In the following exercises, state whether each statement is true, or give an example to show that it is false.
1. If [latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}{x}^{n}[/latex] converges, then [latex]{a}_{n}{x}^{n}\to 0[/latex] as [latex]n\to \infty [/latex].
Show Solution
True. If a series converges then its terms tend to zero.
2. [latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}{x}^{n}[/latex] converges at [latex]x=0[/latex] for any real numbers [latex]{a}_{n}[/latex].
3. Given any sequence [latex]{a}_{n}[/latex], there is always some [latex]R>0[/latex], possibly very small, such that [latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}{x}^{n}[/latex] converges on [latex]\left(\text{-}R,R\right)[/latex].
Show Solution
False. It would imply that [latex]{a}_{n}{x}^{n}\to 0[/latex] for [latex]|x|<R[/latex]. If [latex]{a}_{n}={n}^{n}[/latex], then [latex]{a}_{n}{x}^{n}={\left(nx\right)}^{n}[/latex] does not tend to zero for any [latex]x\ne 0[/latex].
4. If [latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}{x}^{n}[/latex] has radius of convergence [latex]R>0[/latex] and if [latex]|{b}_{n}|\le |{a}_{n}|[/latex] for all n, then the radius of convergence of [latex]\displaystyle\sum _{n=1}^{\infty }{b}_{n}{x}^{n}[/latex] is greater than or equal to R.
5. Suppose that [latex]\displaystyle\sum _{n=0}^{\infty }{a}_{n}{\left(x - 3\right)}^{n}[/latex] converges at [latex]x=6[/latex]. At which of the following points must the series also converge? Use the fact that if [latex]\displaystyle\sum {a}_{n}{\left(x-c\right)}^{n}[/latex] converges at x, then it converges at any point closer to c than x.
- [latex]x=1[/latex]
- [latex]x=2[/latex]
- [latex]x=3[/latex]
- [latex]x=0[/latex]
- [latex]x=5.99[/latex]
- [latex]x=0.000001[/latex]
Show Solution
It must converge on [latex]\left(0,6\right][/latex] and hence at: a. [latex]x=1[/latex]; b. [latex]x=2[/latex]; c. [latex]x=3[/latex]; d. [latex]x=0[/latex]; e. [latex]x=5.99[/latex]; and f. [latex]x=0.000001[/latex].
6. Suppose that [latex]\displaystyle\sum _{n=0}^{\infty }{a}_{n}{\left(x+1\right)}^{n}[/latex] converges at [latex]x=-2[/latex].
At which of the following points must the series also converge? Use the fact that if [latex]\displaystyle\sum {a}_{n}{\left(x-c\right)}^{n}[/latex] converges at x, then it converges at any point closer to c than x.
- [latex]x=2[/latex]
- [latex]x=-1[/latex]
- [latex]x=-3[/latex]
- [latex]x=0[/latex]
- [latex]x=0.99[/latex]
- [latex]x=0.000001[/latex]
In the following exercises, suppose that [latex]|\frac{{a}_{n+1}}{{a}_{n}}|\to 1[/latex] as [latex]n\to \infty [/latex]. Find the radius of convergence for each series.
7. [latex]\displaystyle\sum _{n=0}^{\infty }{a}_{n}{2}^{n}{x}^{n}[/latex]
Show Solution
[latex]|\frac{{a}_{n+1}{2}^{n+1}{x}^{n+1}}{{a}_{n}{2}^{n}{x}^{n}}|=2|x||\frac{{a}_{n+1}}{{a}_{n}}|\to 2|x|[/latex] so [latex]R=\frac{1}{2}[/latex]
8. [latex]\displaystyle\sum _{n=0}^{\infty }\frac{{a}_{n}{x}^{n}}{{2}^{n}}[/latex]
9. [latex]\displaystyle\sum _{n=0}^{\infty }\frac{{a}_{n}{\pi }^{n}{x}^{n}}{{e}^{n}}[/latex]
Show Solution
[latex]|\frac{{a}_{n+1}{\left(\frac{\pi }{e}\right)}^{n+1}{x}^{n+1}}{{a}_{n}{\left(\frac{\pi }{e}\right)}^{n}{x}^{n}}|=\frac{\pi |x|}{e}|\frac{{a}_{n+1}}{{a}_{n}}|\to \frac{\pi |x|}{e}[/latex] so [latex]R=\frac{e}{\pi }[/latex]
10. [latex]\displaystyle\sum _{n=0}^{\infty }\frac{{a}_{n}{\left(-1\right)}^{n}{x}^{n}}{{10}^{n}}[/latex]
11. [latex]\displaystyle\sum _{n=0}^{\infty }{a}_{n}{\left(-1\right)}^{n}{x}^{2n}[/latex]
Show Solution
[latex]|\frac{{a}_{n+1}{\left(-1\right)}^{n+1}{x}^{2n+2}}{{a}_{n}{\left(-1\right)}^{n}{x}^{2n}}|=|{x}^{2}||\frac{{a}_{n+1}}{{a}_{n}}|\to |{x}^{2}|[/latex] so [latex]R=1[/latex]
12. [latex]\displaystyle\sum _{n=0}^{\infty }{a}_{n}{\left(-4\right)}^{n}{x}^{2n}[/latex]
In the following exercises, find the radius of convergence R and interval of convergence for [latex]\displaystyle\sum {a}_{n}{x}^{n}[/latex] with the given coefficients [latex]{a}_{n}[/latex].
13. [latex]\displaystyle\sum _{n=1}^{\infty }\frac{{\left(2x\right)}^{n}}{n}[/latex]
Show Solution
[latex]{a}_{n}=\frac{{2}^{n}}{n}[/latex] so [latex]\frac{{a}_{n+1}x}{{a}_{n}}\to 2x[/latex]. so [latex]R=\frac{1}{2}[/latex]. When [latex]x=\frac{1}{2}[/latex] the series is harmonic and diverges. When [latex]x=-\frac{1}{2}[/latex] the series is alternating harmonic and converges. The interval of convergence is [latex]I=\left[-\frac{1}{2},\frac{1}{2}\right)[/latex].
14. [latex]\displaystyle\sum _{n=1}^{\infty }{\left(-1\right)}^{n}\frac{{x}^{n}}{\sqrt{n}}[/latex]
15. [latex]\displaystyle\sum _{n=1}^{\infty }\frac{n{x}^{n}}{{2}^{n}}[/latex]
Show Solution
[latex]{a}_{n}=\frac{n}{{2}^{n}}[/latex] so [latex]\frac{{a}_{n+1}x}{{a}_{n}}\to \frac{x}{2}[/latex] so [latex]R=2[/latex]. When [latex]x=\pm2[/latex] the series diverges by the divergence test. The interval of convergence is [latex]I=\left(-2,2\right)[/latex].
16. [latex]\displaystyle\sum _{n=1}^{\infty }\frac{n{x}^{n}}{{e}^{n}}[/latex]
17. [latex]\displaystyle\sum _{n=1}^{\infty }\frac{{n}^{2}{x}^{n}}{{2}^{n}}[/latex]
Show Solution
[latex]{a}_{n}=\frac{{n}^{2}}{{2}^{n}}[/latex] so [latex]R=2[/latex]. When [latex]x=\pm[/latex] the series diverges by the divergence test. The interval of convergence is [latex]I=\left(-2,2\right)[/latex].
18. [latex]\displaystyle\sum _{k=1}^{\infty }\frac{{k}^{e}{x}^{k}}{{e}^{k}}[/latex]
19. [latex]\displaystyle\sum _{k=1}^{\infty }\frac{{\pi }^{k}{x}^{k}}{{k}^{\pi }}[/latex]
Show Solution
[latex]{a}_{k}=\frac{{\pi }^{k}}{{k}^{\pi }}[/latex] so [latex]R=\frac{1}{\pi }[/latex]. When [latex]x=\pm\frac{1}{\pi }[/latex] the series is an absolutely convergent p-series. The interval of convergence is [latex]I=\left[-\frac{1}{\pi },\frac{1}{\pi }\right][/latex].
20. [latex]\displaystyle\sum _{n=1}^{\infty }\frac{{x}^{n}}{n\text{!}}[/latex]
21. [latex]\displaystyle\sum _{n=1}^{\infty }\frac{{10}^{n}{x}^{n}}{n\text{!}}[/latex]
Show Solution
[latex]{a}_{n}=\frac{{10}^{n}}{n\text{!}},\frac{{a}_{n+1}x}{{a}_{n}}=\frac{10x}{n+1}\to 0<1[/latex] so the series converges for all x by the ratio test and [latex]I=\left(\text{-}\infty ,\infty \right)[/latex].
22. [latex]\displaystyle\sum _{n=1}^{\infty }{\left(-1\right)}^{n}\frac{{x}^{n}}{\text{ln}\left(2n\right)}[/latex]
In the following exercises, find the radius of convergence of each series.
23. [latex]\displaystyle\sum _{k=1}^{\infty }\frac{{\left(k\text{!}\right)}^{2}{x}^{k}}{\left(2k\right)\text{!}}[/latex]
Show Solution
[latex]{a}_{k}=\frac{{\left(k\text{!}\right)}^{2}}{\left(2k\right)\text{!}}[/latex] so [latex]\frac{{a}_{k+1}}{{a}_{k}}=\frac{{\left(k+1\right)}^{2}}{\left(2k+2\right)\left(2k+1\right)}\to \frac{1}{4}[/latex] so [latex]R=4[/latex]
24. [latex]\displaystyle\sum _{n=1}^{\infty }\frac{\left(2n\right)\text{!}{x}^{n}}{{n}^{2n}}[/latex]
25. [latex]\displaystyle\sum _{k=1}^{\infty }\frac{k\text{!}}{1\cdot 3\cdot 5\text{$\cdots$ }\left(2k - 1\right)}{x}^{k}[/latex]
Show Solution
[latex]{a}_{k}=\frac{k\text{!}}{1\cdot 3\cdot 5\cdots\left(2k - 1\right)}[/latex] so [latex]\frac{{a}_{k+1}}{{a}_{k}}=\frac{k+1}{2k+1}\to \frac{1}{2}[/latex] so [latex]R=2[/latex]
26. [latex]\displaystyle\sum _{k=1}^{\infty }\frac{2\cdot 4\cdot 6\text{$\cdots$ }2k}{\left(2k\right)\text{!}}{x}^{k}[/latex]
27. [latex]\displaystyle\sum _{n=1}^{\infty }\frac{{x}^{n}}{\left(\begin{array}{c}2n\\ n\end{array}\right)}[/latex] where [latex]\left(\begin{array}{c}n\\ k\end{array}\right)=\frac{n\text{!}}{k\text{!}\left(n-k\right)\text{!}}[/latex]
Show Solution
[latex]{a}_{n}=\frac{1}{\left(\begin{array}{c}2n\\ n\end{array}\right)}[/latex] so [latex]\frac{{a}_{n+1}}{{a}_{n}}=\frac{{\left(\left(n+1\right)\text{!}\right)}^{2}}{\left(2n+2\right)\text{!}}\frac{2n\text{!}}{{\left(n\text{!}\right)}^{2}}=\frac{{\left(n+1\right)}^{2}}{\left(2n+2\right)\left(2n+1\right)}\to \frac{1}{4}[/latex] so [latex]R=4[/latex]
28. [latex]\displaystyle\sum _{n=1}^{\infty }{\sin}^{2}n{x}^{n}[/latex]
In the following exercises, use the ratio test to determine the radius of convergence of each series.
29. [latex]\displaystyle\sum _{n=1}^{\infty }\frac{{\left(n\text{!}\right)}^{3}}{\left(3n\right)\text{!}}{x}^{n}[/latex]
Show Solution
[latex]\frac{{a}_{n+1}}{{a}_{n}}=\frac{{\left(n+1\right)}^{3}}{\left(3n+3\right)\left(3n+2\right)\left(3n+1\right)}\to \frac{1}{27}[/latex] so [latex]R=27[/latex]
30. [latex]\displaystyle\sum _{n=1}^{\infty }\frac{{2}^{3n}{\left(n\text{!}\right)}^{3}}{\left(3n\right)\text{!}}{x}^{n}[/latex]
31. [latex]\displaystyle\sum _{n=1}^{\infty }\frac{n\text{!}}{{n}^{n}}{x}^{n}[/latex]
Show Solution
[latex]{a}_{n}=\frac{n\text{!}}{{n}^{n}}[/latex] so [latex]\frac{{a}_{n+1}}{{a}_{n}}=\frac{\left(n+1\right)\text{!}}{n\text{!}}\frac{{n}^{n}}{{\left(n+1\right)}^{n+1}}={\left(\frac{n}{n+1}\right)}^{n}\to \frac{1}{e}[/latex] so [latex]R=e[/latex]
32. [latex]\displaystyle\sum _{n=1}^{\infty }\frac{\left(2n\right)\text{!}}{{n}^{2n}}{x}^{n}[/latex]
In the following exercises, given that [latex]\frac{1}{1-x}=\displaystyle\sum _{n=0}^{\infty }{x}^{n}[/latex] with convergence in [latex]\left(-1,1\right)[/latex], find the power series for each function with the given center a, and identify its interval of convergence.
33. [latex]f\left(x\right)=\frac{1}{x};a=1[/latex] (Hint: [latex]\frac{1}{x}=\frac{1}{1-\left(1-x\right)}[/latex])
Show Solution
[latex]f\left(x\right)=\displaystyle\sum _{n=0}^{\infty }{\left(1-x\right)}^{n}[/latex] on [latex]I=\left(0,2\right)[/latex]
34. [latex]f\left(x\right)=\frac{1}{1-{x}^{2}};a=0[/latex]
35. [latex]f\left(x\right)=\frac{x}{1-{x}^{2}};a=0[/latex]
Show Solution
[latex]\displaystyle\sum _{n=0}^{\infty }{x}^{2n+1}[/latex] on [latex]I=\left(-1,1\right)[/latex]
36. [latex]f\left(x\right)=\frac{1}{1+{x}^{2}};a=0[/latex]
37. [latex]f\left(x\right)=\frac{{x}^{2}}{1+{x}^{2}};a=0[/latex]
Show Solution
[latex]\displaystyle\sum _{n=0}^{\infty }{\left(-1\right)}^{n}{x}^{2n+2}[/latex] on [latex]I=\left(-1,1\right)[/latex]
38. [latex]f\left(x\right)=\frac{1}{2-x};a=1[/latex]
39. [latex]f\left(x\right)=\frac{1}{1 - 2x};a=0[/latex].
Show Solution
[latex]\displaystyle\sum _{n=0}^{\infty }{2}^{n}{x}^{n}[/latex] on [latex]\left(-\frac{1}{2},\frac{1}{2}\right)[/latex]
40. [latex]f\left(x\right)=\frac{1}{1 - 4{x}^{2}};a=0[/latex]
41. [latex]f\left(x\right)=\frac{{x}^{2}}{1 - 4{x}^{2}};a=0[/latex]
Show Solution
[latex]\displaystyle\sum _{n=0}^{\infty }{4}^{n}{x}^{2n+2}[/latex] on [latex]\left(-\frac{1}{2},\frac{1}{2}\right)[/latex]
42. [latex]f\left(x\right)=\frac{{x}^{2}}{5 - 4x+{x}^{2}};a=2[/latex]
Use the next exercise to find the radius of convergence of the given series in the subsequent exercises.
43. Explain why, if [latex]{|{a}_{n}|}^{\frac{1}{n}}\to r>0[/latex], then [latex]{|{a}_{n}{x}^{n}|}^{\frac{1}{n}}\to |x|r<1[/latex] whenever [latex]|x|<\frac{1}{r}[/latex] and, therefore, the radius of convergence of [latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}{x}^{n}[/latex] is [latex]R=\frac{1}{r}[/latex].
Show Solution
[latex]{|{a}_{n}{x}^{n}|}^{\frac{1}{n}}={|{a}_{n}|}^{\frac{1}{n}}|x|\to |x|r[/latex] as [latex]n\to \infty [/latex] and [latex]|x|r<1[/latex] when [latex]|x|<\frac{1}{r}[/latex]. Therefore, [latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}{x}^{n}[/latex] converges when [latex]|x|<\frac{1}{r}[/latex] by the nth root test.
44. [latex]\displaystyle\sum _{n=1}^{\infty }\frac{{x}^{n}}{{n}^{n}}[/latex]
45. [latex]\displaystyle\sum _{k=1}^{\infty }{\left(\frac{k - 1}{2k+3}\right)}^{k}{x}^{k}[/latex]
Show Solution
[latex]{a}_{k}={\left(\frac{k - 1}{2k+3}\right)}^{k}[/latex] so [latex]{\left({a}_{k}\right)}^{\frac{1}{k}}\to \frac{1}{2}<1[/latex] so [latex]R=2[/latex]
46. [latex]\displaystyle\sum _{k=1}^{\infty }{\left(\frac{2{k}^{2}-1}{{k}^{2}+3}\right)}^{k}{x}^{k}[/latex]
47. [latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}={\left({n}^{\frac{1}{n}}-1\right)}^{n}{x}^{n}[/latex]
Show Solution
[latex]{a}_{n}={\left({n}^{\frac{1}{n}}-1\right)}^{n}[/latex] so [latex]{\left({a}_{n}\right)}^{\frac{1}{n}}\to 0[/latex] so [latex]R=\infty [/latex]
48. Suppose that [latex]p\left(x\right)=\displaystyle\sum _{n=0}^{\infty }{a}_{n}{x}^{n}[/latex] such that [latex]{a}_{n}=0[/latex] if n is odd. Explain why [latex]p\left(x\right)=-p\left(\text{-}x\right)[/latex].
49. Suppose that [latex]p\left(x\right)=\displaystyle\sum _{n=0}^{\infty }{a}_{n}{x}^{n}[/latex] such that [latex]{a}_{n}=0[/latex] if n is even. Explain why [latex]p\left(x\right)=p\left(\text{-}x\right)[/latex].
Show Solution
We can rewrite [latex]p\left(x\right)=\displaystyle\sum _{n=0}^{\infty }{a}_{2n+1}{x}^{2n+1}[/latex] and [latex]p\left(x\right)=p\left(\text{-}x\right)[/latex] since [latex]{x}^{2n+1}=\text{-}{\left(\text{-}x\right)}^{2n+1}[/latex].
50. Suppose that [latex]p\left(x\right)=\displaystyle\sum _{n=0}^{\infty }{a}_{n}{x}^{n}[/latex] converges on [latex]\left(-1,1\right][/latex].
Find the interval of convergence of [latex]p\left(Ax\right)[/latex].
51. Suppose that [latex]p\left(x\right)=\displaystyle\sum _{n=0}^{\infty }{a}_{n}{x}^{n}[/latex] converges on [latex]\left(-1,1\right][/latex]. Find the interval of convergence of [latex]p\left(2x - 1\right)[/latex].
Show Solution
If [latex]x\in \left[0,1\right][/latex], then [latex]y=2x - 1\in \left[-1,1\right][/latex] so [latex]p\left(2x - 1\right)=p\left(y\right)=\displaystyle\sum _{n=0}^{\infty }{a}_{n}{y}^{n}[/latex] converges.
In the following exercises, suppose that [latex]p\left(x\right)=\displaystyle\sum _{n=0}^{\infty }{a}_{n}{x}^{n}[/latex] satisfies [latex]\underset{n\to \infty }{\text{lim}}\frac{{a}_{n+1}}{{a}_{n}}=1[/latex] where [latex]{a}_{n}\ge 0[/latex] for each n. State whether each series converges on the full interval [latex]\left(-1,1\right)[/latex], or if there is not enough information to draw a conclusion. Use the comparison test when appropriate.
52. [latex]\displaystyle\sum _{n=0}^{\infty }{a}_{n}{x}^{2n}[/latex]
53. [latex]\displaystyle\sum _{n=0}^{\infty }{a}_{2n}{x}^{2n}[/latex]
Show Solution
Converges on [latex]\left(-1,1\right)[/latex] by the ratio test
54. [latex]\displaystyle\sum _{n=0}^{\infty }{a}_{2n}{x}^{n}\left(Hint\text{:}x=\pm \sqrt{{x}^{2}}\right)[/latex]
55. [latex]\displaystyle\sum _{n=0}^{\infty }{a}_{{n}^{2}}{x}^{{n}^{2}}[/latex] (Hint: Let [latex]{b}_{k}={a}_{k}[/latex] if [latex]k={n}^{2}[/latex] for some n, otherwise [latex]{b}_{k}=0.[/latex])
Show Solution
Consider the series [latex]\displaystyle\sum {b}_{k}{x}^{k}[/latex] where [latex]{b}_{k}={a}_{k}[/latex] if [latex]k={n}^{2}[/latex] and [latex]{b}_{k}=0[/latex] otherwise. Then [latex]{b}_{k}\le {a}_{k}[/latex] and so the series converges on [latex]\left(-1,1\right)[/latex] by the comparison test.
56. Suppose that [latex]p\left(x\right)[/latex] is a polynomial of degree N. Find the radius and interval of convergence of [latex]\displaystyle\sum _{n=1}^{\infty }p\left(n\right){x}^{n}[/latex].
57. [T] Plot the graphs of [latex]\frac{1}{1-x}[/latex] and of the partial sums [latex]{S}_{N}=\displaystyle\sum _{n=0}^{N}{x}^{n}[/latex] for [latex]n=10,20,30[/latex] on the interval [latex]\left[-0.99,0.99\right][/latex]. Comment on the approximation of [latex]\frac{1}{1-x}[/latex] by [latex]{S}_{N}[/latex] near [latex]x=-1[/latex] and near [latex]x=1[/latex] as N increases.
Show Solution
The approximation is more accurate near [latex]x=-1[/latex]. The partial sums follow [latex]\frac{1}{1-x}[/latex] more closely as N increases but are never accurate near [latex]x=1[/latex] since the series diverges there.
58. [T] Plot the graphs of [latex]\text{-}\text{ln}\left(1-x\right)[/latex] and of the partial sums [latex]{S}_{N}=\displaystyle\sum _{n=1}^{N}\frac{{x}^{n}}{n}[/latex] for [latex]n=10,50,100[/latex] on the interval [latex]\left[-0.99,0.99\right][/latex]. Comment on the behavior of the sums near [latex]x=-1[/latex] and near [latex]x=1[/latex] as N increases.
59. [T] Plot the graphs of the partial sums [latex]{S}_{n}=\displaystyle\sum _{n=1}^{N}\frac{{x}^{n}}{{n}^{2}}[/latex] for [latex]n=10,50,100[/latex] on the interval [latex]\left[-0.99,0.99\right][/latex]. Comment on the behavior of the sums near [latex]x=-1[/latex] and near [latex]x=1[/latex] as N increases.
Show Solution
The approximation appears to stabilize quickly near both [latex]x=\pm 1[/latex].
60. [T] Plot the graphs of the partial sums [latex]{S}_{N}=\displaystyle\sum _{n=1}^{N}\sin{n}{x}^{n}[/latex] for [latex]n=10,50,100[/latex] on the interval [latex]\left[-0.99,0.99\right][/latex]. Comment on the behavior of the sums near [latex]x=-1[/latex] and near [latex]x=1[/latex] as N increases.
61. [T] Plot the graphs of the partial sums [latex]{S}_{N}=\displaystyle\sum _{n=0}^{N}{\left(-1\right)}^{n}\frac{{x}^{2n+1}}{\left(2n+1\right)\text{!}}[/latex] for [latex]n=3,5,10[/latex] on the interval [latex]\left[-2\pi ,2\pi \right][/latex]. Comment on how these plots approximate [latex]\sin{x}[/latex] as N increases.
Show Solution
The polynomial curves have roots close to those of [latex]\sin{x}[/latex] up to their degree and then the polynomials diverge from [latex]\sin{x}[/latex].
62. [T] Plot the graphs of the partial sums [latex]{S}_{N}=\displaystyle\sum _{n=0}^{N}{\left(-1\right)}^{n}\frac{{x}^{2n}}{\left(2n\right)\text{!}}[/latex] for [latex]n=3,5,10[/latex] on the interval [latex]\left[-2\pi ,2\pi \right][/latex]. Comment on how these plots approximate [latex]\cos{x}[/latex] as N increases.
