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Mathematics LibreTexts

9.5E: Exercises for Section 9.5

  • Gilbert Strang & Edwin “Jed” Herman
  • OpenStax

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In exercises 1 - 30, determine whether each of the following series converges or diverges.

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 conditionally by AST, but not absolutely, since an=ln(n+1)ln(n) are terms of a telescoping series which add to infinity.

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

Answer
Converges absolutely by comparison with p-series, p=2.

16) n=1(1)n+1ne1+nπ

Answer
Converges conditionally by alternating series test. Does not converge absolutely by limit comparison with p-series, p=πe

Solution:

17) n=1(1)n+121/n

18) n=1(1)n+1n1/n

Answer
Diverges; terms do not tend to zero.

19) n=1(1)n(1n1/n) (Hint: n1/n1+ln(n)/n for large n.)

20) n=1(1)n+1n(1cos(1n)) (Hint: cos(1/n)11/n2 for large n.)

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

21) n=1(1)n+1(n+1n) (Hint: Rationalize the numerator.)

22) n=1(1)n+1(1n1n+1) (Hint: Cross-multiply then rationalize numerator.)

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

23) n=1(1)n+1(ln(n+1)lnn)

24) n=1(1)n+1n(tan1(n+1)tan1n) (Hint: Use Mean Value Theorem.)

Answer
Converges by alternating series test since n(tan1(n+1)tan1n) is decreasing to zero for large n.Does not converge absolutely by limit comparison with harmonic series after applying hint.

25) n=1(1)n+1((n+1)2n2)

26) n=1(1)n+1(1n1n+1)

Answer
Converges absolutely, since an=1n1n+1 are terms of a convergent telescoping series.

27) n=1cos(nπ)n

28) n=1cos(nπ)n1/n

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

29) n=11nsin(nπ2)

30) n=1sin(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 |RN|bN+1 to find a value of N that guarantees that the sum of the first N terms of the alternating series n=1(1)n+1bn differs from the infinite sum by at most the given error. Calculate the partial sum SN for this N.

31) [T] bn=1/n, error <105

32) [T] bn=1/ln(n),n2, error <101

Answer
ln(N+1)>10,N+1>e10,N22026;S22026=0.0257

33) [T] bn=1/n, error <103

34) [T] bn=1/2n, error <106

Answer
2N+1>106 or N+1>6ln(10)/ln(2)=19.93. or N19;S19=0.333333969

35) [T] bn=ln(1+1n), error <103

36) [T] bn=1/n2, error <106

Answer
(N+1)2>106 or N>999;S10000.822466.

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 bn0 is decreasing and limnbn=0, then n=1(b2n1b2n) converges absolutely.

38) If bn0 is decreasing, then n=1(b2n1b2n) converges absolutely.

Answer
True. bn need not tend to zero since if cn=bnlimbn, then c2n1c2n=b2n1b2n.

39) If bn0 and limnbn=0 then n=1(12(b3n2+b3n1)b3n) converges.

40) If bn0 is decreasing and n=1(b3n2+b3n1b3n) converges then n=1b3n2 converges.

Answer
True. b3n1b3n0, so convergence of b3n2 follows from the comparison test.

41) If bn0 is decreasing and n=1(1)n1bn converges conditionally but not absolutely, then bn does not tend to zero.

42) Let a+n=an if an0 and an=an if an<0. (Also, a+n=0 if an<0 and an=0 if an0.) If n=1an converges conditionally but not absolutely, then neither n=1a+n nor n=1an converge.

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

43) Suppose that an is a sequence of positive real numbers and that n=1an converges.

44) Suppose that bn is an arbitrary sequence of ones and minus ones. Does n=1anbn necessarily converge?

45) Suppose that an is a sequence such that n=1anbn converges for every possible sequence bn of zeros and ones. Does n=1an converge absolutely?

Answer
Yes. Take bn=1 if an0 and bn=0 if an<0. Then n=1anbn=n:an0an converges. Similarly, one can show n:an<0an 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) n=1(1)n+1sin2(n)n2

Answer
Not strictly decreasing.  Converges absolutely by limit comparison test since sin2(n)1

47) n=1(1)n+1cos2nn

Answer
Not decreasing. Does not converge absolutely.

48) 1+121314+15+161718+

Answer
Not alternating. Does not converge absolutely.

49) 1+1213+14+1516+17+1819+

Answer
Not alternating. Can be expressed as n=1(13n2+13n113n), which diverges by comparison with n=113n2.

50) Show that the alternating series 112+1214+1316+1418+ does not converge. What hypothesis of the alternating series test is not met?

51) Suppose that an converges absolutely. Show that the series consisting of the positive terms an also converges.

Answer
Let a+n=an if an0 and a+n=0 if an<0. Then a+n|an| for all n so the sequence of partial sums of a+n is increasing and bounded above by the sequence of partial sums of |an|, which converges; hence, n=1a+n converges.

52) Show that the alternating series 2335+4759+ does not converge. What hypothesis of the alternating series test is not met?

53) The formula cosθ=1θ22!+θ44!θ66!+ will be derived in the next chapter. Use the remainder |RN|bN+1 to find a bound for the error in estimating cosθ by the fifth partial sum 1θ2/2!+θ4/4!θ6/6!+θ8/8! for θ=1,θ=π/6, and θ=π.

Answer
For N=5 one has RNb6=θ10/10!. When θ=1,R51/10!2.75×107. When θ=π/6, R5(π/6)10/10!4.26×1010. When θ=π,R5π10/10!=0.0258.

54) The formula sinθ=θθ33!+θ55!θ77!+ will be derived in the next chapter. Use the remainder |RN|bN+1 to find a bound for the error in estimating sinθ by the fifth partial sum θθ3/3!+θ5/5!θ7/7!+θ9/9! for θ=1,θ=π/6, and θ=π.

55) How many terms in cosθ=1θ22!+θ44!θ66!+ are needed to approximate cos1 accurate to an error of at most 0.00001?

Answer
Let bn=1/(2n2)!. Then RN1/(2N)!<0.00001 when (2N)!>105 or N=5 and 112!+14!16!+18!=0.540325, whereas cos1=0.5403023

56) How many terms in sinθ=θθ33!+θ55!θ77!+ are needed to approximate sin1 accurate to an error of at most 0.00001?

Answer
N=4 and 113!+15!17!=0.841468, whereas sin1=0.841471

57) Sometimes the alternating series n=1(1)n1bn converges to a certain fraction of an absolutely convergent series n=1bn at a faster rate. Given that n=11n2=π26, find S=1122+132142+. Which of the series 6n=11n2 and Sn=1(1)n1n2 gives a better estimation of π2 using 1000 terms?

Answer
Let T=1n2. Then TS=12T, so S=T/2. 6×1000n=11/n2=3.140638;12×1000n=1(1)n1/n2=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 Nth 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] π4=n=0(1)n2n+1, error <0.001        (Use Desmos to find the first term less than 0.001)

Answer
N=500,S499=0.784898

59) [T] π12=k=0(3)k2k+1, error <0.0001

Answer
N=6,SN=0.9068

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

61) [T] If Nn=1(1)n11nln2, what is 1+13+15121416+17+19+11118110112+?

Answer
ln(2). The nth partial sum is the same as that for the alternating harmonic series.

62) [T] Plot the series 100n=1cos(2πnx)n for 0x<1. Explain why 100n=1cos(2πnx)n diverges when x=0,1. How does the series behave for other x?

63) [T] Plot the series 100n=1sin(2πnx)n for 0x<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(1/2x).

CNX_Calc_Figure_09_05_202.jpeg

64) [T] Plot the series 100n=1cos(2πnx)n2 for 0x<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 n=1Snn, where sn is a randomly generated sequence of ±1s in which the values ±1 are equally likely to occur. Use a random number generator to produce 1000 random ±1s and plot the partial sums SN=Nn=1snn 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 n=11n2 can be accelerated by writing its partial sums as Nn=11n2=Nn=11n(n+1)+Nn=11n2(n+1) and recalling that Nn=11n(n+1)=11N+1 converges to one as N. Compare the estimate of π2/6 using the sums 1000n=11n2 with the estimate using 1+1000n=11n2(n+1).

67) [T] The Euler transform rewrites S=n=0(1)nbn as S=n=0(1)n2n1nm=0(nm)bnm. For the alternating harmonic series, it takes the form ln(2)=n=1(1)n1n=n=11n2n. Compute partial sums of n=11n2n 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, |SnS|bn+1, so one needs 104 terms of the alternating harmonic series to estimate ln(2) to within 0.0001. The first 10 partial sums of the series n=11n2n 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 an0 is such that an0 as n but n=1an diverges, then, given any number A there is a sequence sn of ±1s such that n=1ansnA. Show this for A>0 as follows.

a. Recursively define sn by sn=1 if Sn1=n1k=1aksk<A and sn=1 otherwise.

b. Explain why eventually SnA, and for any m larger than this n, AamSmA+am.

c. Explain why this implies that SnA as n.


This page titled 9.5E: Exercises for Section 9.5 is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Gilbert Strang & Edwin “Jed” Herman (OpenStax) via source content that was edited to the style and standards of the LibreTexts platform.

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