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4.5E: Exercises for Section 4.5

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

1) n=1(1)n+1nn+3

2) n=1(1)n+1n+1n+3

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

3) n=1(1)n+11n+3

4) n=1(1)n+1n+3n

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

5) n=1(1)n+11n!

6) n=1(1)n+13nn!

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

7) n=1(1)n+1(n1n)n

8) n=1(1)n+1(n+1n)n

Answer
Diverges by divergence test since limn|an|=e and not 0.

9) n=1(1)n+1sin2n

10) n=1(1)n+1cos2n

Answer
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) n=1(1)n+1sin2(1/n)

12) n=1(1)n+1cos2(1/n)

Answer
limncos2(1/n)=1. Diverges by divergence test.

13) n=1(1)n+1ln(1/n)

14) n=1(1)n+1ln(1+1n)

Answer
Converges by alternating series test.

15) n=1(1)n+1n21+n4

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

Answer
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}

Answer
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.)

Answer
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.)

Answer
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.)

Answer
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)

Answer
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}}

Answer
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)

Answer
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}

Answer
\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}

Answer
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}

Answer
(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.

Answer
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.

Answer
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.

Answer
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?

Answer
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}

Answer
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}+⋯

Answer
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.

Answer
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.

Answer
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?

Answer
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?

Answer
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

Answer
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}+⋯?

Answer
\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

Answer

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

CNX_Calc_Figure_09_05_202.jpeg

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.

Answer

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.

CNX_Calc_Figure_09_05_204.jpeg

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).

Answer
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→∞.


This page titled 4.5E: Exercises for Section 4.5 is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform.

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