terms do. The partial sums are therefore squeezed onto $1.5$ .
A concrete case: interleaving two positives to each negative, $1 + \frac{1}{3} - \frac{1}{2} + \frac{1}{5} + \frac{1}{7} - \frac{1}{4} + \dots$ , converges to $\frac{3}{2} ln 2 \approx 1.0397$ — same terms, different sum.

□

Exercises for §2.14 →

2.15Which tests give absolute convergence

What each test certifies decides whether a rearrangement, a product, or a term-by-term operation is permitted afterwards, so it is worth being explicit.

The comparison, limit comparison, ratio, root and integral tests are all tests on $\sum ∣ a_{n} ∣$ : each either assumes non-negative terms or works with $∣ a_{n} ∣$ throughout. When one of them returns "converges" it has proved absolute convergence, and by the theorem Rearrangement: the absolute and the conditional case the sum is then independent of order. The absolute values in the ratio test are not caution, they are the mechanism.

The alternating series test is the outlier: it uses the signs essentially and certifies only convergence, possibly conditional. So the workflow for mixed signs is to test $\sum ∣ a_{n} ∣$ first with whichever positive-term test fits; if it converges, everything is permitted, and if it diverges the alternating test may still deliver convergence — fragile convergence, on which no rearrangement is allowed.

Example 2.37 (Same series, two different verdicts). Classify $n = 1 \sum \infty \frac{( - 1 ) ^{n}}{n ^{2}}$ and $n = 1 \sum \infty \frac{( - 1 ) ^{n}}{n}$ , and say in each case which operations are safe.

Solution.

For the first, $\sum (- 1)^{n} / n^{2} = \sum 1/ n^{2}$ converges by the theorem Integral test and $p$ -series. So the series converges absolutely; its sum is $- π^{2} /12 \approx - 0.8225$ , and every rearrangement gives the same value.
For the second, the absolute version is $\sum n^{- 1/2}$ , a $p$ -series with $p = 1/2 \leq 1$ , which diverges. But $b_{n} = n^{- 1/2}$ decreases to $0$ , so the theorem Alternating series test gives convergence. The convergence is conditional.
Consequences: for the second series rearrangement is forbidden — by Riemann's theorem some rearrangement converges to $47$ — and multiplying two such series term by term in the Cauchy-product sense need not converge either. Only the first series is safe for those manipulations.
Sanity check on the numbers: the second series has partial sums $- 1, - 0.293, - 0.870, - 0.370, \dots$ , converging slowly to about $- 0.6049$ , with the error at step $N$ no worse than $(N + 1)^{- 1/2}$ as the alternating test promises.

□

Exercises for §2.15 →

2.16Power series and the radius of convergence

Theorem 2.38 (Cauchy–Hadamard radius formula). For a power series $\sum_{n \geq 0} c_{n} (x - a)^{n}$ define

\frac{1}{R} = n \to \infty lim sup ∣ c_{n} ∣^{1/ n} \in [0, \infty],

with the conventions $1/0 = \infty$ and $1/\infty = 0$ . Then the series converges absolutely for $∣ x - a ∣ < R$ and diverges for $∣ x - a ∣ > R$ . If $lim ∣ c_{n + 1} / c_{n} ∣$ exists, it equals $1/ R$ .

Proof. Apply the root test from the theorem Ratio and root tests to the series $\sum c_{n} (x - a)^{n}$ , whose $n$ -th term has

n \to \infty lim sup ∣ c_{n} (x - a)^{n} ∣^{1/ n} = ∣ x - a ∣ \cdot n \to \infty lim sup ∣ c_{n} ∣^{1/ n} = \frac{∣ x - a ∣}{R},

the constant $∣ x - a ∣$ passing outside the $lim sup$ because it is a non-negative factor common to every term. If $∣ x - a ∣ < R$ this quantity is $< 1$ and the root test gives absolute convergence; if $∣ x - a ∣ > R$ it is $> 1$ and the root test gives divergence. The two degenerate cases follow from the same computation: if $lim sup ∣ c_{n} ∣^{1/ n} = \infty$ then for every $x \neq = a$ the value is $\infty > 1$ and the series diverges, so $R = 0$ ; if the $lim sup$ is $0$ the value is $0 < 1$ for every $x$ , so $R = \infty$ .

For the last claim, suppose $lim ∣ c_{n + 1} / c_{n} ∣ = ℓ$ . The ratio test applied at a point $x$ gives $lim \frac{c _{n + 1} ( x - a ) ^{n + 1}}{c _{n} ( x - a ) ^{n}} = ℓ ∣ x - a ∣$ , so the series converges for $∣ x - a ∣ < 1/ ℓ$ and diverges for $∣ x - a ∣ > 1/ ℓ$ . Since the theorem just proved gives the same dichotomy at $R$ , and a series cannot both converge and diverge at a point, $1/ ℓ = R$ .∎

The theorem is silent at the two endpoints $x = a \pm R$ , and that silence is not an oversight: all four combinations of endpoint behaviour genuinely occur, so no general theorem is possible there. With $R = 1$ centred at $0$ : $\sum x^{n}$ diverges at both ends; $\sum x^{n} / n$ converges at $- 1$ and diverges at $+ 1$ ; $\sum (- 1)^{n} x^{n} / n$ does the reverse; and $\sum x^{n} / n^{2}$ converges at both. Endpoints are always a separate, hands-on computation.

Figure 2.39. Partial sums of

\sum x^{n}

against its sum

1/ (1 - x)

. The degree-

3

and degree-

7

partial sums (dashed) track the limit near the centre and peel away as

∣ x ∣

approaches the radius

1

, where convergence fails. Inside the radius more terms help everywhere; at and beyond

\pm 1

no number of terms helps.

Example 2.40 (Three radii and their endpoints). Find the radius and the exact domain of convergence of (a) $\sum_{n \geq 1} \frac{x ^{n}}{n}$ , (b) $\sum_{n \geq 0} \frac{x ^{n}}{n !}$ , (c) $\sum_{n \geq 0} n! x^{n}$ , and (d) $\sum_{n \geq 1} \frac{( x - 2 ) ^{n}}{n ^{2} 3 ^{n}}$ .

Solution.

(a) $∣ c_{n + 1} / c_{n} ∣ = n / (n + 1) \to 1$ , so $R = 1$ . At $x = 1$ the series is the harmonic series, divergent; at $x = - 1$ it is the alternating harmonic series, convergent. The domain is $[- 1, 1)$ .
(b) $∣ c_{n + 1} / c_{n} ∣ = n! / (n + 1)! = 1/ (n + 1) \to 0$ , so $R = \infty$ : the series converges for every real $x$ . There are no endpoints to check.
(c) $∣ c_{n + 1} / c_{n} ∣ = (n + 1)! / n! = n + 1 \to \infty$ , so $R = 0$ and the series converges only at $x = 0$ , where it is the single term $1$ .
(d) Here the centre is $a = 2$ and $c_{n} = 1/ (n^{2} 3^{n})$ . Then $∣ c_{n + 1} / c_{n} ∣ = \frac{n ^{2}}{3 ( n + 1 ) ^{2}} \to \frac{1}{3}$ , so $R = 3$ and the series converges for $∣ x - 2∣ < 3$ , that is $- 1 < x < 5$ . At both endpoints $∣ x - 2∣ = 3$ the series becomes $\sum (\pm 1)^{n} / n^{2}$ , absolutely convergent by the $p$ -series with $p = 2$ . The domain is the closed interval $[- 1, 5]$ .
Sanity check on the pattern: factorials in the denominator make $R$ infinite, factorials in the numerator make it zero, and polynomial coefficients leave $R$ determined entirely by the geometric part — $n^{2}$ in (d) changed nothing about $R = 3$ , only the endpoint verdict.

□

Exercises for §2.16 →

2.17Differentiating and integrating term by term

Inside its radius a power series behaves like a polynomial of infinite degree, and the precise statement of that is the following theorem. What has to be proved is that two limiting processes — the infinite sum and the derivative — may be exchanged, which is false for general series of functions and true here because the convergence is uniform on every closed subinterval strictly inside the radius.

Theorem 2.41 (Term-by-term differentiation and integration). Let $f (x) = \sum_{n = 0}^{\infty} c_{n} (x - a)^{n}$ have radius of convergence $R > 0$ . Then the differentiated and integrated series

n = 1 \sum \infty n c_{n} (x - a)^{n - 1}, n = 0 \sum \infty \frac{c _{n}}{n + 1} (x - a)^{n + 1}

have the same radius $R$ ; $f$ is differentiable on $(a - R, a + R)$ with $f^{'} (x)$ given by the first series; and for $∣ x - a ∣ < R$ , $\int_{a}^{x} f (t) d t$ is given by the second.

Proof. The radius is unchanged. By the theorem Cauchy–Hadamard radius formula the differentiated series has radius determined by $lim sup ∣ n c_{n} ∣^{1/ n} = (lim n^{1/ n}) \cdot lim sup ∣ c_{n} ∣^{1/ n} = 1 \cdot \frac{1}{R}$ , using the proposition Four standard limits (iii) for $n^{1/ n} \to 1$ ; the factor $(x - a)^{- 1}$ in front changes no radius. The same computation with $∣ c_{n} / (n + 1) ∣^{1/ n}$ handles the integrated series.

Uniform convergence on compact subintervals. Fix $r < R$ and let $ρ$ satisfy $r < ρ < R$ . Since $\sum ∣ c_{n} ∣ ρ^{n}$ converges (absolute convergence holds at every point inside the radius), its terms are bounded, say $∣ c_{n} ∣ ρ^{n} \leq B$ . Then for $∣ x - a ∣ \leq r$ ,

n c_{n} (x - a)^{n - 1} \leq \frac{n B}{ρ} (\frac{r}{ρ})^{n - 1},

and $\sum n t^{n - 1}$ converges for $t = r / ρ < 1$ by the ratio test. So the differentiated series is dominated by a convergent series of constants on $[a - r, a + r]$ , which is the Weierstrass $M$ -test: it converges uniformly there.

Exchanging the limits. A uniformly convergent series of continuous functions has a continuous sum, and it may be integrated term by term over a bounded interval, because the uniform bound controls the error of every partial sum simultaneously. Let $g (x)$ denote the sum of the differentiated series on $[a - r, a + r]$ . Integrating $g$ term by term from $a$ to $x$ recovers $\sum_{n \geq 1} c_{n} (x - a)^{n} = f (x) - c_{0}$ . Since $g$ is continuous, the Fundamental Theorem of Calculus applies to $\int_{a}^{x} g$ , giving $f^{'} (x) = g (x)$ . As $r < R$ was arbitrary, this holds on all of $(a - R, a + R)$ . Since $f^{'}$ is again a power series with radius $R$ , the same argument applies to it, and induction shows $f$ is $C^{\infty}$ on $(a - R, a + R)$ with every derivative given by repeated term-by-term differentiation.

The integration statement is the same exchange used directly: $\int_{a}^{x} f = \int_{a}^{x} \sum c_{n} (t - a)^{n} = \sum \int_{a}^{x} c_{n} (t - a)^{n} d t = \sum \frac{c _{n}}{n + 1} (x - a)^{n + 1}$ , legitimate on $[a - r, a + r]$ by the uniform convergence of the original series, proved exactly as above with $B (r / ρ)^{n}$ as the dominating terms.∎

Pitfall. Everything here is stated on the open interval. The series $\sum_{n \geq 1} x^{n} / n^{2}$ converges at $x = 1$ , but its term-by-term derivative $\sum_{n \geq 1} x^{n - 1} / n$ diverges there. Nothing in the theorem is violated — $x = 1$ is an endpoint, and the uniform bound used in the proof requires $r < R$ strictly. Differentiating at an endpoint needs a separate argument, typically Abel's theorem, and is not covered by anything above.

Example 2.42 (Building new series from the geometric one). Starting from $\sum_{n = 0}^{\infty} x^{n} = \frac{1}{1 - x}$ on $∣ x ∣ < 1$ , derive series for $\frac{1}{( 1 - x ) ^{2}}$ and for $- ln (1 - x)$ , and use one of them to evaluate $\sum_{n = 1}^{\infty} \frac{n}{2 ^{n}}$ .

Solution.

Differentiate. By the theorem Term-by-term differentiation and integration, on $∣ x ∣ < 1$

\frac{1}{( 1 - x ) ^{2}} = \frac{d}{d x} \frac{1}{1 - x} = n = 1 \sum \infty n x^{n - 1} .

Integrate from $0$ to $x$ , with $∣ x ∣ < 1$ :

- ln (1 - x) = \int_{0}^{x} \frac{d t}{1 - t} = n = 0 \sum \infty \frac{x ^{n + 1}}{n + 1} = m = 1 \sum \infty \frac{x ^{m}}{m} .

Evaluate the numerical series. Multiply the identity in step 1 by $x$ to get $\sum_{n \geq 1} n x^{n} = \frac{x}{( 1 - x ) ^{2}}$ , then set $x = 1/2$ , legitimate since $∣1/2∣ < 1$ :

n = 1 \sum \infty \frac{n}{2 ^{n}} = \frac{1/2}{( 1/2 ) ^{2}} = 2.

Sanity check by direct addition: $0.5 + 0.5 + 0.375 + 0.25 + 0.15625 + 0.09375 = 1.875$ , with the remaining terms contributing the last $0.125$ .
Step 2 at $x = - 1$ yields the alternating harmonic series and the value $ln 2$ — a value that is right but not justified here, since $x = - 1$ is an endpoint and needs Abel's theorem.

□

Exercises for §2.17 →

2.18Taylor series and analytic functions

The theorem Term-by-term differentiation and integration says that a convergent power series is as smooth as a function can be. Differentiating it repeatedly and evaluating at the centre extracts the coefficients, and that computation proves the sum can have only one power series expansion.

Theorem 2.43 (A power series is its own Taylor series). If $f (x) = \sum_{n \geq 0} c_{n} (x - a)^{n}$ with radius $R > 0$ , then $f$ has derivatives of every order at $a$ and

c_{n} = \frac{f ^{(n)} ( a )}{n !} .

In particular, two power series centred at the same point that agree as functions on a neighbourhood of that point have identical coefficients.

Proof. Differentiating $k$ times term by term, which the theorem Term-by-term differentiation and integration licenses inside the radius,

f^{(k)} (x) = n = k \sum \infty n (n - 1) \dots (n - k + 1) c_{n} (x - a)^{n - k} .

Every term with $n > k$ carries a positive power of $(x - a)$ and so vanishes at $x = a$ , leaving only the $n = k$ term, whose coefficient is $k (k - 1) \dots 1 = k!$ . Hence $f^{(k)} (a) = k! c_{k}$ , which rearranges to the claim.

For uniqueness, if $\sum c_{n} (x - a)^{n} = \sum d_{n} (x - a)^{n}$ near $a$ then both sides define the same function $f$ , and the formula forces $c_{n} = f^{(n)} (a) / n! = d_{n}$ for every $n$ .∎

Definition 2.44 (Analytic function). A function $f$ is analytic at $a$ if there is $δ > 0$ and a power series centred at $a$ converging to $f (x)$ for all $∣ x - a ∣ < δ$ . It is analytic on an open set if it is analytic at each of its points.

Every analytic function is $C^{\infty}$ , by the theorem Term-by-term differentiation and integration. The converse is false, and the standard counterexample is worth carrying around because it marks the exact boundary between real and complex analysis.

Example 2.45 (A smooth function that is not analytic). Let $f (x) = e^{- 1/ x^{2}}$ for $x \neq = 0$ and $f (0) = 0$ . Show that $f$ is $C^{\infty}$ with $f^{(n)} (0) = 0$ for every $n$ , and conclude that $f$ is not analytic at $0$ .

Solution.

For $x \neq = 0$ , repeated differentiation by the chain and product rules gives $f^{(n)} (x) = P_{n} (1/ x) e^{- 1/ x^{2}}$ for some polynomial $P_{n}$ ; this is immediate by induction, since differentiating such an expression produces another of the same shape.
At $x = 0$ use the definition of the derivative together with $t^{k} e^{- t^{2}} \to 0$ as $t \to \infty$ , which is the proposition Four standard limits (iv) in continuous form — exponentials beat polynomials. Hence $P_{n} (1/ x) e^{- 1/ x^{2}} \to 0$ as $x \to 0$ , and an induction on $n$ gives $f^{(n)} (0) = 0$ for all $n$ .
So $f$ is $C^{\infty}$ and its Taylor series at $0$ is $\sum_{n} 0 \cdot x^{n}$ , the zero series, which converges everywhere — to the zero function.
But $f (x) > 0$ for every $x \neq = 0$ . So the Taylor series converges, and converges to the wrong function, at every point except the centre. By the definition Analytic function, $f$ is not analytic at $0$ .
The moral: Taylor coefficients record the behaviour of $f$ at the single point $a$ , and for a real function that need not determine $f$ anywhere else. Here $e^{- 1/ x^{2}}$ flattens against the axis faster than any polynomial, so it is invisible to differentiation at the origin. Over $C$ this cannot happen — one complex derivative forces analyticity — which is also why $\frac{1}{1 + x ^{2}}$ , smooth on all of $R$ , has radius exactly $1$ : its singularities sit at $\pm i$ .

□

Exercises for §2.18 →

Common Mistakes

Reversing the quantifiers in the $ε$ – $N$ definition. The order is: $ε$ first, then $N$ , then all $n > N$ . Letting $ε$ depend on $n$ , or $N$ depend on $n$ , produces a statement that every sequence satisfies and that proves nothing.
Using the algebra of limits before knowing the limits exist. The theorem Algebra of limits converts known limits into new ones. Applied to $a_{n} = n$ and $b_{n} = 1/ n$ it "proves" $1 = \infty \cdot 0$ ; the hypothesis that both limits exist in $R$ is what blocks this.
Treating $lim a_{n} = \infty$ as a limit. There is no real number $\infty$ . A sequence with $a_{n} \to + \infty$ diverges; the arrow notation abbreviates "eventually exceeds every bound" and no theorem about convergent sequences applies to it.
Reading "Cauchy" as "consecutive terms get close". $∣ a_{n + 1} - a_{n} ∣ \to 0$ does not imply Cauchy: the harmonic partial sums satisfy it and diverge. The condition must hold for all pairs $m, n > N$ , however far apart.
Expecting Bolzano–Weierstrass to give convergence. It gives one convergent subsequence. $(- 1)^{n}$ is bounded with subsequential limits $1$ and $- 1$ and converges to neither.
Concluding convergence from $a_{n} \to 0$ . The theorem Term test runs one way only. The harmonic series has terms tending to zero and unbounded partial sums.
Continuing to use the ratio test when $L = 1$ . The value $1$ is not a verdict, it is the absence of one — $\sum 1/ n$ and $\sum 1/ n^{2}$ both produce it with opposite outcomes. Switch to comparison or the integral test.
Applying the alternating series test without checking monotonicity. Terms tending to zero with alternating signs is not sufficient; a non-monotone $(b_{n})$ can give a divergent alternating series.
Rearranging a conditionally convergent series. Permitted only under absolute convergence. By Riemann's theorem the alternating harmonic series can be rearranged to converge to any real number at all, so an order-dependent "sum" is not a number the series possesses.
Forgetting the endpoints of a power series. The radius describes the open interval only. All four endpoint patterns occur, and each end must be tested by hand as an ordinary numerical series.
Differentiating a power series at an endpoint. Term-by-term operations are justified strictly inside the radius, where convergence is uniform on compact subintervals. At $∣ x - a ∣ = R$ that argument is unavailable.
Assuming every $C^{\infty}$ function equals its Taylor series. $e^{- 1/ x^{2}}$ (extended by $0$ ) has every derivative zero at the origin and is positive elsewhere, so it is smooth but not analytic there.

About the practice questions. They check that you can carry out this chapter's computations correctly, and each one is graded on a single answer. They are not proof exercises: working through them confirms the mechanics, not that you could prove the results yourself. For that, re-read the statements above and try to reconstruct their proofs with the page closed.

Practice this chapter

Introduction

2.1Sequences as functions on the integers

2.2Monotonicity, boundedness, and subsequences

2.3Convergence and divergence, informally

2.4The ε –N definition

2.5Uniqueness of the limit and the algebra of limits

2.6The squeeze theorem, monotone convergence, and Bolzano–Weierstrass

2.7Cauchy sequences

2.8The Cauchy criterion

2.9Cauchy sequences and completeness

2.10Series as sequences of partial sums

2.11The term test and the harmonic series

2.12The convergence tests

2.13Absolute convergence

2.14Rearrangements and Riemann's theorem

2.15Which tests give absolute convergence

2.16Power series and the radius of convergence

2.17Differentiating and integrating term by term

2.18Taylor series and analytic functions

2.4The $ε$ – $N$ definition