State whether each of the following series converges absolutely, conditionally, or not at all.
1. [latex]\displaystyle\sum _{n=1}^{\infty }{\left(-1\right)}^{n+1}\frac{n}{n+3}[/latex]
2. [latex]\displaystyle\sum _{n=1}^{\infty }{\left(-1\right)}^{n+1}\frac{\sqrt{n}+1}{\sqrt{n}+3}[/latex]
Show Solution
Does not converge by divergence test. Terms do not tend to zero.
3. [latex]\displaystyle\sum _{n=1}^{\infty }{\left(-1\right)}^{n+1}\frac{1}{\sqrt{n+3}}[/latex]
4. [latex]\displaystyle\sum _{n=1}^{\infty }{\left(-1\right)}^{n+1}\frac{\sqrt{n+3}}{n}[/latex]
Show Solution
Converges conditionally by alternating series test, since [latex]\frac{\sqrt{n+3}}{n}[/latex] is decreasing. Does not converge absolutely by comparison with p-series, [latex]p=\frac{1}{2}[/latex].
5. [latex]\displaystyle\sum _{n=1}^{\infty }{\left(-1\right)}^{n+1}\frac{1}{n\text{!}}[/latex]
6. [latex]\displaystyle\sum _{n=1}^{\infty }{\left(-1\right)}^{n+1}\frac{{3}^{n}}{n\text{!}}[/latex]
Show Solution
Converges absolutely by limit comparison to [latex]\frac{{3}^{n}}{{4}^{n}}[/latex], for example.
7. [latex]\displaystyle\sum _{n=1}^{\infty }{\left(-1\right)}^{n+1}{\left(\frac{n - 1}{n}\right)}^{n}[/latex]
8. [latex]\displaystyle\sum _{n=1}^{\infty }{\left(-1\right)}^{n+1}{\left(\frac{n+1}{n}\right)}^{n}[/latex]
Show Solution
Diverges by divergence test since [latex]\underset{n\to \infty }{\text{lim}}|{a}_{n}|=e[/latex].
9. [latex]\displaystyle\sum _{n=1}^{\infty }{\left(-1\right)}^{n+1}{\sin}^{2}n[/latex]
10. [latex]\displaystyle\sum _{n=1}^{\infty }{\left(-1\right)}^{n+1}{\cos}^{2}n[/latex]
Show Solution
Does not converge. Terms do not tend to zero.
11. [latex]\displaystyle\sum _{n=1}^{\infty }{\left(-1\right)}^{n+1}{\sin}^{2}\left(\frac{1}{n}\right)[/latex]
12. [latex]\displaystyle\sum _{n=1}^{\infty }{\left(-1\right)}^{n+1}{\cos}^{2}\left(\frac{1}{n}\right)[/latex]
Show Solution
[latex]\underset{n\to \infty }{\text{lim}}{\cos}^{2}\left(\frac{1}{n}\right)=1[/latex]. Diverges by divergence test.
13. [latex]\displaystyle\sum _{n=1}^{\infty }{\left(-1\right)}^{n+1}\text{ln}\left(\frac{1}{n}\right)[/latex]
14. [latex]\displaystyle\sum _{n=1}^{\infty }{\left(-1\right)}^{n+1}\text{ln}\left(1+\frac{1}{n}\right)[/latex]
Show Solution
Converges by alternating series test.
15. [latex]\displaystyle\sum _{n=1}^{\infty }{\left(-1\right)}^{n+1}\frac{{n}^{2}}{1+{n}^{4}}[/latex]
16. [latex]\displaystyle\sum _{n=1}^{\infty }{\left(-1\right)}^{n+1}\frac{{n}^{e}}{1+{n}^{\pi }}[/latex]
Show Solution
Converges conditionally by alternating series test. Does not converge absolutely by limit comparison with p-series, [latex]p=\pi -e[/latex]
17. [latex]\displaystyle\sum _{n=1}^{\infty }{\left(-1\right)}^{n+1}{2}^{\frac{1}{n}}[/latex]
18. [latex]\displaystyle\sum _{n=1}^{\infty }{\left(-1\right)}^{n+1}{n}^{\frac{1}{n}}[/latex]
Show Solution
Diverges; terms do not tend to zero.
19. [latex]\displaystyle\sum _{n=1}^{\infty }{\left(-1\right)}^{n}\left(1-{n}^{\frac{1}{n}}\right)[/latex] (Hint: [latex]{n}^{\frac{1}{n}}\approx 1+\frac{\text{ln}\left(n\right)}{n}[/latex] for large [latex]n.[/latex])
20. [latex]\displaystyle\sum _{n=1}^{\infty }{\left(-1\right)}^{n+1}n\left(1-\cos\left(\frac{1}{n}\right)\right)[/latex] (Hint: [latex]\cos\left(\frac{1}{n}\right)\approx 1 - \frac{1}{{n}^{2}}[/latex] for large [latex]n.[/latex])
Show Solution
Converges by alternating series test. Does not converge absolutely by limit comparison with harmonic series.
21. [latex]\displaystyle\sum _{n=1}^{\infty }{\left(-1\right)}^{n+1}\left(\sqrt{n+1}-\sqrt{n}\right)[/latex] (Hint: Rationalize the numerator.)
22. [latex]\displaystyle\sum _{n=1}^{\infty }{\left(-1\right)}^{n+1}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)[/latex] (Hint: Find common denominator then rationalize numerator.)
Show Solution
Converges absolutely by limit comparison with p-series, [latex]p=\frac{3}{2}[/latex], after applying the hint.
23. [latex]\displaystyle\sum _{n=1}^{\infty }{\left(-1\right)}^{n+1}\left(\text{ln}\left(n+1\right)-\text{ln}n\right)[/latex]
24. [latex]\displaystyle\sum _{n=1}^{\infty }{\left(-1\right)}^{n+1}n\left({\tan}^{-1}\left(n+1\right)-{\tan}^{-1}n\right)[/latex] (Hint: Use Mean Value Theorem.)
Show Solution
Converges by alternating series test since [latex]n\left({\tan}^{-1}\left(n+1\right)\text{-}{\tan}^{-1}n\right)[/latex] is decreasing to zero for large [latex]n[/latex]. Does not converge absolutely by limit comparison with harmonic series after applying hint.
25. [latex]\displaystyle\sum _{n=1}^{\infty }{\left(-1\right)}^{n+1}\left({\left(n+1\right)}^{2}-{n}^{2}\right)[/latex]
26. [latex]\displaystyle\sum _{n=1}^{\infty }{\left(-1\right)}^{n+1}\left(\frac{1}{n}-\frac{1}{n+1}\right)[/latex]
Show Solution
Converges absolutely, since [latex]{a}_{n}=\frac{1}{n}-\frac{1}{n+1}[/latex] are terms of a telescoping series.
27. [latex]\displaystyle\sum _{n=1}^{\infty }\frac{\cos\left(n\pi \right)}{n}[/latex]
28. [latex]\displaystyle\sum _{n=1}^{\infty }\frac{\cos\left(n\pi \right)}{{n}^{\frac{1}{n}}}[/latex]
Show Solution
Terms do not tend to zero. Series diverges by divergence test.
29. [latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{n}\sin\left(\frac{n\pi }{2}\right)[/latex]
30. [latex]\displaystyle\sum _{n=1}^{\infty }\sin\left(\frac{n\pi}{2}\right)\sin\left(\frac{1}{n}\right)[/latex]
Show Solution
Converges by alternating series test. Does not converge absolutely by limit comparison with harmonic series.
In each of the following problems, use the estimate [latex]|{R}_{N}|\le {b}_{N+1}[/latex] to find a value of [latex]N[/latex] that guarantees that the sum of the first [latex]N[/latex] terms of the alternating series [latex]\displaystyle\sum _{n=1}^{\infty }{\left(-1\right)}^{n+1}{b}_{n}[/latex] differs from the infinite sum by at most the given error. Calculate the partial sum [latex]{S}_{N}[/latex] for this [latex]N[/latex].
31. [T] [latex]{b}_{n}=\frac{1}{n}[/latex], error [latex]<{10}^{-5}[/latex]
32. [T] [latex]{b}_{n}=\frac{1}{{ln}\left(n\right)}[/latex], [latex]n\ge 2[/latex], error [latex]<{10}^{-1}[/latex]
Show Solution
[latex]\text{ln}\left(N+1\right)>10[/latex], [latex]N+1>{e}^{10}[/latex], [latex]N\ge 22026[/latex]; [latex]{S}_{22026}=0.0257\text{$\ldots$ }[/latex]
33. [T] [latex]{b}_{n}=\frac{1}{\sqrt{n}}[/latex], error [latex]<{10}^{-3}[/latex]
34. [T] [latex]{b}_{n}=\frac{1}{{2}^{n}}[/latex], error [latex]<{10}^{-6}[/latex]
Show Solution
[latex]{2}^{N+1}>{10}^{6}[/latex] or [latex]N+1>\frac{6\text{ln}\left(10\right)}{\text{ln}\left(2\right)}=19.93[/latex]. or [latex]N\ge 19[/latex]; [latex]{S}_{19}=0.333333969\text{$\ldots$ }[/latex]
35. [T] [latex]{b}_{n}=\text{ln}\left(1+\frac{1}{n}\right)[/latex], error [latex]<{10}^{-3}[/latex]
36. [T] [latex]{b}_{n}=\frac{1}{{n}^{2}}[/latex], error [latex]<{10}^{-6}[/latex]
Show Solution
[latex]{\left(N+1\right)}^{2}>{10}^{6}[/latex] or [latex]N>999[/latex]; [latex]{S}_{1000}\approx 0.822466[/latex].
For the following exercises, 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 [latex]{b}_{n}\ge 0[/latex] is decreasing and [latex]\underset{n\to \infty }{\text{lim}}{b}_{n}=0[/latex], then [latex]\displaystyle\sum _{n=1}^{\infty }\left({b}_{2n - 1}-{b}_{2n}\right)[/latex] converges absolutely.
38. If [latex]{b}_{n}\ge 0[/latex] is decreasing, then [latex]\displaystyle\sum _{n=1}^{\infty }\left({b}_{2n - 1}-{b}_{2n}\right)[/latex] converges absolutely.
Show Solution
True. [latex]{b}_{n}[/latex] need not tend to zero since if [latex]{c}_{n}={b}_{n}-\text{lim}{b}_{n}[/latex], then [latex]{c}_{2n - 1}-{c}_{2n}={b}_{2n - 1}-{b}_{2n}[/latex].
39. If [latex]{b}_{n}\ge 0[/latex] and [latex]\underset{n\to \infty }{\text{lim}}{b}_{n}=0[/latex] then [latex]\displaystyle\sum _{n=1}^{\infty }\left(\frac{1}{2}\left({b}_{3n - 2}+{b}_{3n - 1}\right)\text{-}{b}_{3n}\right)[/latex] converges.
40. If [latex]{b}_{n}\ge 0[/latex] is decreasing and [latex]\displaystyle\sum _{n=1}^{\infty }\left({b}_{3n - 2}+{b}_{3n - 1}-{b}_{3n}\right)[/latex] converges then [latex]\displaystyle\sum _{n=1}^{\infty }{b}_{3n - 2}[/latex] converges.
Show Solution
True. [latex]{b}_{3n - 1}-{b}_{3n}\ge 0[/latex], so convergence of [latex]\displaystyle\sum {b}_{3n - 2}[/latex] follows from the comparison test.
41. If [latex]{b}_{n}\ge 0[/latex] is decreasing and [latex]\displaystyle\sum _{n=1}^{\infty }{\left(-1\right)}^{n - 1}{b}_{n}[/latex] converges conditionally but not absolutely, then [latex]{b}_{n}[/latex] does not tend to zero.
42. Let [latex]{a}_{n}^{+}={a}_{n}[/latex] if [latex]{a}_{n}\ge 0[/latex] and [latex]{a}_{n}^{-}=\text{-}{a}_{n}[/latex] if [latex]{a}_{n}<0[/latex]. (Also, [latex]{a}_{n}^{+}=0\text{ if }{a}_{n}<0[/latex] and [latex]{a}_{n}^{-}=0\text{ if }{a}_{n}\ge 0.[/latex]) If [latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}[/latex] converges conditionally but not absolutely, then neither [latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}^{+}[/latex] nor [latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}^{-}[/latex] converge.
Show Solution
True. If one converges, then so must the other, implying absolute convergence.
43. Suppose that [latex]{a}_{n}[/latex] is a sequence of positive real numbers and that [latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}[/latex] converges.
Suppose that [latex]{b}_{n}[/latex] is an arbitrary sequence of ones and minus ones. Does [latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}{b}_{n}[/latex] necessarily converge?
44. Suppose that [latex]{a}_{n}[/latex] is a sequence such that [latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}{b}_{n}[/latex] converges for every possible sequence [latex]{b}_{n}[/latex] of zeros and ones. Does [latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}[/latex] converge absolutely?
Show Solution
Yes. Take [latex]{b}_{n}=1[/latex] if [latex]{a}_{n}\ge 0[/latex] and [latex]{b}_{n}=0[/latex] if [latex]{a}_{n}<0[/latex]. Then [latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}{b}_{n}=\displaystyle\sum _{n:{a}_{n}\ge 0}{a}_{n}[/latex] converges. Similarly, one can show [latex]\displaystyle\sum _{n:{a}_{n}<0}{a}_{n}[/latex] converges. Since both series converge, the series must converge absolutely.
The following 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.
45. [latex]\displaystyle\sum _{n=1}^{\infty }{\left(-1\right)}^{n+1}\frac{{\sin}^{2}n}{n}[/latex]
46. [latex]\displaystyle\sum _{n=1}^{\infty }{\left(-1\right)}^{n+1}\frac{{\cos}^{2}n}{n}[/latex]
Show Solution
Not decreasing. Does not converge absolutely.
47. [latex]1+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}+\frac{1}{5}+\frac{1}{6}-\frac{1}{7}-\frac{1}{8}+\text{$\cdots$ }[/latex]
48. [latex]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}+\text{$\cdots$ }[/latex]
Show Solution
Not alternating. Can be expressed as [latex]\displaystyle\sum _{n=1}^{\infty }\left(\frac{1}{3n - 2}+\frac{1}{3n - 1}-\frac{1}{3n}\right)[/latex], which diverges by comparison with [latex]\displaystyle\sum \frac{1}{3n - 2}[/latex].
49. Show that the alternating series [latex]1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{6}+\frac{1}{4}-\frac{1}{8}+\text{$\cdots$ }[/latex] does not converge. What hypothesis of the alternating series test is not met?
50. Suppose that [latex]\displaystyle\sum {a}_{n}[/latex] converges absolutely. Show that the series consisting of the positive terms [latex]{a}_{n}[/latex] also converges.
Show Solution
Let [latex]{a}^{+}{}_{n}={a}_{n}[/latex] if [latex]{a}_{n}\ge 0[/latex] and [latex]{a}^{+}{}_{n}=0[/latex] if [latex]{a}_{n}<0[/latex]. Then [latex]{a}^{+}{}_{n}\le |{a}_{n}|[/latex] for all [latex]n[/latex] so the sequence of partial sums of [latex]{a}^{+}{}_{n}[/latex] is increasing and bounded above by the sequence of partial sums of [latex]|{a}_{n}|[/latex], which converges; hence, [latex]\displaystyle\sum _{n=1}^{\infty }{a}^{+}{}_{n}[/latex] converges.
51. Show that the alternating series [latex]\frac{2}{3}-\frac{3}{5}+\frac{4}{7}-\frac{5}{9}+\text{$\cdots$ }[/latex] does not converge. What hypothesis of the alternating series test is not met?
52. The formula [latex]\cos\theta =1-\frac{{\theta }^{2}}{2\text{!}}+\frac{{\theta }^{4}}{4\text{!}}-\frac{{\theta }^{6}}{6\text{!}}+\text{$\cdots$ }[/latex] will be derived in the next chapter. Use the remainder [latex]|{R}_{N}|\le {b}_{N+1}[/latex] to find a bound for the error in estimating [latex]\cos\theta[/latex] by the fifth partial sum [latex]1-\frac{{\theta }^{2}}{2\text{!}}+\frac{{\theta }^{4}}{4\text{!}}\frac{\text{-}{\theta }^{6}}{6\text{!}}+\frac{{\theta }^{8}}{8\text{!}}[/latex] for [latex]\theta =1[/latex], [latex]\theta =\frac{\pi}{6}[/latex], and [latex]\theta =\pi[/latex].
Show Solution
For [latex]N=5[/latex] one has [latex]|{R}_{N}|{b}_{6}=\frac{{\theta }^{10}}{10\text{!}}[/latex]. When [latex]\theta =1[/latex], [latex]{R}_{5}\le \frac{1}{10\text{!}}\approx 2.75\times {10}^{-7}[/latex]. When [latex]\theta =\frac{\pi}{6}[/latex], [latex]{R}_{5}\le {\left(\frac{\pi}{6}\right)}^\frac{{10}}{10\text{!}}\approx 4.26\times {10}^{-10}[/latex]. When [latex]\theta =\pi[/latex], [latex]{R}_{5}\le \frac{{\pi }^{10}}{10\text{!}}=0.0258[/latex].
53. The formula [latex]\sin\theta =\theta -\frac{{\theta }^{3}}{3\text{!}}+\frac{{\theta }^{5}}{5\text{!}}-\frac{{\theta }^{7}}{7\text{!}}+\text{$\cdots$ }[/latex] will be derived in the next chapter. Use the remainder [latex]|{R}_{N}|\le {b}_{N+1}[/latex] to find a bound for the error in estimating [latex]\sin\theta[/latex] by the fifth partial sum [latex]\theta -{\theta }^\frac{{3}}{3\text{!}}+\frac{{\theta }^{5}}{5\text{!}}\frac{\text{-}{\theta }^{7}}{7\text{!}}+\frac{{\theta }^{9}}{9\text{!}}[/latex] for [latex]\theta =1[/latex], [latex]\theta =\frac{\pi}{6}[/latex], and [latex]\theta =\pi[/latex].
54. How many terms in [latex]\cos\theta =1-\frac{{\theta }^{2}}{2\text{!}}+\frac{{\theta }^{4}}{4\text{!}}-\frac{{\theta }^{6}}{6\text{!}}+\text{$\cdots$ }[/latex] are needed to approximate [latex]\cos1[/latex] accurate to an error of at most [latex]0.00001\text{?}[/latex]
Show Solution
Let [latex]{b}_{n}=\frac{1}{\left(2n - 2\right)}\text{!}[/latex]. Then [latex]{R}_{N}\le \frac{1}{\left(2N\right)\text{!}}<0.00001[/latex] when [latex]\left(2N\right)\text{!}>{10}^{5}[/latex] or [latex]N=5[/latex] and [latex]1-\frac{1}{2\text{!}}+\frac{1}{4\text{!}}-\frac{1}{6\text{!}}+\frac{1}{8\text{!}}=0.540325\text{$\ldots$ }[/latex], whereas [latex]\cos1=0.5403023\text{$\ldots$ }[/latex]
55. How many terms in [latex]\sin\theta =\theta -\frac{{\theta }^{3}}{3\text{!}}+\frac{{\theta }^{5}}{5\text{!}}-\frac{{\theta }^{7}}{7\text{!}}+\text{$\cdots$ }[/latex] are needed to approximate [latex]\sin1[/latex] accurate to an error of at most [latex]0.00001\text{?}[/latex]
56. Sometimes the alternating series [latex]\displaystyle\sum _{n=1}^{\infty }{\left(-1\right)}^{n - 1}{b}_{n}[/latex] converges to a certain fraction of an absolutely convergent series [latex]\displaystyle\sum _{n=1}^{\infty }{b}_{n}[/latex] at a faster rate. Given that [latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{{n}^{2}}=\frac{{\pi }^{2}}{6}[/latex], find [latex]12=1-\frac{1}{{2}^{2}}+\frac{1}{{3}^{2}}-\frac{1}{{4}^{2}}+\text{$\cdots$ }[/latex]. Which of the series [latex]6\displaystyle\sum _{n=1}^{\infty }\frac{1}{{n}^{2}}[/latex] and [latex]S\displaystyle\sum _{n=1}^{\infty }\frac{{\left(-1\right)}^{n - 1}}{{n}^{2}}[/latex] gives a better estimation of [latex]{\pi }^{2}[/latex] using [latex]1000[/latex] terms?
Show Solution
Let [latex]T=\displaystyle\sum \frac{1}{{n}^{2}}[/latex]. Then [latex]T-S=\frac{1}{2}T[/latex], so [latex]S=\frac{T}{2}[/latex]. [latex]\sqrt{6\times \displaystyle\sum _{n=1}^{1000}\frac{1}{{n}^{2}}}=3.140638\text{$\ldots$ }[/latex]; [latex]\sqrt{12\times \displaystyle\sum _{n=1}^{1000}\frac{{\left(-1\right)}^{n - 1}}{{n}^{2}}}=3.141591\text{$\ldots$ }[/latex]; [latex]\pi =3.141592\text{$\ldots$ }[/latex]. The alternating series is more accurate for [latex]1000[/latex] terms.
The following alternating series converge to given multiples of [latex]\pi[/latex]. Find the value of [latex]N[/latex] predicted by the remainder estimate such that the [latex]N\text{th}[/latex] partial sum of the series accurately approximates the left-hand side to within the given error. Find the minimum [latex]N[/latex] for which the error bound holds, and give the desired approximate value in each case. Up to [latex]15[/latex] decimals places, [latex]\pi =3.141592653589793\text{$\ldots$ }[/latex].
57. [T] [latex]\frac{\pi }{4}=\displaystyle\sum _{n=0}^{\infty }\frac{{\left(-1\right)}^{n}}{2n+1}[/latex], error [latex]<0.0001[/latex]
58. [T] [latex]\frac{\pi }{\sqrt{12}}=\displaystyle\sum _{k=0}^{\infty }\frac{{\left(-3\right)}^{\text{-}k}}{2k+1}[/latex], error [latex]<0.0001[/latex]
Show Solution
[latex]N=6[/latex], [latex]{S}_{N}=0.9068[/latex]
59. [T] The series [latex]\displaystyle\sum _{n=0}^{\infty }\frac{\sin\left(x+\pi n\right)}{x+\pi n}[/latex] plays an important role in signal processing. Show that [latex]\displaystyle\sum _{n=0}^{\infty }\frac{\sin\left(x+\pi n\right)}{x+\pi n}[/latex] converges whenever [latex]0Hint: Use the formula for the sine of a sum of angles.)
60. [T] If [latex]\displaystyle\sum _{n=1}^{N}{\left(-1\right)}^{n - 1}\frac{1}{n}\to \text{ln}2[/latex], what is [latex]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}+\text{$\cdots$ }\text{?}[/latex]
Show Solution
[latex]\text{ln}\left(2\right)[/latex]. The [latex]3n\text{th}[/latex] partial sum is the same as that for the alternating harmonic series.
61. [T] Plot the series [latex]\displaystyle\sum _{n=1}^{100}\frac{\cos\left(2\pi nx\right)}{n}[/latex] for [latex]0\le x<1[/latex]. Explain why [latex]\displaystyle\sum _{n=1}^{100}\frac{\cos\left(2\pi nx\right)}{n}[/latex] diverges when [latex]x=0,1[/latex]. How does the series behave for other [latex]x\text{?}[/latex]
62. [T] Plot the series [latex]\displaystyle\sum _{n=1}^{100}\frac{\sin\left(2\pi nx\right)}{n}[/latex] for [latex]0\le x<1[/latex] and comment on its behavior
Show Solution
The series jumps rapidly near the endpoints. For [latex]x[/latex] away from the endpoints, the graph looks like [latex]\pi \left(\frac{1}{2}-x\right)[/latex].
63. [T] Plot the series [latex]\displaystyle\sum _{n=1}^{100}\frac{\cos\left(2\pi nx\right)}{{n}^{2}}[/latex] for [latex]0\le x<1[/latex] and describe its graph.
64. [T] The alternating harmonic series converges because of cancellation among its terms. Its sum is known because the cancellation can be described explicitly. A random harmonic series is one of the form [latex]\displaystyle\sum _{n=1}^{\infty }\frac{{S}_{n}}{n}[/latex], where [latex]{s}_{n}[/latex] is a randomly generated sequence of [latex]\pm 1\text{'s}[/latex] in which the values [latex]\pm 1[/latex] are equally likely to occur. Use a random number generator to produce [latex]1000[/latex] random [latex]\pm 1\text{s}[/latex] and plot the partial sums [latex]{S}_{N}=\displaystyle\sum _{n=1}^{N}\frac{{s}_{n}}{n}[/latex] of your random harmonic sequence for [latex]N=1[/latex] to [latex]1000[/latex]. Compare to a plot of the first [latex]1000[/latex] partial sums of the harmonic series.
Show Solution
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.
65. [T] Estimates of [latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{{n}^{2}}[/latex] can be accelerated by writing its partial sums as [latex]\displaystyle\sum _{n=1}^{N}\frac{1}{{n}^{2}}=\displaystyle\sum _{n=1}^{N}\frac{1}{n\left(n+1\right)}+\displaystyle\sum _{n=1}^{N}\frac{1}{{n}^{2}\left(n+1\right)}[/latex] and recalling that [latex]\displaystyle\sum _{n=1}^{N}\frac{1}{n\left(n+1\right)}=1-\frac{1}{N+1}[/latex] converges to one as [latex]N\to \infty[/latex]. Compare the estimate of [latex]\frac{{\pi }^{2}}{6}[/latex] using the sums [latex]\displaystyle\sum _{n=1}^{1000}\frac{1}{{n}^{2}}[/latex] with the estimate using [latex]1+\displaystyle\sum _{n=1}^{1000}\frac{1}{{n}^{2}\left(n+1\right)}[/latex].
66. [T] The Euler transform rewrites [latex]S=\displaystyle\sum _{n=0}^{\infty }{\left(-1\right)}^{n}{b}_{n}[/latex] as [latex]S=\displaystyle\sum _{n=0}^{\infty }{\left(-1\right)}^{n}{2}^{\text{-}n - 1}\displaystyle\sum _{m=0}^{n}\left(\begin{array}{c}n\\ m\end{array}\right){b}_{n-m}[/latex]. For the alternating harmonic series, it takes the form [latex]\text{ln}\left(2\right)=\displaystyle\sum _{n=1}^{\infty }\frac{{\left(-1\right)}^{n - 1}}{n}=\displaystyle\sum _{n=1}^{\infty }\frac{1}{n{2}^{n}}[/latex]. Compute partial sums of [latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{n{2}^{n}}[/latex] until they approximate [latex]\text{ln}\left(2\right)[/latex] accurate to within [latex]0.0001[/latex]. How many terms are needed? Compare this answer to the number of terms of the alternating harmonic series are needed to estimate [latex]\text{ln}\left(2\right)[/latex].
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By the alternating series test, [latex]|{S}_{n}-S|\le {b}_{n+1}[/latex], so one needs [latex]{10}^{4}[/latex] terms of the alternating harmonic series to estimate [latex]\text{ln}\left(2\right)[/latex] to within [latex]0.0001[/latex]. The first [latex]10[/latex] partial sums of the series [latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{n{2}^{n}}[/latex] are (up to four decimals) [latex]0.5000,0.6250,0.6667,0.6823,0.6885,0.6911,0.6923,0.6928,0.6930,0.6931[/latex] and the tenth partial sum is within [latex]0.0001[/latex] of [latex]\text{ln}\left(2\right)=0.6931\text{$\ldots$ }[/latex].
67. [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 [latex]{a}_{n}\ge 0[/latex] is such that [latex]{a}_{n}\to 0[/latex] as [latex]n\to \infty[/latex] but [latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}[/latex] diverges, then, given any number [latex]A[/latex] there is a sequence [latex]{s}_{n}[/latex] of [latex]\pm 1\text{'s}[/latex] such that [latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}{s}_{n}\to A[/latex]. Show this for [latex]A>0[/latex] as follows.
- Recursively define [latex]{s}_{n}[/latex] by [latex]{s}_{n}=1[/latex] if [latex]{S}_{n - 1}=\displaystyle\sum _{k=1}^{n - 1}{a}_{k}{s}_{k}
- Explain why eventually [latex]{S}_{n}\ge A[/latex], and for any [latex]m[/latex] larger than this [latex]n[/latex], [latex]A-{a}_{m}\le {S}_{m}\le A+{a}_{m}[/latex].
- Explain why this implies that [latex]{S}_{n}\to A[/latex] as [latex]n\to \infty[/latex].
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