Series & Sequences

A sequence $\{a_n\}$ is an ordered list: $a_1, a_2, a_3, \ldots$ The sequence converges if $\lim_{n\to\infty} a_n = L$ for some finite $L$. If no such $L$ exists, the sequence diverges.

Common sequences: $a_n = 1/n \to 0$, $a_n = (1+1/n)^n \to e$ (the definition of $e$), $a_n = (-1)^n$ diverges (oscillates between $-1$ and $1$, never settling), $a_n = n^2$ diverges to $\infty$.

Tools for evaluating sequence limits: L'Hôpital's Rule (treat $n$ as a continuous variable and apply to $f(x)$), the Squeeze Theorem (trap the sequence between two that converge to the same limit), and the growth-rate hierarchy: $\ln n \ll n^p \ll a^n \ll n! \ll n^n$ for $a > 1$, $p > 0$. This hierarchy means, for example, that $n^{100}/2^n \to 0$ — any exponential eventually crushes any polynomial.

A sequence is monotonic if it is always increasing ($a_{n+1} \geq a_n$) or always decreasing ($a_{n+1} \leq a_n$). A bounded monotonic sequence always converges (Monotone Convergence Theorem). This is useful for proving convergence when you can't find the limit directly.

A sequence is bounded if there exist numbers $m$ and $M$ such that $m \leq a_n \leq M$ for all $n$. Bounded alone doesn't guarantee convergence ($(-1)^n$ is bounded but divergent). You need bounded plus monotonic.

A series $\sum_{n=1}^{\infty} a_n$ is the limit of its partial sums: $S_N = \sum_{n=1}^{N} a_n$. If $\lim_{N\to\infty} S_N$ exists and is finite, the series converges.

Key distinction: a sequence $\{a_n\}$ converging to $0$ is necessary for the series to converge, but not sufficient. $a_n = 1/n \to 0$, yet $\sum 1/n$ diverges.

The harmonic series $\sum 1/n$ diverges. This is one of the most important facts in the theory of series. It shows that terms going to zero isn't enough.

A geometric series has the form $\sum_{n=0}^{\infty} ar^n$ where $a$ is the first term and $r$ is the common ratio.

It converges if and only if $|r| < 1$, and the sum is $\frac{a}{1-r}$.

Example: $\sum_{n=0}^{\infty} \frac{3}{2^n} = \frac{3}{1-1/2} = 6$.

This is the only common series where we can easily compute the exact sum. Many convergence tests compare other series to geometric ones.

A telescoping series is one where consecutive terms in the partial sum cancel, leaving only a few surviving terms — like a collapsing telescope.

The classic example: $\sum_{n=1}^{\infty} \frac{1}{n(n+1)}$. Use partial fractions: $\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$.

Write out the partial sum: $S_N = \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \cdots + \left(\frac{1}{N} - \frac{1}{N+1}\right)$. Almost everything cancels! What survives: $S_N = 1 - \frac{1}{N+1}$. As $N \to \infty$, $S_N \to 1$.

To recognize telescoping: use partial fractions to split the general term into a difference $f(n) - f(n+1)$ (or $f(n) - f(n+k)$ for wider telescoping). If consecutive terms cancel, the partial sum collapses to a simple expression.

Telescoping series are one of the few types where you can find the exact sum, not just determine convergence. Always check for this pattern when partial fractions reveal a clean difference.

If $\lim_{n\to\infty} a_n \neq 0$, then $\sum a_n$ diverges. Full stop. If the terms don't approach zero, they can't possibly add up to a finite sum — the partial sums keep jumping by non-negligible amounts.

This should always be your first check for any series. It's quick and catches obvious divergence. Example: $\sum \frac{n}{n+1}$ diverges because $a_n = \frac{n}{n+1} \to 1 \neq 0$.

Critical limitation: if $\lim_{n\to\infty} a_n = 0$, the test tells you nothing. The series might converge ($\sum 1/n^2$ converges) or diverge ($\sum 1/n$ diverges). Terms going to zero is necessary for convergence but not sufficient. You need another test to decide.

Think of this test as a 'quick rejection filter.' It can rule out convergence (if terms don't go to zero), but it can never confirm convergence.

A p-series has the form $\sum_{n=1}^{\infty} \frac{1}{n^p}$ where $p$ is a positive constant. These are the most important benchmark series in all of convergence testing.

The rule: converges if $p > 1$, diverges if $p \leq 1$. The critical boundary is $p = 1$.

Key examples: $p=1$: the harmonic series $\sum 1/n$ diverges (this is the most important single fact about series). $p=2$: $\sum 1/n^2 = \pi^2/6$ (Euler proved this in 1734 — a celebrated result). $p=1/2$: $\sum 1/\sqrt{n}$ diverges (the terms shrink, but too slowly).

Why $p=1$ diverges: group terms in powers of 2. $1 + \frac{1}{2} + (\frac{1}{3}+\frac{1}{4}) + (\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}) + \cdots > 1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \cdots = \infty$. Each group contributes at least $1/2$.

p-series serve as the go-to comparison: when you encounter a series with terms that 'look like' $1/n^p$ for large $n$, use the limit comparison test against the appropriate p-series.

If $f(x)$ is positive, continuous, and decreasing for $x \geq N$, and $a_n = f(n)$, then $\sum_{n=N}^{\infty} a_n$ and $\int_N^{\infty} f(x)\,dx$ either both converge or both diverge. The series and the integral live or die together.

Why it works: the sum $\sum a_n$ is a left Riemann sum for $\int f(x)\,dx$ (or a right Riemann sum, depending on the direction). Since $f$ is decreasing, these Riemann sums sandwich the integral. If one is finite, the other must be too.

The integral test does not give the exact sum — just convergence or divergence. The integral's value is related to but not equal to the series sum.

Example: does $\sum_{n=2}^{\infty} \frac{1}{n(\ln n)^2}$ converge? Let $f(x) = \frac{1}{x(\ln x)^2}$. It's positive, continuous, and decreasing for $x \geq 2$. $\int_2^\infty \frac{1}{x(\ln x)^2}\,dx$: substitute $u = \ln x$, giving $\int_{\ln 2}^{\infty} u^{-2}\,du = \frac{1}{\ln 2}$. The integral converges, so the series converges.

Remainder estimate: $\int_{N+1}^{\infty} f(x)\,dx \leq R_N \leq \int_N^{\infty} f(x)\,dx$ where $R_N = \sum_{n=N+1}^{\infty} a_n$ is the error from truncating at $N$ terms. This gives concrete error bounds.

Direct comparison: if $0 \leq a_n \leq b_n$ for all sufficiently large $n$ and $\sum b_n$ converges, then $\sum a_n$ converges (a smaller series is forced to converge if a larger one does). Conversely, if $\sum a_n$ diverges, then $\sum b_n$ diverges.

Direct comparison requires an inequality that holds term by term. This can be tricky to establish. The limit comparison test removes that difficulty.

Limit comparison: if $\lim_{n\to\infty} a_n / b_n = L$ where $0 < L < \infty$, then $\sum a_n$ and $\sum b_n$ either both converge or both diverge. The series behave the same because their terms are proportional for large $n$.

Strategy: look at your series for large $n$ and simplify by dropping lower-order terms. $\frac{n}{n^3+1}$ behaves like $\frac{n}{n^3} = \frac{1}{n^2}$ for large $n$. Use $b_n = 1/n^2$ in the limit comparison test. Since $\sum 1/n^2$ converges ($p = 2 > 1$), the original series converges.

Choosing $b_n$: almost always a p-series or geometric series. Keep only the highest-power terms in the numerator and denominator, cancel, and you'll get $b_n = 1/n^p$ for some $p$. Then the p-series rule tells you the answer.

Ratio test: compute $L = \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right|$. If $L < 1$: converges absolutely. $L > 1$: diverges. $L = 1$: inconclusive.

Root test: compute $L = \lim_{n\to\infty} \sqrt[n]{|a_n|}$. Same criteria as the ratio test.

The ratio test excels for series with factorials (like $n!/3^n$) or products. The root test is ideal for terms raised to the $n$th power (like $(2/3)^n$).

Warning: both tests are inconclusive when $L=1$. This happens for all p-series, so you can't use ratio/root tests on $\sum 1/n^p$.

An alternating series has terms that switch sign: $\sum (-1)^n a_n$ or $\sum (-1)^{n+1} a_n$ where $a_n > 0$. The positive and negative terms partially cancel each other.

The Alternating Series Test (Leibniz's test): the series converges if two conditions hold: (1) $\{a_n\}$ is eventually decreasing ($a_{n+1} \leq a_n$ for large enough $n$), and (2) $\lim_{n\to\infty} a_n = 0$.

Example: the alternating harmonic series $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} = 1 - 1/2 + 1/3 - 1/4 + \cdots = \ln 2$. The regular harmonic series diverges, but the alternation saves it — the positive and negative terms cancel just enough to give a finite sum.

Alternating series estimation theorem: the error from truncating after $N$ terms satisfies $|R_N| \leq a_{N+1}$ (the first omitted term). This is an exceptionally convenient error bound — you know exactly how accurate your partial sum is. If $a_{N+1} = 0.001$, your approximation is within $0.001$ of the true sum.

This estimation theorem is why alternating series are 'friendly' — they come with a built-in accuracy guarantee, unlike most other convergent series.

A series $\sum a_n$ converges absolutely if $\sum |a_n|$ converges. It converges conditionally if $\sum a_n$ converges but $\sum |a_n|$ diverges.

Absolute convergence implies convergence. So if the absolute value series converges, you're done.

Example: $\sum (-1)^n/n^2$ converges absolutely because $\sum 1/n^2$ converges.

Example: $\sum (-1)^n/n$ converges conditionally. It converges by the alternating series test, but $\sum 1/n$ diverges.

Why it matters: absolutely convergent series can be rearranged freely without changing the sum. Conditionally convergent series cannot (Riemann rearrangement theorem).

A power series centered at $a$ is $\sum_{n=0}^{\infty} c_n(x-a)^n = c_0 + c_1(x-a) + c_2(x-a)^2 + \cdots$. It's an 'infinite polynomial' with infinitely many terms.

The radius of convergence $R$ determines where it converges: $|x-a| < R$ (converges absolutely), $|x-a| > R$ (diverges). At $|x-a| = R$ (the endpoints), you must check separately — the series may converge at one, both, or neither endpoint.

Three possible scenarios: $R = 0$ (converges only at $x = a$), $0 < R < \infty$ (converges on a finite interval), or $R = \infty$ (converges for all $x$, like the series for $e^x$).

Find $R$ using the ratio test: $R = \lim_{n\to\infty} \left|\frac{c_n}{c_{n+1}}\right|$ (or equivalently, $1/R = \lim_{n\to\infty}|c_{n+1}/c_n|$). The root test also works: $1/R = \limsup_{n\to\infty} |c_n|^{1/n}$.

Inside the radius, power series behave like polynomials: you can differentiate and integrate them term by term, and the radius of convergence stays the same (endpoints may change).

The simplest power series: $\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n$ for $|x| < 1$ ($R = 1$). From this, many other series can be derived by substitution, differentiation, or integration.

The Taylor series of $f(x)$ centered at $a$ is: $f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n$.

A Maclaurin series is a Taylor series centered at $a=0$.

Essential Maclaurin series to memorize:

$e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$ for all $x$.

$\sin x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots$ for all $x$.

$\cos x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots$ for all $x$.

$\ln(1+x) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^n}{n} = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots$ for $-1 < x \leq 1$.

$\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n$ for $|x| < 1$.

The $n$th degree Taylor polynomial $T_n(x)$ approximates $f(x)$ near $x = a$. The error (remainder) is $R_n(x) = f(x) - T_n(x)$ — the difference between the true function and the polynomial approximation.

Taylor's inequality (Lagrange error bound): $|R_n(x)| \leq \frac{M}{(n+1)!}|x-a|^{n+1}$, where $M$ is an upper bound for $|f^{(n+1)}(t)|$ on the interval between $a$ and $x$. The $(n+1)!$ in the denominator grows extremely fast, which is why Taylor polynomials get accurate quickly.

For alternating Taylor series (like $\sin x$, $\cos x$, $\ln(1+x)$), the alternating series estimation applies: $|R_n| \leq |a_{n+1}|$ (the first omitted term). This is usually easier to use than the Lagrange bound.

Practical use: 'How many terms of the Maclaurin series for $e^x$ do I need to approximate $e^{0.5}$ to within $0.0001$?' Find the smallest $n$ such that $\frac{(0.5)^{n+1}}{(n+1)!} < 0.0001$. With $n = 5$: $\frac{0.5^6}{720} \approx 0.0000217 < 0.0001$. So 6 terms suffice (through the $x^5$ term).

This is how calculators and computers evaluate functions like $\sin$, $\cos$, and $e^x$ — they use Taylor polynomials with enough terms to guarantee the desired precision.

Instead of computing Taylor coefficients from scratch (which requires evaluating $n$ derivatives), you can derive new series from known ones using three operations: substitution, differentiation, and integration.

Substitution: replace $x$ in a known series with an expression. $e^{-x^2}$? Start with $e^u = \sum u^n/n!$ and substitute $u = -x^2$: $e^{-x^2} = \sum \frac{(-1)^n x^{2n}}{n!}$. No derivatives needed.

Term-by-term differentiation: differentiate a known series to get a new one. $\frac{1}{1-x} = \sum x^n$ for $|x|<1$. Differentiate both sides: $\frac{1}{(1-x)^2} = \sum n x^{n-1}$. The radius of convergence is preserved.

Term-by-term integration: integrate a known series. $\frac{1}{1+t^2} = \sum (-1)^n t^{2n}$ for $|t|<1$. Integrate from $0$ to $x$: $\arctan x = \sum \frac{(-1)^n x^{2n+1}}{2n+1}$. This gives the series for $\arctan$ without ever differentiating $\arctan$.

Multiplication and addition: you can add or multiply series. $\frac{x}{1-x^2} = x \cdot \frac{1}{1-x^2} = x \sum x^{2n} = \sum x^{2n+1}$.

These techniques are immensely powerful. In practice, almost every Taylor series you encounter can be derived from the five basic series ($e^x$, $\sin x$, $\cos x$, $\ln(1+x)$, $1/(1-x)$) using these operations.

1. Divergence test first: if $a_n \not\to 0$, done (diverges).

2. Is it geometric? Check if $a_n = ar^n$.

3. Is it a p-series? Check if $a_n = 1/n^p$.

4. Does it telescope? Try partial fractions.

5. Are there factorials or $n$th powers? Try ratio or root test.

6. Does it look like a known series? Try comparison or limit comparison.

7. Is it alternating? Try the alternating series test.

8. Is $f(n) = a_n$ easy to integrate? Try the integral test.