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# 9.5E: Exercises for Alternating Series

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In exercises 1 - 30, state whether each of the following series converges absolutely, conditionally, or not at all.

1) $$\displaystyle \sum^∞_{n=1}(−1)^{n+1}\frac{n}{n+3}$$

2) $$\displaystyle \sum^∞_{n=1}(−1)^{n+1}\frac{\sqrt{n}+1}{\sqrt{n}+3}$$

This series diverges by the divergence test. Terms do not tend to zero.

3) $$\displaystyle \sum^∞_{n=1}(−1)^{n+1}\frac{1}{\sqrt{n+3}}$$

4) $$\displaystyle \sum^∞_{n=1}(−1)^{n+1}\frac{\sqrt{n+3}}{n}$$

Converges conditionally by alternating series test, since $$\sqrt{n+3}/n$$ is decreasing and its limit is 0. Does not converge absolutely by comparison with $$p$$-series, $$p=1/2$$.

5) $$\displaystyle \sum^∞_{n=1}(−1)^{n+1}\frac{1}{n!}$$

6) $$\displaystyle \sum^∞_{n=1}(−1)^{n+1}\frac{3^n}{n!}$$

Converges absolutely by limit comparison to $$3^n/4^n,$$ for example.

7) $$\displaystyle \sum^∞_{n=1}(−1)^{n+1}\left(\frac{n−1}{n}\right)^n$$

8) $$\displaystyle \sum^∞_{n=1}(−1)^{n+1}\left(\frac{n+1}{n}\right)^n$$

Diverges by divergence test since $$\displaystyle \lim_{n→∞}|a_n|=e$$ and not $$0$$.

9) $$\displaystyle \sum^∞_{n=1}(−1)^{n+1}\sin^2n$$

10) $$\displaystyle \sum^∞_{n=1}(−1)^{n+1}\cos^2n$$

Diverges by the divergence test, since its terms do not tend to zero. The limit of the sequence of its terms does not exist.

11) $$\displaystyle \sum^∞_{n=1}(−1)^{n+1}\sin^2(1/n)$$

12) $$\displaystyle \sum^∞_{n=1}(−1)^{n+1}\cos^2(1/n)$$

$$\displaystyle \lim_{n→∞}\cos^2(1/n)=1.$$ Diverges by divergence test.

13) $$\displaystyle \sum^∞_{n=1}(−1)^{n+1}\ln(1/n)$$

14) $$\displaystyle \sum^∞_{n=1}(−1)^{n+1}\ln(1+\frac{1}{n})$$

Converges by alternating series test.

15) $$\displaystyle \sum^∞_{n=1}(−1)^{n+1}\frac{n^2}{1+n^4}$$

16) $$\displaystyle \sum^∞_{n=1}(−1)^{n+1}\frac{n^e}{1+n^π}$$

Converges conditionally by alternating series test. Does not converge absolutely by limit comparison with $$p$$-series, $$p=π−e$$

Solution:

17) $$\displaystyle \sum^∞_{n=1}(−1)^{n+1}2^{1/n}$$

18) $$\displaystyle \sum^∞_{n=1}(−1)^{n+1}n^{1/n}$$

Diverges; terms do not tend to zero.

19) $$\displaystyle \sum^∞_{n=1}(−1)^n(1−n^{1/n})$$ (Hint: $$n^{1/n}≈1+\ln(n)/n$$ for large $$n$$.)

20) $$\displaystyle \sum^∞_{n=1}(−1)^{n+1}n\left(1−\cos\left(\frac{1}{n}\right)\right)$$ (Hint: $$\cos(1/n)≈1−1/n^2$$ for large $$n$$.)

Converges by alternating series test. Does not converge absolutely by limit comparison with harmonic series.

21) $$\displaystyle \sum^∞_{n=1}(−1)^{n+1}(\sqrt{n+1}−\sqrt{n})$$ (Hint: Rationalize the numerator.)

22) $$\displaystyle \sum^∞_{n=1}(−1)^{n+1}\left(\frac{1}{\sqrt{n}}−\frac{1}{\sqrt{n+1}}\right)$$ (Hint: Cross-multiply then rationalize numerator.)

Converges absolutely by limit comparison with $$p$$-series, $$p=3/2$$, after applying the hint.

23) $$\displaystyle \sum^∞_{n=1}(−1)^{n+1}(\ln(n+1)−\ln n)$$

24) $$\displaystyle \sum^∞_{n=1}(−1)^{n+1}n(\tan^{−1}(n+1)−\tan^{−1}n)$$ (Hint: Use Mean Value Theorem.)

Converges by alternating series test since $$n(\tan^{−1}(n+1)−\tan^{−1}n)$$ is decreasing to zero for large $$n$$.Does not converge absolutely by limit comparison with harmonic series after applying hint.

25) $$\displaystyle \sum^∞_{n=1}(−1)^{n+1}((n+1)^2−n^2)$$

26) $$\displaystyle \sum^∞_{n=1}(−1)^{n+1}\left(\frac{1}{n}−\frac{1}{n+1}\right)$$

Converges absolutely, since $$a_n=\dfrac{1}{n}−\dfrac{1}{n+1}$$ are terms of a telescoping series.

27) $$\displaystyle \sum^∞_{n=1}\frac{\cos(nπ)}{n}$$

28) $$\displaystyle \sum^∞_{n=1}\frac{\cos(nπ)}{n^{1/n}}$$

Terms do not tend to zero. Series diverges by divergence test.

29) $$\displaystyle \sum^∞_{n=1}\frac{1}{n}\sin(\frac{nπ}{2})$$

30) $$\displaystyle \sum^∞_{n=1}\sin(nπ/2)\sin(1/n)$$

Converges by alternating series test. Does not converge absolutely by limit comparison with harmonic series.

In exercises 31 - 36, use the estimate $$|R_N|≤b_{N+1}$$ to find a value of $$N$$ that guarantees that the sum of the first $$N$$ terms of the alternating series $$\displaystyle \sum^∞_{n=1}(−1)^{n+1}b_n$$ differs from the infinite sum by at most the given error. Calculate the partial sum $$S_N$$ for this $$N$$.

31) [T] $$b_n=1/n,$$ error $$<10^{−5}$$

32) [T] $$b_n=1/\ln(n), n≥2,$$ error $$<10^{−1}$$

$$\ln(N+1)>10, N+1>e^{10}, N≥22026; S_{22026}=0.0257…$$

33) [T] $$b_n=1/\sqrt{n},$$ error $$<10^{−3}$$

34) [T] $$b_n=1/2^n$$, error $$<10^{−6}$$

$$2^{N+1}>10^6$$ or $$N+1>6\ln(10)/\ln(2)=19.93.$$ or $$N≥19; S_{19}=0.333333969…$$

35) [T] $$b_n=ln(1+\dfrac{1}{n}),$$ error $$<10^{−3}$$

36) [T] $$b_n=1/n^2,$$ error $$<10^{−6}$$

$$(N+1)^2>10^6$$ or $$N>999; S_{1000}≈0.822466.$$

For exercises 37 - 45, indicate whether each of the following statements is true or false. If the statement is false, provide an example in which it is false.

37) If $$b_n≥0$$ is decreasing and $$\displaystyle \lim_{n→∞}b_n=0$$, then $$\displaystyle \sum_{n=1}^∞(b_{2n−1}−b_{2n})$$ converges absolutely.

38) If $$b_n≥0$$ is decreasing, then $$\displaystyle \sum_{n=1}^∞(b_{2n−1}−b_{2n})$$ converges absolutely.

True. $$b_n$$ need not tend to zero since if $$\displaystyle c_n=b_n−\lim b_n$$, then $$c_{2n−1}−c_{2n}=b_{2n−1}−b_{2n}.$$

39) If $$b_n≥0$$ and $$\displaystyle \lim_{n→∞}b_n=0$$ then $$\displaystyle \sum_{n=1}^∞(\frac{1}{2}(b_{3n−2}+b_{3n−1})−b_{3n})$$ converges.

40) If $$b_n≥0$$ is decreasing and $$\displaystyle \sum_{n=1}^∞(b_{3n−2}+b_{3n−1}−b_{3n})$$ converges then $$\displaystyle \sum_{n=1}^∞b_{3n−2}$$ converges.

True. $$b_{3n−1}−b_{3n}≥0,$$ so convergence of $$\displaystyle \sum b_{3n−2}$$ follows from the comparison test.

41) If $$b_n≥0$$ is decreasing and $$\displaystyle \sum_{n=1}^∞(−1)^{n−1}b_n$$ converges conditionally but not absolutely, then $$b_n$$ does not tend to zero.

42) Let $$a^+_n=a_n$$ if $$a_n≥0$$ and $$a^−_n=−a_n$$ if $$a_n<0$$. (Also, $$a^+_n=0$$ if $$a_n<0$$ and $$a^−_n=0$$ if $$a_n≥0$$.) If $$\displaystyle \sum_{n=1}^∞a_n$$ converges conditionally but not absolutely, then neither $$\displaystyle \sum_{n=1}^∞a^+_n$$ nor $$\displaystyle \sum_{n=1}^∞a^−_n$$ converge.

True. If one converges, then so must the other, implying absolute convergence.

43) Suppose that $$a_n$$ is a sequence of positive real numbers and that $$\displaystyle \sum_{n=1}^∞a_n$$ converges.

44) Suppose that $$b_n$$ is an arbitrary sequence of ones and minus ones. Does $$\displaystyle \sum_{n=1}^∞a_nb_n$$ necessarily converge?

45) Suppose that $$a_n$$ is a sequence such that $$\displaystyle \sum_{n=1}^∞a_nb_n$$ converges for every possible sequence $$b_n$$ of zeros and ones. Does $$\displaystyle \sum_{n=1}^∞a_n$$ converge absolutely?

Yes. Take $$b_n=1$$ if $$a_n≥0$$ and $$b_n=0$$ if $$a_n<0$$. Then $$\displaystyle \sum_{n=1}^∞a_nb_n=\sum_{n:a_n≥0}a_n$$ converges. Similarly, one can show $$\displaystyle \sum_{n:a_n<0}a_n$$ converges. Since both series converge, the series must converge absolutely.

In exercises 46 - 49, the series do not satisfy the hypotheses of the alternating series test as stated. In each case, state which hypothesis is not satisfied. State whether the series converges absolutely.

46) $$\displaystyle \sum_{n=1}^∞(−1)^{n+1}\frac{\sin^2n}{n}$$

47) $$\displaystyle \sum_{n=1}^∞(−1)^{n+1}\frac{\cos^2n}{n}$$

Not decreasing. Does not converge absolutely.

48) $$\displaystyle 1+\frac{1}{2}−\frac{1}{3}−\frac{1}{4}+\frac{1}{5}+\frac{1}{6}−\frac{1}{7}−\frac{1}{8}+⋯$$

49) $$\displaystyle 1+\frac{1}{2}−\frac{1}{3}+\frac{1}{4}+\frac{1}{5}−\frac{1}{6}+\frac{1}{7}+\frac{1}{8}−\frac{1}{9}+⋯$$

Not alternating. Can be expressed as $$\displaystyle \sum_{n=1}^∞\left(\frac{1}{3n−2}+\frac{1}{3n−1}−\frac{1}{3n}\right),$$ which diverges by comparison with $$\displaystyle \sum_{n=1}^∞\frac{1}{3n−2}.$$

50) Show that the alternating series $$\displaystyle 1−\frac{1}{2}+\frac{1}{2}−\frac{1}{4}+\frac{1}{3}−\frac{1}{6}+\frac{1}{4}−\frac{1}{8}+⋯$$ does not converge. What hypothesis of the alternating series test is not met?

51) Suppose that $$\displaystyle \sum a_n$$ converges absolutely. Show that the series consisting of the positive terms $$a_n$$ also converges.

Let $$a^+_n=a_n$$ if $$a_n≥0$$ and $$a^+_n=0$$ if $$a_n<0$$. Then $$a^+_n≤|a_n|$$ for all $$n$$ so the sequence of partial sums of $$a^+_n$$ is increasing and bounded above by the sequence of partial sums of $$|a_n|$$, which converges; hence, $$\displaystyle \sum_{n=1}^∞a^+_n$$ converges.

52) Show that the alternating series $$\displaystyle \frac{2}{3}−\frac{3}{5}+\frac{4}{7}−\frac{5}{9}+⋯$$ does not converge. What hypothesis of the alternating series test is not met?

53) The formula $$\displaystyle \cos θ=1−\frac{θ^2}{2!}+\frac{θ^4}{4!}−\frac{θ^6}{6!}+⋯$$ will be derived in the next chapter. Use the remainder $$|R_N|≤b_{N+1}$$ to find a bound for the error in estimating $$\cos θ$$ by the fifth partial sum $$1−θ^2/2!+θ^4/4!−θ^6/6!+θ^8/8!$$ for $$θ=1, θ=π/6,$$ and $$θ=π.$$

For $$N=5$$ one has $$∣R_N∣b_6=θ^{10}/10!$$. When $$θ=1, R_5≤1/10!≈2.75×10^{−7}$$. When $$θ=π/6,$$ $$R_5≤(π/6)^{10}/10!≈4.26×10^{−10}$$. When $$θ=π, R_5≤π^{10}/10!=0.0258.$$

54) The formula $$\sin θ=θ−\dfrac{θ^3}{3!}+\dfrac{θ^5}{5!}−\dfrac{θ^7}{7!}+⋯$$ will be derived in the next chapter. Use the remainder $$|R_N|≤b_{N+1}$$ to find a bound for the error in estimating $$\sin θ$$ by the fifth partial sum $$θ−θ^3/3!+θ^5/5!−θ^7/7!+θ^9/9!$$ for $$θ=1, θ=π/6,$$ and $$θ=π.$$

55) How many terms in $$\cos θ=1−\dfrac{θ^2}{2!}+\dfrac{θ^4}{4!}−\dfrac{θ^6}{6!}+⋯$$ are needed to approximate $$\cos 1$$ accurate to an error of at most $$0.00001$$?

Let $$b_n=1/(2n−2)!.$$ Then $$R_N≤1/(2N)!<0.00001$$ when $$(2N)!>10^5$$ or $$N=5$$ and $$\displaystyle 1−\frac{1}{2!}+\frac{1}{4!}−\frac{1}{6!}+\frac{1}{8!}=0.540325…$$, whereas $$\cos 1=0.5403023…$$

56) How many terms in $$\sin θ=θ−\dfrac{θ^3}{3!}+\dfrac{θ^5}{5!}−\dfrac{θ^7}{7!}+⋯$$ are needed to approximate $$\sin 1$$ accurate to an error of at most $$0.00001?$$

57) Sometimes the alternating series $$\displaystyle \sum_{n=1}^∞(−1)^{n−1}b_n$$ converges to a certain fraction of an absolutely convergent series $$\displaystyle \sum_{n=1}^∞b_n$$ at a faster rate. Given that $$\displaystyle \sum_{n=1}^∞\frac{1}{n^2}=\frac{π^2}{6}$$, find $$\displaystyle S=1−\frac{1}{2^2}+\frac{1}{3^2}−\frac{1}{4^2}+⋯$$. Which of the series $$\displaystyle 6\sum_{n=1}^∞\frac{1}{n^2}$$ and $$\displaystyle S\sum_{n=1}^∞\frac{(−1)^{n−1}}{n^2}$$ gives a better estimation of $$π^2$$ using $$1000$$ terms?

Let $$\displaystyle T=\sum\frac{1}{n^2}.$$ Then $$T−S=\dfrac{1}{2}T$$, so $$S=T/2$$. $$\displaystyle \sqrt{6×\sum_{n=1}^{1000}1/n^2}=3.140638…; \sqrt{\frac{1}{2}×\sum_{n=1}^{1000}(−1)^{n−1}/n^2}=3.141591…; π=3.141592….$$ The alternating series is more accurate for $$1000$$ terms.

The alternating series in exercises 58 & 59 converge to given multiples of $$π$$. Find the value of $$N$$ predicted by the remainder estimate such that the $$N^{\text{th}}$$ partial sum of the series accurately approximates the left-hand side to within the given error. Find the minimum $$N$$ for which the error bound holds, and give the desired approximate value in each case. Up to $$15$$ decimals places, $$π=3.141592653589793….$$

58) [T] $$\displaystyle \frac{π}{4}=\sum_{n=0}^∞\frac{(−1)^n}{2n+1},$$ error $$<0.0001$$

59) [T] $$\displaystyle \frac{π}{\sqrt{12}}=\sum_{k=0}^∞\frac{(−3)^{−k}}{2k+1},$$ error $$<0.0001$$

$$N=6, S_N=0.9068$$

60) [T] The series $$\displaystyle \sum_{n=0}^∞\frac{\sin(x+πn)}{x+πn}$$ plays an important role in signal processing. Show that $$\displaystyle \sum_{n=0}^∞\frac{\sin(x+πn)}{x+πn}$$ converges whenever $$0<x<π$$. (Hint: Use the formula for the sine of a sum of angles.)

61) [T] If $$\displaystyle \sum_{n=1}^N(−1)^{n−1}\frac{1}{n}→ln2,$$ what is $$\displaystyle 1+\frac{1}{3}+\frac{1}{5}−\frac{1}{2}−\frac{1}{4}−\frac{1}{6}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}−\frac{1}{8}−\frac{1}{10}−\frac{1}{12}+⋯?$$

$$\ln(2).$$ The $$n^{\text{th}}$$ partial sum is the same as that for the alternating harmonic series.

62) [T] Plot the series $$\displaystyle \sum_{n=1}^{100}\frac{\cos(2πnx)}{n}$$ for $$0≤x<1$$. Explain why $$\displaystyle \sum_{n=1}^{100}\frac{\cos(2πnx)}{n}$$ diverges when $$x=0,1$$. How does the series behave for other $$x$$?

63) [T] Plot the series $$\displaystyle \sum_{n=1}^{100}\frac{\sin(2πnx)}{n}$$ for $$0≤x<1$$ and comment on its behavior

The series jumps rapidly near the endpoints. For $$x$$ away from the endpoints, the graph looks like $$π(1/2−x)$$.

64) [T] Plot the series $$\displaystyle \sum_{n=1}^{100}\frac{\cos(2πnx)}{n^2}$$ for $$0≤x<1$$ and describe its graph.

65) [T] The alternating harmonic series converges because of cancelation among its terms. Its sum is known because the cancelation can be described explicitly. A random harmonic series is one of the form $$\displaystyle \sum_{n=1}^∞\frac{S_n}{n}$$, where $$s_n$$ is a randomly generated sequence of $$±1's$$ in which the values $$±1$$ are equally likely to occur. Use a random number generator to produce $$1000$$ random $$±1's$$ and plot the partial sums $$\displaystyle S_N=\sum_{n=1}^N\frac{s_n}{n}$$ of your random harmonic sequence for $$N=1$$ to $$1000$$. Compare to a plot of the first $$1000$$ partial sums of the harmonic series.

Here is a typical result. The top curve consists of partial sums of the harmonic series. The bottom curve plots partial sums of a random harmonic series.

66) [T] Estimates of $$\displaystyle \sum_{n=1}^∞\frac{1}{n^2}$$ can be accelerated by writing its partial sums as $$\displaystyle \sum_{n=1}^N\frac{1}{n^2}=\sum_{n=1}^N\frac{1}{n(n+1)}+\sum_{n=1}^N\frac{1}{n^2(n+1)}$$ and recalling that $$\displaystyle \sum_{n=1}^N\frac{1}{n(n+1)}=1−\frac{1}{N+1}$$ converges to one as $$N→∞.$$ Compare the estimate of $$π^2/6$$ using the sums $$\displaystyle \sum_{n=1}^{1000}\frac{1}{n^2}$$ with the estimate using $$\displaystyle 1+\sum_{n=1}^{1000}\frac{1}{n^2(n+1)}$$.

67) [T] The Euler transform rewrites $$\displaystyle S=\sum_{n=0}^∞(−1)^nb_n$$ as $$\displaystyle S=\sum_{n=0}^∞(−1)^n2^{−n−1}\sum_{m=0}^n(^n_m)b_{n−m}$$. For the alternating harmonic series, it takes the form $$\displaystyle \ln(2)=\sum_{n=1}^∞\frac{(−1)^{n−1}}{n}=\sum_{n=1}^∞\frac{1}{n2^n}$$. Compute partial sums of $$\displaystyle \sum_{n=1}^∞\frac{1}{n2^n}$$ until they approximate $$\ln(2)$$ accurate to within $$0.0001$$. How many terms are needed? Compare this answer to the number of terms of the alternating harmonic series are needed to estimate $$\ln(2)$$.

By the alternating series test, $$|S_n−S|≤b_{n+1},$$ so one needs $$10^4$$ terms of the alternating harmonic series to estimate $$\ln(2)$$ to within $$0.0001$$. The first $$10$$ partial sums of the series $$\displaystyle \sum_{n=1}^∞\frac{1}{n2^n}$$ are (up to four decimals) $$0.5000,0.6250,0.6667,0.6823,0.6885,0.6911,0.6923,0.6928,0.6930,0.6931$$ and the tenth partial sum is within $$0.0001$$ of $$\ln(2)=0.6931….$$
68) [T] In the text it was stated that a conditionally convergent series can be rearranged to converge to any number. Here is a slightly simpler, but similar, fact. If $$a_n≥0$$ is such that $$a_n→0$$ as $$n→∞$$ but $$\displaystyle \sum_{n=1}^∞a_n$$ diverges, then, given any number $$A$$ there is a sequence $$s_n$$ of $$±1's$$ such that $$\displaystyle \sum_{n=1}^∞a_ns_n→A.$$ Show this for $$A>0$$ as follows.
a. Recursively define $$s_n$$ by $$s_n=1$$ if $$\displaystyle S_{n−1}=\sum_{k=1}^{n−1}a_ks_k<A$$ and $$s_n=−1$$ otherwise.
b. Explain why eventually $$S_n≥A,$$ and for any $$m$$ larger than this $$n$$, $$A−a_m≤S_m≤A+a_m$$.
c. Explain why this implies that $$S_n→A$$ as $$n→∞.$$