Print this page and work through it on paper. Revealed answers are never printed, so you can check one question and still print a clean sheet.

1. 1.Computation · 8 min

   Find the limit of each sequence, or show that it diverges. Name the result that licenses each step (a limit law, the Squeeze Theorem, the corollary on absolute values, the Monotone Sequence Theorem).

   1. (a)$a_n=\frac{3n^2-n}{2n^2+5}$
   2. (b)$b_n=\sqrt{n^2+4n}-n$
   3. (c)$c_n=\frac{(-1{)}^n\sqrt{n}}{n+1}$
   4. (d)$a_1=1$ and $a_{n+1}=\sqrt{3a_n+4}$. Prove that the sequence converges before you find its limit.
   Show the solution to question 1

   (a) Divide by the dominant power. Dividing numerator and denominator by $n^2$,

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

   Since $\frac{1}{n}\to 0$ and $\frac{5}{n^2}\to 0$, the sum and constant-multiple laws give numerator $\to 3$ and denominator $\to 2\ne 0$, so the quotient law gives $a_n\to \frac{3}{2}$.

   (b) Rationalise. The form is $\infty -\infty$, so nothing may be concluded from the pieces. Multiply and divide by the conjugate:

   $$\sqrt{n^2+4n}-n=\frac{(n^2+4n)-n^2}{\sqrt{n^2+4n}+n}=\frac{4n}{\sqrt{n^2+4n}+n}=\frac{4}{\sqrt{1+\frac{4}{n}}+1},$$

   where the last step divides top and bottom by $n>0$ (so $\sqrt{n^2}=n$). The denominator tends to $1+1=2$ by continuity of the square root, so $b_n\to 2$.

   Sanity check: $b_{100}=\sqrt{10400}-100\approx 101.98-100=1.98$.

   (c) Absolute values. The sign flips make $c_n$ itself hard to handle directly, so look at its size:

   $$0\le |c_n|=\frac{\sqrt{n}}{n+1}<\frac{\sqrt{n}}{n}=\frac{1}{\sqrt{n}}⟶0.$$

   By the Squeeze Theorem $|c_n|\to 0$, and by the corollary on absolute values $c_n\to 0$. (Contrast $(-1{)}^n$, whose absolute value tends to $1$: there the corollary says nothing and the sequence diverges.)

   (d) Monotone Sequence Theorem. Compute a few terms to see what to prove: $a_1=1$, $a_2=\sqrt{7}\approx 2.65$, $a_3=\sqrt{3\sqrt{7}+4}\approx 3.46$. The terms seem to rise toward $4$.

   Bounded above by $4$. True for $a_1=1$. If $a_n<4$ then $3a_n+4<16$, so $a_{n+1}=\sqrt{3a_n+4}<4$. By induction every term is below $4$. Every term is also positive, since $a_1>0$ and square roots are nonnegative.

   Increasing. Because both sides are positive, $a_{n+1}>a_n$ is equivalent to $3a_n+4>a_n^2$, that is $a_n^2-3a_n-4<0$, that is $(a_n-4)(a_n+1)<0$. With $0<a_n<4$ the first factor is negative and the second positive, so the inequality holds.

   The sequence is increasing and bounded above, so by the Monotone Sequence Theorem it converges to some $L$. Now, and only now, take limits in the recursion: $a_{n+1}\to L$ as well, and the square root is continuous, so

   $$L=\sqrt{3L+4}⟹L^2-3L-4=0⟹(L-4)(L+1)=0.$$

   Since every term is at least $a_1=1$, so is $L$, which rules out $-1$. The limit is $4$. Solving the equation before proving convergence would have been a bluff: for a divergent sequence the equation can still have a solution, and that solution means nothing.

2. 2.Computation · 8 min

   Sum each series exactly. Identify the first term actually present and the common ratio, and state why the series converges before you use the formula. In (d), find every $x$ for which the series converges before summing.

   1. (a)$\sum _{n=2}^{\infty }\frac{3^{n+1}}{5^n}$
   2. (b)The repeating decimal $0.4\overline{63}=0.4636363\dots$, written as a ratio of integers in lowest terms.
   3. (c)$\sum _{n=1}^{\infty }\frac{2^n+(-1{)}^n}{3^n}$
   4. (d)$\sum _{n=0}^{\infty }\frac{(x-3{)}^n}{2^n}$
   Show the solution to question 2

   (a) Read off the first term present. The term is $3\cdot {\left(\frac{3}{5}\right)}^n$, so the ratio is $r=\frac{3}{5}$ with $|r|<1$: the series converges. The first term present is at $n=2$: $\frac{3^3}{5^2}=\frac{27}{25}$. Then

   $$\sum _{n=2}^{\infty }\frac{3^{n+1}}{5^n}=\frac{\text{first term}}{1-r}=\frac{27/25}{1-3/5}=\frac{27/25}{2/5}=\frac{27}{10}.$$

   Sanity check: $1.08+0.648+0.389\approx 2.12$ after three terms, and the remaining tail is $0.233/(1-0.6)\approx 0.58$, giving $2.70$.

   (b) Separate the non-repeating part.

   $$0.4636363\dots =\frac{4}{10}+\frac{63}{10^3}+\frac{63}{10^5}+\frac{63}{10^7}+\cdots$$

   After $\frac{4}{10}$ the terms are geometric with first term $a=\frac{63}{1000}$ and ratio $r=\frac{1}{100}$, so they sum to

   $$\frac{63/1000}{1-1/100}=\frac{63/1000}{99/100}=\frac{63}{990}=\frac{7}{110}.$$

   Hence $0.4\overline{63}=\frac{4}{10}+\frac{7}{110}=\frac{44}{110}+\frac{7}{110}=\frac{51}{110}$. Lowest terms: $51=3\cdot 17$ and $110=2\cdot 5\cdot 11$ share no factor. Check: $110\times 0.4636\approx 51.0$.

   (c) Split, because both pieces converge. Term by term,

   $$\frac{2^n+(-1{)}^n}{3^n}={\left(\frac{2}{3}\right)}^n+{\left(-\frac{1}{3}\right)}^n.$$

   Both are geometric with $|r|<1$, so each series converges on its own, and the algebra of convergent series allows the split. First: ${\sum }_{n\ge 1}{\left(\frac{2}{3}\right)}^n=\frac{2/3}{1-2/3}=2$. Second: ${\sum }_{n\ge 1}{\left(-\frac{1}{3}\right)}^n=\frac{-1/3}{1+1/3}=\frac{-1/3}{4/3}=-\frac{1}{4}$. Total $2-\frac{1}{4}=\frac{7}{4}$.

   Sanity check: partial sums $\frac{1}{3},\frac{8}{9},1.15,1.36,1.49,\dots$ climbing toward $1.75$.

   (d) A geometric series with a variable ratio. Here $a=1$ and $r=\frac{x-3}{2}$. It converges exactly when $\left|\frac{x-3}{2}\right|<1$, that is $|x-3|<2$, that is $1<x<5$. At $x=1$ and $x=5$ the ratio is $\mp 1$, the terms do not tend to $0$, and the series diverges. For $1<x<5$,

   $$\sum _{n=0}^{\infty }\frac{(x-3{)}^n}{2^n}=\frac{1}{1-\frac{x-3}{2}}=\frac{2}{2-(x-3)}=\frac{2}{5-x}.$$

   Check at $x=3$: the series is $1+0+0+\cdots =1$ and the formula gives $\frac{2}{2}=1$. This is a power series centred at $3$ with radius $2$, summed by the one method that gives a closed form.

3. 3.Computation · 8 min

   Each series telescopes. In every case write the partial sum $s_N$ explicitly, showing which terms survive the cancellation, before you take the limit.

   1. (a)$\sum _{n=1}^{\infty }\frac{2}{(2n-1)(2n+1)}$
   2. (b)$\sum _{n=1}^{\infty }\frac{1}{n(n+3)}$
   3. (c)$\sum _{n=1}^{\infty }\ln \left(\frac{n(n+2)}{(n+1{)}^2}\right)$
   Show the solution to question 3

   (a) Partial fractions, gap of one. Write $\frac{2}{(2n-1)(2n+1)}=\frac{A}{2n-1}+\frac{B}{2n+1}$, so $2=A(2n+1)+B(2n-1)$. Setting $n=\frac{1}{2}$ gives $A=1$; setting $n=-\frac{1}{2}$ gives $B=-1$. So $a_n=\frac{1}{2n-1}-\frac{1}{2n+1}$, which is $b_n-b_{n+1}$ with $b_n=\frac{1}{2n-1}$, and

   $$s_N=\left(1-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\cdots +\left(\frac{1}{2N-1}-\frac{1}{2N+1}\right)=1-\frac{1}{2N+1}.$$

   As $N\to \infty$, $s_N\to 1$. Check: $s_2=\frac{2}{3}+\frac{2}{15}=\frac{4}{5}=1-\frac{1}{5}$.

   (b) Gap of three: three survivors at each end. Partial fractions give $\frac{1}{n(n+3)}=\frac{1}{3}\left(\frac{1}{n}-\frac{1}{n+3}\right)$. Write out enough of $3s_N$ to see the pattern:

   $$3s_N=\left(1-\frac{1}{4}\right)+\left(\frac{1}{2}-\frac{1}{5}\right)+\left(\frac{1}{3}-\frac{1}{6}\right)+\left(\frac{1}{4}-\frac{1}{7}\right)+\cdots +\left(\frac{1}{N}-\frac{1}{N+3}\right).$$

   Every negative $-\frac{1}{k}$ with $4\le k\le N$ is cancelled by the positive $\frac{1}{k}$ three brackets later. What survives is the three leading positives and the three trailing negatives:

   $$s_N=\frac{1}{3}\left(1+\frac{1}{2}+\frac{1}{3}-\frac{1}{N+1}-\frac{1}{N+2}-\frac{1}{N+3}\right)=\frac{1}{3}\left(\frac{11}{6}-\frac{1}{N+1}-\frac{1}{N+2}-\frac{1}{N+3}\right).$$

   The three tail fractions tend to $0$, so the sum is $\frac{1}{3}\cdot \frac{11}{6}=\frac{11}{18}$. Check at $N=1$: the formula gives $\frac{1}{3}\left(\frac{11}{6}-\frac{1}{2}-\frac{1}{3}-\frac{1}{4}\right)=\frac{1}{3}\cdot \frac{22-6-4-3}{12}=\frac{1}{3}\cdot \frac{9}{12}=\frac{1}{4}$, which is $a_1$. Answering $\frac{1}{3}\cdot 1$ by cancelling “everything” would be wrong: the gap leaves three survivors, not one.

   (c) A logarithm that telescopes. By the laws of logarithms,

   $$\ln \frac{n(n+2)}{(n+1{)}^2}=\ln \frac{n}{n+1}-\ln \frac{n+1}{n+2},$$

   so with $b_n=\ln \frac{n}{n+1}$ the term is $b_n-b_{n+1}$ and $s_N=b_1-b_{N+1}=\ln \frac{1}{2}-\ln \frac{N+1}{N+2}$. Since $\frac{N+1}{N+2}\to 1$ and $\ln$ is continuous at $1$, the second logarithm tends to $\ln 1=0$, and

   $$\sum _{n=1}^{\infty }\ln \frac{n(n+2)}{(n+1{)}^2}=\ln \frac{1}{2}=-\ln 2.$$

   Check: $s_3=\ln \frac{3}{4}+\ln \frac{8}{9}+\ln \frac{15}{16}=\ln \frac{3\cdot 8\cdot 15}{4\cdot 9\cdot 16}=\ln \frac{360}{576}=\ln \frac{5}{8}$, and the formula gives $\ln \frac{1}{2}-\ln \frac{4}{5}=\ln \frac{5}{8}$ as well. A negative sum is fine: every term is the logarithm of a number below $1$.

4. 4.Critique · 7 min

   Three students' arguments follow. Each reaches a false conclusion. For each: quote the sentence that is wrong, name the theorem or hypothesis being misused, and give the correct verdict with a correct reason.

   1. (a)“For $\sum _{n=1}^{\infty }\frac{1}{\sqrt{n}}$: the terms $1/\sqrt{n}$ tend to $0$, so by the Divergence Test the series converges.”
   2. (b)“For $\sum _{n=1}^{\infty }(-1{)}^n\frac{n+1}{n}$: the signs alternate and $b_n=\frac{n+1}{n}=1+\frac{1}{n}$ is decreasing, so the Alternating Series Test shows the series converges.”
   3. (c)“For $\sum _{n=1}^{\infty }{\left(1-\frac{1}{n}\right)}^n$: the base $1-\frac{1}{n}$ is less than $1$, so ${\left(1-\frac{1}{n}\right)}^n$ decays like $r^n$ with $r<1$, and the series converges by comparison with a geometric series.”
   Show the solution to question 4

   (a) The wrong sentence is “so by the Divergence Test the series converges.” The Divergence Test runs in one direction only: if $a_n\nrightarrow 0$ then $\sum a_n$ diverges. When $a_n\to 0$ it says nothing at all; it never proves convergence. Terms tending to $0$ is necessary for convergence, not sufficient.

   Correct verdict: this is a $p$-series with $p=\frac{1}{2}\le 1$, so it diverges. (Via the Integral Test: ${\int }_1^{\infty }{x}^{-1/2}dx={\lim }_{t\to \infty }\left(2\sqrt{t}-2\right)=\infty$.) The student's conclusion is not just unjustified but false.

   (b) The wrong sentence is “so the Alternating Series Test shows the series converges.” The test has two hypotheses, and the student checked only one. The other is $b_n\to 0$, and here $b_n=1+\frac{1}{n}\to 1\ne 0$. The test does not apply.

   Correct verdict: the terms $(-1{)}^n\left(1+\frac{1}{n}\right)$ approach $-1,1,-1,1,\dots$ and have no limit, so by the Divergence Test the series diverges. Check: the partial sums run $-2,-0.5,-1.83,-0.58,\dots$ and never settle.

   (c) The wrong sentence is “decays like $r^n$ with $r<1$.” Comparison with a geometric series needs a fixed ratio $r<1$. Here the base $1-\frac{1}{n}$ is below $1$ but climbs toward $1$ as the exponent grows, and the two effects compete; there is no single $r$ with ${\left(1-\frac{1}{n}\right)}^n\le r^n$ for all large $n$.

   Compute the limit of the terms instead. Take logarithms: $\ln {\left(1-\frac{1}{n}\right)}^n=n\ln \left(1-\frac{1}{n}\right)$. Pass to the function $x\ln (1-1/x)=\frac{\ln (1-1/x)}{1/x}$, which is of the form $\frac{0}{0}$ as $x\to \infty$. L'Hôpital's Rule (both parts differentiable for $x>1$, denominator's derivative $-1/x^2\ne 0$) gives

   $$\underset{x\to \infty }{\lim }\frac{\ln (1-1/x)}{1/x}=\underset{x\to \infty }{\lim }\frac{\frac{1/x^2}{1-1/x}}{-1/x^2}=\underset{x\to \infty }{\lim }\frac{-1}{1-1/x}=-1.$$

   By the theorem on sequence limits from function limits, $n\ln (1-\frac{1}{n})\to -1$, and since $\exp$ is continuous, ${\left(1-\frac{1}{n}\right)}^n\to e^{-1}\ne 0$. By the Divergence Test the series diverges. Sanity check: $0.9^{10}\approx 0.349$, close to $1/e\approx 0.368$ and nowhere near $0$.

5. 5.Sketch · 9 min

   Let $H_N=1+\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{N}$. This question asks for the picture behind the Integral Test, and then makes the picture pay.

   1. (a)On one set of axes draw $y=1/x$ for $1\le x\le 5$ together with four rectangles of width $1$ whose areas are $\frac{1}{2},\frac{1}{3},\frac{1}{4},\frac{1}{5}$ and which lie entirely below the curve. On a second set of axes draw the curve with four rectangles of areas $1,\frac{1}{2},\frac{1}{3},\frac{1}{4}$ lying entirely above it. For each picture say exactly where each rectangle touches the curve, and which hypothesis of the Integral Test guarantees that the rectangle is on the correct side.
   2. (b)Read off from the two pictures, for a general $N$, the inequalities $\ln (N+1)\le H_N\le 1+\ln N$, computing the integral that appears.
   3. (c)Deduce that the harmonic series diverges, and show that $H_N<20$ for every $N<e^{19}$.
   Show the solution to question 5

   (a) The rectangles below the curve. Draw $y=1/x$ falling from $(1,1)$ to $(5,\frac{1}{5})$. For $k=2,3,4,5$ put a rectangle on the base $[k-1,k]$ with height $\frac{1}{k}$: on $[1,2]$ height $\frac{1}{2}$, on $[2,3]$ height $\frac{1}{3}$, on $[3,4]$ height $\frac{1}{4}$, on $[4,5]$ height $\frac{1}{5}$. Each has area $\frac{1}{k}$, and its top-right corner $(k,\frac{1}{k})$ sits on the curve. Its top edge runs to the left from there, and stays under the curve because $f(x)=1/x$ is decreasing: for $x\le k$ we have $1/x\ge 1/k$. The picture is a staircase tucked under the curve, touching it at the right end of every step. Its total area $\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}=H_5-1$ is at most ${\int }_1^5\frac{dx}{x}$.

   The rectangles above the curve. Redraw the curve, and for $k=1,2,3,4$ put a rectangle on the base $[k,k+1]$ with height $\frac{1}{k}$: on $[1,2]$ height $1$, on $[2,3]$ height $\frac{1}{2}$, on $[3,4]$ height $\frac{1}{3}$, on $[4,5]$ height $\frac{1}{4}$. Now the top-left corner $(k,\frac{1}{k})$ is on the curve, and the top edge runs to the right above the curve, again because $f$ is decreasing: for $x\ge k$, $1/x\le 1/k$. Its total area $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}=H_4$ is at least ${\int }_1^5\frac{dx}{x}$.

   The other two hypotheses do quieter jobs: positivity makes every area a genuine area and the partial sums increasing, and continuity makes the integral meaningful. Decreasing is what decides the side.

   (b) The general inequalities. Below the curve, use the $N-1$ rectangles of heights $\frac{1}{2},\dots ,\frac{1}{N}$ on $[1,N]$:

   $$H_N-1=\frac{1}{2}+\cdots +\frac{1}{N}\le {\int }_1^N\frac{dx}{x}=[\ln x{]}_1^N=\ln N,\text{so}{H}_N\le 1+\ln N.$$

   Above the curve, use the $N$ rectangles of heights $1,\dots ,\frac{1}{N}$ on $[1,N+1]$:

   $$H_N=1+\frac{1}{2}+\cdots +\frac{1}{N}\ge {\int }_1^{N+1}\frac{dx}{x}=\ln (N+1).$$

   Check at $N=4$: $\ln 5\approx 1.609\le H_4=\frac{25}{12}\approx 2.083\le 1+\ln 4\approx 2.386$.

   (c) Divergence, and how slow it is. From the lower bound, $H_N\ge \ln (N+1)\to \infty$, so the partial sums of $\sum \frac{1}{n}$ are unbounded and the harmonic series diverges. (This is the Integral Test's proof, with the integral computed.)

   From the upper bound, $H_N\le 1+\ln N$. If $N<e^{19}$ then $\ln N<19$, so $H_N<1+19=20$. Since $e^{19}$ is roughly $1.8\times 10^8$, none of the first hundred and eighty million partial sums reaches $20$: the series diverges, but at the pace of $\ln N$, and no amount of adding by hand will ever show it.

6. 6.Computation · 9 min

   Decide whether each series converges. Use a comparison test each time: name the comparison series, say whether you are using direct or limit comparison, and exhibit either the inequality (with its direction) or the limit (with its value).

   1. (a)$\sum _{n=1}^{\infty }\frac{2n^2+1}{n^4+3n}$
   2. (b)$\sum _{n=1}^{\infty }\frac{1}{\sqrt{n^2-n+1}}$
   3. (c)$\sum _{n=1}^{\infty }\frac{3^n}{4^n-2^n}$
   4. (d)$\sum _{n=1}^{\infty }\frac{1}{n^{1+1/n}}$ — you may quote $\sqrt[n]{n}\to 1$.
   Show the solution to question 6

   (a) Limit comparison with $1/n^2$. Keeping the dominant powers, the term looks like $\frac{2n^2}{n^4}=\frac{2}{n^2}$. With $b_n=\frac{1}{n^2}$,

   $$\frac{a_n}{b_n}=\frac{(2n^2+1)n^2}{n^4+3n}=\frac{2+\frac{1}{n^2}}{1+\frac{3}{n^3}}⟶2,$$

   which is finite and positive. Since $\sum 1/n^2$ converges ($p=2>1$), the series converges by the Limit Comparison Test. (Direct comparison also works: $\frac{2n^2+1}{n^4+3n}\le \frac{3n^2}{n^4}=\frac{3}{n^2}$.)

   (b) Direct comparison with the harmonic series. For $n\ge 1$, $n^2-n+1\le n^2$ because $1-n\le 0$; and $n^2-n+1>0$ always (its discriminant $1-4$ is negative). Taking square roots of positive quantities, $\sqrt{n^2-n+1}\le n$, so

   $$\frac{1}{\sqrt{n^2-n+1}}\ge \frac{1}{n}>0.$$

   The series dominates $\sum \frac{1}{n}$, which diverges, so it diverges by the Direct Comparison Test. Direction check: to prove divergence one must bound below by a divergent series, which is what was done.

   (c) Comparison with a geometric series. For large $n$ the $2^n$ is negligible next to $4^n$, so the term behaves like ${\left(\frac{3}{4}\right)}^n$. Limit comparison with $b_n={\left(\frac{3}{4}\right)}^n$:

   $$\frac{a_n}{b_n}=\frac{3^n}{4^n-2^n}\cdot \frac{4^n}{3^n}=\frac{1}{1-{\left(\frac{1}{2}\right)}^n}⟶1.$$

   Since $\sum {\left(\frac{3}{4}\right)}^n$ is geometric with ratio below $1$, the series converges. A direct comparison is available too: for $n\ge 1$, $2^n\le \frac{1}{2}\cdot 4^n$ (equivalently $2\le 2^n$), so $4^n-2^n\ge \frac{1}{2}\cdot 4^n$ and $0<a_n\le 2{\left(\frac{3}{4}\right)}^n$.

   (d) Not a $p$-series. The tempting argument “the exponent exceeds $1$, so it converges” is wrong: the $p$-series rule needs a constant exponent, and $1+\frac{1}{n}$ shrinks toward $1$. Limit comparison with $b_n=\frac{1}{n}$:

   $$\frac{a_n}{b_n}=\frac{n}{n^{1+1/n}}=\frac{1}{n^{1/n}}⟶\frac{1}{1}=1,$$

   using $\sqrt[n]{n}\to 1$ (which follows from $\ln \sqrt[n]{n}=\frac{\ln n}{n}\to 0$ and continuity of $\exp$). The harmonic series diverges, so this series diverges. Its terms are only a factor $n^{1/n}$, which tends to $1$, smaller than $\frac{1}{n}$ — not nearly enough.

7. 7.Computation · 8 min

   Both parts concern alternating series. State both hypotheses of the Alternating Series Test each time you use it and verify them; where the terms are not obviously decreasing, use a derivative.

   1. (a)Show that $\sum _{n=1}^{\infty }(-1{)}^n\frac{\sqrt{n}}{n+4}$ converges, then decide whether the convergence is absolute or conditional.
   2. (b)Let $S=\sum _{n=1}^{\infty }\frac{(-1{)}^n}{n!}$. Find the smallest $N$ for which the Alternating Series Estimation Theorem guarantees $|S-s_N|<\frac{1}{5000}$, compute that $s_N$ exactly as a fraction in lowest terms, and say on which side of $s_N$ the true sum lies.
   Show the solution to question 7

   (a) The two hypotheses. Here $b_n=\frac{\sqrt{n}}{n+4}>0$.

   Limit. Dividing by $n$: $b_n=\frac{1/\sqrt{n}}{1+4/n}\to \frac{0}{1}=0$.

   Decreasing. Let $f(x)=\frac{\sqrt{x}}{x+4}$. By the quotient rule,

   $$f^{\prime }(x)=\frac{\frac{1}{2\sqrt{x}}(x+4)-\sqrt{x}}{(x+4{)}^2}=\frac{(x+4)-2x}{2\sqrt{x}(x+4{)}^2}=\frac{4-x}{2\sqrt{x}(x+4{)}^2},$$

   which is negative for $x>4$. So $b_{n+1}\le b_n$ for all $n\ge 4$, and “for all large $n$” is all the test asks (the first few terms never affect convergence). By the Alternating Series Test the series converges.

   Absolute or conditional? $|a_n|=\frac{\sqrt{n}}{n+4}$ behaves like $\frac{\sqrt{n}}{n}=\frac{1}{\sqrt{n}}$. Limit comparison with $\frac{1}{\sqrt{n}}$:

   $$\frac{\sqrt{n}}{n+4}\cdot \sqrt{n}=\frac{n}{n+4}⟶1,$$

   and $\sum 1/\sqrt{n}$ diverges ($p=\frac{1}{2}$), so $\sum |a_n|$ diverges. The series converges conditionally: the cancellation between signs is doing all the work.

   (b) Error control. Here $b_n=\frac{1}{n!}$: positive, decreasing (since $\frac{1}{(n+1)!}=\frac{1}{n+1}\cdot \frac{1}{n!}$) and tending to $0$, so the estimation theorem applies: $|S-s_N|\le b_{N+1}=\frac{1}{(N+1)!}$. We need $(N+1)!>5000$. Since $6!=720$ and $7!=5040$, the smallest choice is $N+1=7$, that is $N=6$; with $N=5$ the guarantee is only $\frac{1}{720}$.

   Now add six terms over the common denominator $720$:

   $$s_6=-1+\frac{1}{2}-\frac{1}{6}+\frac{1}{24}-\frac{1}{120}+\frac{1}{720}=\frac{-720+360-120+30-6+1}{720}=\frac{-455}{720}=-\frac{91}{144},$$

   dividing by $5$; and $91=7\cdot 13$, $144=2^4\cdot 3^2$ share no factor.

   The sign of $S-s_6$ is the sign of the first omitted term, $\frac{(-1{)}^7}{7!}<0$, so the true sum lies below $s_6$:

   $$-\frac{91}{144}-\frac{1}{5040}<S<-\frac{91}{144}.$$

   A check that needs no calculator's trust: the series is the exponential series at $x=-1$ with the $n=0$ term removed, so $S=e^{-1}-1\approx -0.63212$, while $-\frac{91}{144}\approx -0.63194$. The true value is indeed below, by about $0.00018<\frac{1}{5040}\approx 0.000198$.

8. 8.Proof · 9 min

   Recall that $\sum a_n$ converges absolutely if $\sum |a_n|$ converges, and conditionally if $\sum a_n$ converges but $\sum |a_n|$ does not.

   1. (a)Classify $\sum _{n=2}^{\infty }(-1{)}^n\frac{\ln n}{n}$ as absolutely convergent, conditionally convergent or divergent.
   2. (b)Prove: if $\sum a_n$ converges absolutely, then $\sum a_n^2$ converges.
   3. (c)Show by example that neither converse holds: give a series $\sum a_n$ with $\sum a_n^2$ convergent but $\sum a_n$ not absolutely convergent, and a series $\sum a_n$ that converges while $\sum a_n^2$ diverges.
   Show the solution to question 8

   (a) Let $b_n=\frac{\ln n}{n}$, positive for $n\ge 2$.

   Limit. The form is $\frac{\infty }{\infty }$; pass to $f(x)=\frac{\ln x}{x}$ and apply L'Hôpital's Rule: $\frac{1/x}{1}\to 0$. By the theorem on sequence limits from function limits, $b_n\to 0$.

   Decreasing. $f^{\prime }(x)=\frac{1-\ln x}{x^2}<0$ for $x>e$, so $b_n$ decreases from $n=3$ onward, which is enough.

   By the Alternating Series Test the series converges. For absolute convergence look at $\sum \frac{\ln n}{n}$: for $n\ge 3$ we have $\ln n\ge 1$, hence $\frac{\ln n}{n}\ge \frac{1}{n}>0$, and the harmonic series diverges, so $\sum \frac{\ln n}{n}$ diverges by the Direct Comparison Test. The series is conditionally convergent.

   (b) Proof. Suppose $\sum |a_n|$ converges. The terms of a convergent series tend to $0$, so $|a_n|\to 0$, and therefore there is an index $N$ with $|a_n|<1$ for every $n>N$. For such $n$,

   $$0\le a_n^2=|a_n|\cdot |a_n|\le |a_n|\cdot 1=|a_n|.$$

   So beyond $N$ the nonnegative terms $a_n^2$ are dominated by the terms of the convergent series $\sum |a_n|$, and the Direct Comparison Test (whose hypothesis is only required for all $n$ beyond some index) shows that $\sum a_n^2$ converges. $■$

   Note exactly where absolute convergence was used: to know that $|a_n|$ is eventually below $1$, so that squaring makes the terms smaller rather than larger. That is the only ingredient.

   (c) Both converses fail.

   Squares converge, absolute values do not. Take $a_n=\frac{1}{n}$. Then $\sum a_n^2=\sum \frac{1}{n^2}$ converges ($p=2$), while $\sum |a_n|=\sum \frac{1}{n}$ diverges.

   The series converges, its squares do not. Take $a_n=\frac{(-1{)}^n}{\sqrt{n}}$. Since $\frac{1}{\sqrt{n}}$ decreases to $0$, the Alternating Series Test gives convergence (conditional, as $\sum 1/\sqrt{n}$ diverges). But $a_n^2=\frac{1}{n}$, and $\sum \frac{1}{n}$ diverges.

   The second example shows why the hypothesis in (b) had to be absolute convergence: squaring destroys the signs, and a series that converged only because of cancellation between signs loses its reason to converge.

9. 9.Computation · 9 min

   For each series choose the Ratio Test or the Root Test, say why that one, compute the limit $L$ exactly, and state the verdict. You may quote ${\left(1+\frac{1}{n}\right)}^n\to e$.

   1. (a)$\sum _{n=1}^{\infty }\frac{(n!{)}^2}{(2n)!}$
   2. (b)$\sum _{n=1}^{\infty }{\left(\frac{n}{n+1}\right)}^{n^2}$
   3. (c)$\sum _{n=1}^{\infty }\frac{n^n}{n!2^n}$
   4. (d)$\sum _{n=1}^{\infty }\frac{1}{n^2+n}$ — report what the Ratio Test says, then settle the series by another method and find its sum.
   Show the solution to question 9

   (a) Ratio Test, because of the factorials. Consecutive factorials collapse:

   $$\frac{a_{n+1}}{a_n}=\frac{((n+1)!{)}^2}{(2n+2)!}\cdot \frac{(2n)!}{(n!{)}^2}=\frac{(n+1{)}^2}{(2n+2)(2n+1)}=\frac{n+1}{2(2n+1)}⟶\frac{1}{4}<1.$$

   The series converges (absolutely). Sanity check: $a_1=\frac{1}{2}$, $a_2=\frac{4}{24}=\frac{1}{6}$, $a_3=\frac{36}{720}=\frac{1}{20}$; the ratios $\frac{1}{3},\frac{3}{10},\dots$ are heading to $\frac{1}{4}$.

   (b) Root Test, because the whole term is an $n$th power. The exponent is $n^2$, so an $n$th root leaves an exponent of $n$:

   $$\sqrt[n]{a_n}={\left(\frac{n}{n+1}\right)}^n=\frac{1}{{\left(1+\frac{1}{n}\right)}^n}⟶\frac{1}{e}<1.$$

   The series converges. (The Ratio Test would face ${\left(\frac{n+1}{n+2}\right)}^{(n+1{)}^2}/{\left(\frac{n}{n+1}\right)}^{n^2}$, which is far more work for the same answer.)

   (c) Ratio Test, because of the factorial.

   $$\frac{a_{n+1}}{a_n}=\frac{(n+1{)}^{n+1}}{(n+1)!2^{n+1}}\cdot \frac{n!2^n}{n^n}=\frac{(n+1{)}^{n+1}}{(n+1)\cdot 2\cdot n^n}=\frac{1}{2}{\left(\frac{n+1}{n}\right)}^n=\frac{1}{2}{\left(1+\frac{1}{n}\right)}^n⟶\frac{e}{2}>1.$$

   The series diverges; its terms grow, since $n^n$ outruns $n!2^n$.

   (d) The Ratio Test is silent.

   $$\frac{a_{n+1}}{a_n}=\frac{n^2+n}{(n+1{)}^2+(n+1)}=\frac{n(n+1)}{(n+1)(n+2)}=\frac{n}{n+2}⟶1,$$

   so $L=1$ and the test decides nothing — as it never does for a rational function of $n$. Reporting “diverges” here would be an error. Instead, partial fractions: $\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$, so the series telescopes,

   $$s_N=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\cdots +\left(\frac{1}{N}-\frac{1}{N+1}\right)=1-\frac{1}{N+1}⟶1.$$

   The series converges, with sum $1$. (Limit comparison with $\frac{1}{n^2}$ would also prove convergence, but not give the sum.)

10. 10.Computation · 11 min

    Find the radius and the interval of convergence of each power series. Test every endpoint with a named test. You may use without proof that ${\left(1+\frac{1}{n}\right)}^n$ increases to $e$, so that ${\left(1+\frac{1}{n}\right)}^n<e$ for every $n$.

    1. (a)$\sum _{n=1}^{\infty }\frac{(-1{)}^n(2x+1{)}^n}{n{4}^n}$
    2. (b)$\sum _{n=1}^{\infty }\frac{x^{2n}}{n{9}^n}$
    3. (c)$\sum _{n=1}^{\infty }\frac{n!x^n}{n^n}$
    Show the solution to question 10

    (a) Ratio Test on absolute values.

    $$\left|\frac{a_{n+1}}{a_n}\right|=\frac{|2x+1{|}^{n+1}}{(n+1)4^{n+1}}\cdot \frac{n{4}^n}{|2x+1{|}^n}=\frac{|2x+1|}{4}\cdot \frac{n}{n+1}⟶\frac{|2x+1|}{4}.$$

    Convergence needs $|2x+1|<4$, that is $\left|x+\frac{1}{2}\right|<2$. The centre is $-\frac{1}{2}$ and the radius is $R=2$ — not $4$: the factor of $2$ inside $2x+1$ halves it. The candidate interval is $\left(-\frac{5}{2},\frac{3}{2}\right)$.

    At $x=\frac{3}{2}$: $2x+1=4$ and the term is $\frac{(-1{)}^n{4}^n}{n{4}^n}=\frac{(-1{)}^n}{n}$, the alternating harmonic series, which converges by the Alternating Series Test ($\frac{1}{n}$ decreases to $0$).

    At $x=-\frac{5}{2}$: $2x+1=-4$ and $(-1{)}^n(-4{)}^n=4^n$, so the term is $\frac{1}{n}$: the harmonic series, which diverges.

    Interval of convergence: $\left(-\frac{5}{2},\frac{3}{2}\right]$.

    (b) Only even powers. Consecutive terms differ by a factor involving $x^2$, not $x$:

    $$\left|\frac{a_{n+1}}{a_n}\right|=\frac{|x{|}^{2n+2}}{(n+1)9^{n+1}}\cdot \frac{n{9}^n}{|x{|}^{2n}}=\frac{x^2}{9}\cdot \frac{n}{n+1}⟶\frac{x^2}{9}.$$

    This is below $1$ exactly when $x^2<9$, that is $|x|<3$: centre $0$, $R=3$. At $x=\pm 3$ the term is $\frac{9^n}{n{9}^n}=\frac{1}{n}$ in both cases, and the harmonic series diverges. Interval of convergence: $(-3,3)$.

    (c) Ratio Test with the given fact.

    $$\left|\frac{a_{n+1}}{a_n}\right|=\frac{(n+1)!|x{|}^{n+1}}{(n+1{)}^{n+1}}\cdot \frac{n^n}{n!|x{|}^n}=|x|\cdot \frac{(n+1)n^n}{(n+1{)}^{n+1}}=|x|{\left(\frac{n}{n+1}\right)}^n=\frac{|x|}{{\left(1+\frac{1}{n}\right)}^n}⟶\frac{|x|}{e}.$$

    Convergence for $|x|<e$, so $R=e$ about the centre $0$.

    At the endpoints $x=\pm e$ the Ratio Test returns $L=1$ and is silent, so look at the terms themselves: $|a_n|=\frac{n!e^n}{n^n}$. From the computation above with $|x|=e$,

    $$\frac{|a_{n+1}|}{|a_n|}=\frac{e}{{\left(1+\frac{1}{n}\right)}^n}>1,$$

    because ${\left(1+\frac{1}{n}\right)}^n<e$. So $|a_n|$ is strictly increasing, hence $|a_n|\ge |a_1|=e$ for every $n$, and the terms cannot tend to $0$. By the Divergence Test the series diverges at $x=e$ and at $x=-e$ alike (the sign pattern does not matter; the terms fail to shrink).

    Interval of convergence: $(-e,e)$.

11. 11.Proof · 12 min

    Quote Taylor's Theorem in Lagrange form, or Taylor's Inequality, by name each time you use one, and say what $n$, $a$ and $z$ (or $M$) are. In (c) you may use that $e=\sum _{n=0}^{\infty }\frac{1}{n!}$.

    1. (a)Prove that for every $x>0$, $$\frac{x^3}{6}<e^x-1-x-\frac{x^2}{2}<\frac{x^3}{6}{e}^x.$$
    2. (b)Deduce the value of $\underset{x\to 0^{+}}{\lim }\frac{e^x-1-x-x^2/2}{x^3}$ without L'Hôpital's Rule and without series.
    3. (c)Prove that $e$ is irrational. (Suppose $e=p/q$ with positive integers $p,q$, multiply by $q!$, and show that the tail $q!{\sum }_{n>q}1/n!$ would be an integer strictly between $0$ and $1$.)
    Show the solution to question 11

    (a) The Lagrange remainder, two-sided. Take $f(x)=e^x$, centre $a=0$ and $n=2$. The Taylor polynomial is $T_2(x)=1+x+\frac{x^2}{2}$, so the middle quantity is exactly the remainder $R_2(x)=f(x)-T_2(x)$. Since $f$ has continuous derivatives of every order, Taylor's Theorem in Lagrange form gives a number $z$ strictly between $0$ and $x$ with

    $$R_2(x)=\frac{f^{\prime \prime \prime }(z)}{3!}{x}^3=\frac{e^z}{6}{x}^3.$$

    For $x>0$ we have $0<z<x$, and $\exp$ is strictly increasing, so $1=e^0<e^z<e^x$. Multiplying through by $\frac{x^3}{6}>0$ preserves the strict inequalities:

    $$\frac{x^3}{6}<R_2(x)<\frac{x^3}{6}{e}^x.■$$

    Taylor's Inequality with $M=e^x$ would give only the upper bound, and only with $\le$; the Lagrange form is what gives the lower bound and the strictness. Check at $x=1$: $e-2.5\approx 0.218$, between $\frac{1}{6}\approx 0.167$ and $\frac{e}{6}\approx 0.453$.

    (b) Squeeze. Divide (a) by $x^3>0$:

    $$\frac{1}{6}<\frac{e^x-1-x-x^2/2}{x^3}<\frac{e^x}{6}.$$

    As $x\to 0^{+}$ the right-hand bound tends to $\frac{e^0}{6}=\frac{1}{6}$ by continuity of $\exp$, and the left-hand bound is the constant $\frac{1}{6}$. By the Squeeze Theorem the limit is $\frac{1}{6}$. (This is the coefficient of $x^3$ in the exponential series, as it should be.)

    (c) Proof that $e$ is irrational. Suppose, for contradiction, that $e=\frac{p}{q}$ with positive integers $p$ and $q$. Multiply the series for $e$ by $q!$ and split it at $n=q$:

    $$q!e=\sum _{n=0}^q\frac{q!}{n!}+\sum _{n=q+1}^{\infty }\frac{q!}{n!}.$$

    (The split is legitimate: the first piece is a finite sum and the second is the tail of a convergent series, and both converge.) The left side is $q!\cdot \frac{p}{q}=p(q-1)!$, an integer. In the first sum every $n\le q$ gives $\frac{q!}{n!}=(n+1)(n+2)\cdots q$, an integer; call the first sum $A$. Therefore the tail

    $$T=q!e-A=\sum _{n=q+1}^{\infty }\frac{q!}{n!}$$

    is an integer.

    Now bound $T$. Every term is positive, so $T>0$. For $n\ge q+1$,

    $$\frac{q!}{n!}=\frac{1}{(q+1)(q+2)\cdots n},$$

    a product of $n-q$ factors each at least $q+1$, so $\frac{q!}{n!}\le \frac{1}{(q+1{)}^{n-q}}$, with strict inequality as soon as $n\ge q+2$. Summing over $n\ge q+1$ and writing $k=n-q$,

    $$T<\sum _{k=1}^{\infty }\frac{1}{(q+1{)}^k}=\frac{1/(q+1)}{1-1/(q+1)}=\frac{1}{q}\le 1,$$

    the middle step being a geometric series with ratio $\frac{1}{q+1}<1$ and first term $\frac{1}{q+1}$. So $T$ is an integer with $0<T<1$, which is impossible. Hence $e$ is irrational. $■$

    The whole proof rests on one comparison: the factorial tail ${\sum }_{n>q}\frac{q!}{n!}$ is beaten by a geometric tail, and a geometric tail can be summed exactly. (With $q=1$ the same argument shows $e-2\in (0,1)$, that is, $e$ is not even an integer.)

12. 12.Computation · 10 min

    Every part is a computation with series. Give exact values where they exist; in (b) give the bound that guarantees the accuracy, with the theorem that supplies it.

    1. (a)$\underset{x\to 0}{\lim }\frac{\ln (1+x)-\sin x}{x^2}$
    2. (b)Approximate ${\int }_0^{1/2}\frac{\sin x}{x}dx$ with error below $10^{-4}$, giving the approximation as an exact fraction and saying on which side of it the true value lies.
    3. (c)Evaluate $\sum _{n=0}^{\infty }\frac{(-1{)}^n}{(2n+1)3^n}$ exactly, by recognising it as a standard Maclaurin series at a particular point and checking that the point lies in the interval of convergence.
    Show the solution to question 12

    (a) Expand both pieces to one order past the cancellation. For $|x|<1$,

    $$\ln (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots ,\sin x=x-\frac{x^3}{6}+\cdots$$

    The linear terms cancel, and the numerator is

    $$\ln (1+x)-\sin x=-\frac{x^2}{2}+\left(\frac{1}{3}+\frac{1}{6}\right)x^3+\cdots =-\frac{x^2}{2}+\frac{x^3}{2}+\cdots$$

    Dividing by $x^2$ gives $-\frac{1}{2}+\frac{x}{2}+\cdots$, a power series, hence continuous inside its interval of convergence, so the limit is $-\frac{1}{2}$. Check at $x=0.01$: $\ln 1.01\approx 0.0099503$ and $\sin 0.01\approx 0.0099998$; the difference $-0.0000495$ divided by $10^{-4}$ is $-0.495$.

    (b) Integrate the series term by term. Dividing the sine series by $x$,

    $$\frac{\sin x}{x}=\sum _{n=0}^{\infty }\frac{(-1{)}^n{x}^{2n}}{(2n+1)!}=1-\frac{x^2}{6}+\frac{x^4}{120}-\cdots$$

    for all $x\ne 0$, and the series takes the value $1$ at $x=0$, so it represents the integrand (extended continuously) on all of $[0,\frac{1}{2}]$ with infinite radius. Term-by-term integration is therefore legal:

    $${\int }_0^{1/2}\frac{\sin x}{x}dx=\sum _{n=0}^{\infty }\frac{(-1{)}^n(1/2{)}^{2n+1}}{(2n+1)(2n+1)!}=\frac{1}{2}-\frac{1}{8\cdot 3\cdot 6}+\frac{1}{32\cdot 5\cdot 120}-\frac{1}{128\cdot 7\cdot 5040}+\cdots$$

    $$=\frac{1}{2}-\frac{1}{144}+\frac{1}{19200}-\frac{1}{4515840}+\cdots$$

    This is an alternating series whose terms decrease to $0$, so by the Alternating Series Estimation Theorem the error after $N$ terms is at most the first omitted term. The bound $\frac{1}{144}\approx 0.0069$ is too large, but $\frac{1}{19200}\approx 0.000052<10^{-4}$, so two terms suffice:

    $${\int }_0^{1/2}\frac{\sin x}{x}dx\approx \frac{1}{2}-\frac{1}{144}=\frac{72-1}{144}=\frac{71}{144}\approx \mathrm{0.49306.}$$

    The first omitted term $+\frac{1}{19200}$ is positive, so the true value lies above the estimate:

    $$\frac{71}{144}<{\int }_0^{1/2}\frac{\sin x}{x}dx<\frac{71}{144}+\frac{1}{19200}.$$

    (The true value is $0.49311\dots$, inside the bracket.)

    (c) Recognise the arctangent series. The standard series is

    $$\arctan x=\sum _{n=0}^{\infty }\frac{(-1{)}^n{x}^{2n+1}}{2n+1},|x|\le 1.$$

    To match $\frac{1}{(2n+1)3^n}$ we need $x^{2n+1}$ to produce $3^{-n}$: take $x=\frac{1}{\sqrt{3}}$, for which $x^{2n+1}=\frac{1}{3^n\sqrt{3}}$. Then

    $$\arctan \frac{1}{\sqrt{3}}=\frac{1}{\sqrt{3}}\sum _{n=0}^{\infty }\frac{(-1{)}^n}{(2n+1)3^n},\text{so}\sum _{n=0}^{\infty }\frac{(-1{)}^n}{(2n+1)3^n}=\sqrt{3}\arctan \frac{1}{\sqrt{3}}=\sqrt{3}\cdot \frac{\pi }{6}=\frac{\pi \sqrt{3}}{6}.$$

    The point $x=\frac{1}{\sqrt{3}}$ satisfies $|x|<1$, strictly inside the interval of convergence, so the substitution is legitimate (and the series in fact converges absolutely, since its terms are dominated by the geometric $3^{-n}$). Check: $1-\frac{1}{9}+\frac{1}{45}-\frac{1}{189}+\frac{1}{729}\approx 0.9072$, against $\frac{\pi \sqrt{3}}{6}\approx 0.9069$.