Chapter 5 Solutions

1. Write the terms as $(-1{)}^{n-1}{b}_n$ (or with the opposite initial sign), with $b_n\ge 0$. This identifies the alternating sign pattern.
2. Check that $b_{n+1}\le b_n$ eventually and that ${\lim }_{n\to \infty }{b}_n=0$. The alternating series test then gives convergence. A finite initial segment does not change convergence.

1. For $\sum a_n$, form the series of magnitudes $\sum \left|a_n\right|$. This removes cancellation.
2. Use a valid convergence test on $\sum \left|a_n\right|$. If it converges, the original series is absolutely convergent and therefore convergent.

(a)

1. For integer $n$, ${\sin }^2(n\pi /2)$ is 1 for odd n and 0 for even n.
2. The terms are $1,0,9,0,25,0,\dots$, all nonnegative. The nonzero signs do not alternate, so this is not an alternating series.

(b)

1. If n is even, $\sin (n\pi /2)=0$. If n is odd, n² is odd and $\cos (n^2\pi /2)=0$.
2. Every summand is therefore zero. The zero series converges to zero, but has no nonzero alternating sign pattern.

(a)

1. The identity $(-1{)}^{n^2}=(-1{)}^n$ follows because n² and n have the same parity.
2. The first eleven terms are finite numbers even though some are large. For $n\ge 11$, $(n-10{)}^3\ge n-10$, so .
3. The geometric comparison converges. Thus the sum of magnitudes converges and the original series converges absolutely.

(b)

1. The series is geometric: ${\sum }_{n=0}^{\infty }(-1{)}^{n+1}{2}^{-n}=-{\sum }_{n=0}^{\infty }(-1/2{)}^n$.
2. Its ratio has magnitude 1/2, so it converges absolutely. Its sum is $-1/(1+1/2)=-2/3$.

(a)

1. The terms $e^{-n}$ are positive and form a geometric series with ratio .
2. The magnitudes have the same convergent sum, $1/(1-e^{-1})$. This is absolute convergence.

(b)

1. Use $\left|\sin n\right|\le 1$ to get $\left|\right(-1/3{)}^n\sin n|\le (1/3{)}^n$.
2. The comparison geometric series converges, so this series is absolutely convergent.

(c)

1. The magnitudes are $(2/5{)}^n$, a convergent geometric series.
2. Therefore the original geometric series is absolutely convergent; its sum is $1/(1+2/5)=5/7$.

(d)

1. The magnitudes $1/n$ decrease to zero, so the alternating series test gives convergence.
2. The absolute-value series is the harmonic series, which diverges. Therefore convergence is conditional.

1. Use the alternating harmonic terms, but place two unused positive terms before each unused negative term: $1+1/3-1/2+1/5+1/7-1/4+\dots$. Every odd reciprocal occurs positively and every even reciprocal occurs negatively exactly once.
2. After N complete blocks, the sum is $H_{4N}-(1/2)H_{2N}-(1/2)H_N$, where $H_m={\sum }_{k=1}^{m}1/k$.
3. Since $H_m-\ln m$ tends to the same constant as m grows, the constant terms cancel. The limit is $\ln 4-(1/2)\ln 2=(3/2)\ln 2$.
4. The one or two leftover terms between block endpoints tend to zero, so the whole sequence of partial sums has this limit. It differs from both $\ln 2$ and the example’s $(1/2)\ln 2$.

1. The magnitudes $1/\mathrm{sqrt}\left(n\right)$ decrease to zero. The alternating series test therefore gives convergence.
2. The absolute-value series is a p-series with $p=1/2\le 1$, so it diverges.

1. Taking absolute values gives ${\sum }_{n=1}^{\infty }1/n^2$.
2. This p-series converges because $p=2>1$, so the original series is absolutely convergent.

1. The magnitudes satisfy $n/(n+1)\to 1$.
2. The terms therefore alternate near −1 and 1 and do not tend to zero. The necessary term test for convergence fails.

1. For $x>e$, the derivative of $\ln x/x$ is , so the magnitudes eventually decrease. Also $\ln n/n\to 0$. The alternating series test gives convergence.
2. For $n\ge 3$, $\ln n\ge 1$, so $\ln n/n\ge 1/n$. Harmonic comparison proves divergence of the magnitudes.

1. The positive magnitudes $1/(2n+1)$ decrease to zero, so the alternating series test applies.
2. Because $1/(2n+1)\ge 1/\left(3n\right)$ for $n\ge 1$, the magnitude series diverges by comparison with the harmonic series.

1. The alternating harmonic magnitudes decrease to zero, so the remainder after N terms satisfies $\left|R_N\right|\le 1/(N+1)$.
2. Require . This gives $N+1>100$, so $N=100$ is the smallest integer guaranteed by this bound.

1. The first four terms give $S_4=1-1/2+1/3-1/4=7/12$.
2. The next magnitude is $1/5$. The alternating error bound gives $|S-S_4|\le 1/5$. Since the fifth term is positive, $S_4\le S\le S_4+1/5$.

1. Here $b_n=1/n^2$ decreases to zero. Thus $\left|R_N\right|\le 1/(N+1{)}^2$.
2. To have , require $N+1>100$, so the smallest bound-based choice is $N=100$.

1. Alternating signs alone do not ensure that the terms tend to zero. In $1-1+1-1+\dots$, they do not.
2. Its partial sums alternate between 1 and 0, so no limit exists. The zero-limit and monotonicity conditions in the alternating series test must be checked.

1. The magnitude series is geometric with ratio $3/4$, whose magnitude is less than 1.
2. The original sum is $1/(1-(-3/4\left)\right)=4/7$. The convergent magnitude series makes convergence absolute.

1. Since $\left|\cos n\right|\le 1$, the term magnitude is at most $1/n^2$.
2. The comparison p-series converges, so this series is absolutely convergent.

1. For the magnitude $b_n=n/2^n$, $b_{n+1}/b_n=(n+1)/\left(2n\right)\to 1/2$.
2. Because the ratio-test limit is less than 1, the magnitude series converges and the original series converges absolutely.

1. Take absolute values and set $b_n=3^n/n!$. The ratio of consecutive magnitudes is $b_{n+1}/b_n=3/(n+1)$.
2. As n grows, this ratio tends to zero, which is less than 1. The ratio test therefore proves that the magnitude series converges, so the original series is absolutely convergent.

1. Absolute convergence means $\sum \left|a_n\right|$ has a finite sum. Every rearranged partial sum is controlled by the same summable tail of magnitudes.
2. After enough of the original initial terms have appeared, omitted or newly included tail terms have arbitrarily small total magnitude. Thus every rearrangement has the original sum.

1. The alternating harmonic series converges by the alternating series test.
2. Its magnitude series $\sum 1/n$ diverges. This gives a convergent series that is not absolutely convergent.

1. For $n\ge 2$, $1/\left(n\ln n\right)$ decreases to zero, so the alternating series test gives convergence.
2. The integral ${\int }_2^R\mathrm{dx}/\left(x\ln x\right)=\ln \left(\ln R\right)-\ln \left(\ln 2\right)$ grows without bound. Thus the magnitude series diverges by the integral test.

1. For $n\ge 2$, the positive function $1/\left(x\right(\ln x{)}^2)$ decreases.
2. The improper integral equals ${\int }_2^{\infty }\mathrm{dx}/\left(x\right(\ln x{)}^2)=1/\ln 2$, which is finite. Thus the magnitude series converges by the integral test.

1. Conditional convergence means the signed sum converges but the magnitude sum does not. Its positive and negative subseries must both have infinite total magnitude.
2. The Riemann rearrangement theorem then permits a rearrangement converging to any prescribed real number. In particular, zero is allowed.

1. Take unused positive terms in their original order until the partial sum first exceeds the target $L$. Then take unused negative terms until it first falls below L. Repeat.
2. Both sign subseries have infinite total magnitude, so each crossing is possible. Terms tend to zero, so the crossing errors shrink to zero.
3. Every term is eventually used: both sign indices increase without bound. Between crossings, partial sums stay between neighboring crossings, so all partial sums tend to L.

1. The first N blocks of one positive term followed by two negatives use positive reciprocals through $1/(2N-1)$ and negative even reciprocals through $-1/\left(4N\right)$.
2. Their sum is $H_{2N}-(1/2)H_N-(1/2)H_{2N}=(1/2)(H_{2N}-H_N)$.
3. Since $H_{2N}-H_N\to \ln 2$, the block sums tend to $(1/2)\ln 2$. The intervening individual terms vanish, so this is the full sum.

1. The magnitudes of $\sum (-1{)}^n/n^2$ form a convergent p-series.
2. Absolute convergence preserves the sum under every rearrangement. The Riemann freedom to change the sum applies to conditional convergence.

1. For $b_n=n/(n^2+1)$, $b_n\to 0$. The derivative of $x/(x^2+1)$ is $(1-x^2)/(x^2+1{)}^2$, nonpositive for $x\ge 1$, so the magnitudes decrease.
2. The alternating series test gives convergence. For $n\ge 1$, $n/(n^2+1)\ge 1/\left(2n\right)$, so the magnitude series diverges.

1. The series is geometric with ratio $-1/2$, and its sum is $2/3$.
2. The first three terms sum to $1-1/2+1/4=3/4$. The actual error is $|3/4-2/3|=1/12$.
3. The alternating bound is the next magnitude, $1/8$, which is larger than the actual error.

1. The signed geometric ratio is $-1/3$, so its sum is $1/(1+1/3)=3/4$.
2. The magnitude ratio is $1/3$, so its sum is $1/(1-1/3)=3/2$. Both are finite, proving absolute convergence.