Problem Set: Power Series and Functions

In the following exercises, state whether each statement is true, or give an example to show that it is false.

1. If $\displaystyle\sum _{n=1}^{\infty }{a}_{n}{x}^{n}$ converges, then ${a}_{n}{x}^{n}\to 0$ as $n\to \infty$ .

Show Solution

\infty \sum n = 1 a_{n} x^{n}

$\displaystyle\sum _{n=1}^{\infty }{a}_{n}{x}^{n}$ converges at

x = 0

$x=0$ for any real numbers

a_{n}

${a}_{n}$ .

3. Given any sequence ${a}_{n}$ , there is always some $R>0$ , possibly very small, such that $\displaystyle\sum _{n=1}^{\infty }{a}_{n}{x}^{n}$ converges on $\left(\text{-}R,R\right)$ .

Show Solution

False. It would imply that

a_{n} x^{n} \to 0

${a}_{n}{x}^{n}\to 0$ for [latex]|x|

4. If

\infty \sum n = 1 a_{n} x^{n}

$\displaystyle\sum _{n=1}^{\infty }{a}_{n}{x}^{n}$ has radius of convergence

R > 0

$R>0$ and if

| b_{n} | \leq | a_{n} |

$|{b}_{n}|\le |{a}_{n}|$ for all n, then the radius of convergence of

\infty \sum n = 1 b_{n} x^{n}

$\displaystyle\sum _{n=1}^{\infty }{b}_{n}{x}^{n}$ is greater than or equal to R.

5. Suppose that $\displaystyle\sum _{n=0}^{\infty }{a}_{n}{\left(x - 3\right)}^{n}$ converges at $x=6$ . At which of the following points must the series also converge? Use the fact that if $\displaystyle\sum {a}_{n}{\left(x-c\right)}^{n}$ converges at x, then it converges at any point closer to c than x.

$x=1$
$x=2$
$x=3$
$x=0$
$x=5.99$
$x=0.000001$

Show Solution

It must converge on

(0, 6]

$\left(0,6\right]$ and hence at: a.

x = 1

$x=1$ ; b.

x = 2

$x=2$ ; c.

x = 3

$x=3$ ; d.

x = 0

$x=0$ ; e.

x = 5.99

$x=5.99$ ; and f.

x = 0.000001

$x=0.000001$ .

6. Suppose that $\displaystyle\sum _{n=0}^{\infty }{a}_{n}{\left(x+1\right)}^{n}$ converges at $x=-2$ .
At which of the following points must the series also converge? Use the fact that if $\displaystyle\sum {a}_{n}{\left(x-c\right)}^{n}$ converges at x, then it converges at any point closer to c than x.

$x=2$
$x=-1$
$x=-3$
$x=0$
$x=0.99$
$x=0.000001$

In the following exercises, suppose that $|\frac{{a}_{n+1}}{{a}_{n}}|\to 1$ as $n\to \infty$ . Find the radius of convergence for each series.

7. $\displaystyle\sum _{n=0}^{\infty }{a}_{n}{2}^{n}{x}^{n}$

Show Solution

| \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 |

$|\frac{{a}_{n+1}{2}^{n+1}{x}^{n+1}}{{a}_{n}{2}^{n}{x}^{n}}|=2|x||\frac{{a}_{n+1}}{{a}_{n}}|\to 2|x|$ so

R = \frac{1}{2}

$R=\frac{1}{2}$

\infty \sum n = 0 \frac{a_{n} x^{n}}{2^{n}}

$\displaystyle\sum _{n=0}^{\infty }\frac{{a}_{n}{x}^{n}}{{2}^{n}}$

9. $\displaystyle\sum _{n=0}^{\infty }\frac{{a}_{n}{\pi }^{n}{x}^{n}}{{e}^{n}}$

Show Solution

| \frac{a_{n + 1} {(\frac{π}{e})}^{n + 1} x^{n + 1}}{a_{n} {(\frac{π}{e})}^{n} x^{n}} | = \frac{π | x |}{e} | \frac{a_{n + 1}}{a_{n}} | \to \frac{π | x |}{e}

$|\frac{{a}_{n+1}{\left(\frac{\pi }{e}\right)}^{n+1}{x}^{n+1}}{{a}_{n}{\left(\frac{\pi }{e}\right)}^{n}{x}^{n}}|=\frac{\pi |x|}{e}|\frac{{a}_{n+1}}{{a}_{n}}|\to \frac{\pi |x|}{e}$ so

R = \frac{e}{π}

$R=\frac{e}{\pi }$

10.

\infty \sum n = 0 \frac{a_{n} {(- 1)}^{n} x^{n}}{10^{n}}

$\displaystyle\sum _{n=0}^{\infty }\frac{{a}_{n}{\left(-1\right)}^{n}{x}^{n}}{{10}^{n}}$

11. $\displaystyle\sum _{n=0}^{\infty }{a}_{n}{\left(-1\right)}^{n}{x}^{2n}$

Show Solution

| \frac{a_{n + 1} {(- 1)}^{n + 1} x^{2 n + 2}}{a_{n} {(- 1)}^{n} x^{2 n}} | = | x^{2} | | \frac{a_{n + 1}}{a_{n}} | \to | x^{2} |

$|\frac{{a}_{n+1}{\left(-1\right)}^{n+1}{x}^{2n+2}}{{a}_{n}{\left(-1\right)}^{n}{x}^{2n}}|=|{x}^{2}||\frac{{a}_{n+1}}{{a}_{n}}|\to |{x}^{2}|$ so

R = 1

$R=1$

12.

\infty \sum n = 0 a_{n} {(- 4)}^{n} x^{2 n}

$\displaystyle\sum _{n=0}^{\infty }{a}_{n}{\left(-4\right)}^{n}{x}^{2n}$

In the following exercises, find the radius of convergence R and interval of convergence for $\displaystyle\sum {a}_{n}{x}^{n}$ with the given coefficients ${a}_{n}$ .

13. $\displaystyle\sum _{n=1}^{\infty }\frac{{\left(2x\right)}^{n}}{n}$

Show Solution

a_{n} = \frac{2^{n}}{n}

${a}_{n}=\frac{{2}^{n}}{n}$ so

\frac{a_{n + 1} x}{a_{n}} \to 2 x

$\frac{{a}_{n+1}x}{{a}_{n}}\to 2x$ . so

R = \frac{1}{2}

$R=\frac{1}{2}$ . When

x = \frac{1}{2}

$x=\frac{1}{2}$ the series is harmonic and diverges. When

x = - \frac{1}{2}

$x=-\frac{1}{2}$ the series is alternating harmonic and converges. The interval of convergence is

I = [- \frac{1}{2}, \frac{1}{2})

$I=\left[-\frac{1}{2},\frac{1}{2}\right)$ .

14.

\infty \sum n = 1 {(- 1)}^{n} \frac{x^{n}}{\sqrt{n}}

$\displaystyle\sum _{n=1}^{\infty }{\left(-1\right)}^{n}\frac{{x}^{n}}{\sqrt{n}}$

15. $\displaystyle\sum _{n=1}^{\infty }\frac{n{x}^{n}}{{2}^{n}}$

Show Solution

a_{n} = \frac{n}{2^{n}}

${a}_{n}=\frac{n}{{2}^{n}}$ so

\frac{a_{n + 1} x}{a_{n}} \to \frac{x}{2}

$\frac{{a}_{n+1}x}{{a}_{n}}\to \frac{x}{2}$ so

R = 2

$R=2$ . When

x = \pm 2

$x=\pm2$ the series diverges by the divergence test. The interval of convergence is

I = (- 2, 2)

$I=\left(-2,2\right)$ .

16.

\infty \sum n = 1 \frac{n x^{n}}{e^{n}}

$\displaystyle\sum _{n=1}^{\infty }\frac{n{x}^{n}}{{e}^{n}}$

17. $\displaystyle\sum _{n=1}^{\infty }\frac{{n}^{2}{x}^{n}}{{2}^{n}}$

Show Solution

a_{n} = \frac{n^{2}}{2^{n}}

${a}_{n}=\frac{{n}^{2}}{{2}^{n}}$ so

R = 2

$R=2$ . When

x = \pm

$x=\pm$ the series diverges by the divergence test. The interval of convergence is

I = (- 2, 2)

$I=\left(-2,2\right)$ .

18.

\infty \sum k = 1 \frac{k^{e} x^{k}}{e^{k}}

$\displaystyle\sum _{k=1}^{\infty }\frac{{k}^{e}{x}^{k}}{{e}^{k}}$

19. $\displaystyle\sum _{k=1}^{\infty }\frac{{\pi }^{k}{x}^{k}}{{k}^{\pi }}$

Show Solution

a_{k} = \frac{π^{k}}{k^{π}}

${a}_{k}=\frac{{\pi }^{k}}{{k}^{\pi }}$ so

R = \frac{1}{π}

$R=\frac{1}{\pi }$ . When

x = \pm \frac{1}{π}

$x=\pm\frac{1}{\pi }$ the series is an absolutely convergent p-series. The interval of convergence is

I = [- \frac{1}{π}, \frac{1}{π}]

$I=\left[-\frac{1}{\pi },\frac{1}{\pi }\right]$ .

20.

\infty \sum n = 1 \frac{x^{n}}{n!}

$\displaystyle\sum _{n=1}^{\infty }\frac{{x}^{n}}{n\text{!}}$

21. $\displaystyle\sum _{n=1}^{\infty }\frac{{10}^{n}{x}^{n}}{n\text{!}}$

Show Solution

${a}_{n}=\frac{{10}^{n}}{n\text{!}},\frac{{a}_{n+1}x}{{a}_{n}}=\frac{10x}{n+1}\to 0<1$ so the series converges for all x by the ratio test and

$I=\left(\text{-}\infty ,\infty \right)$ .

22.

$\displaystyle\sum _{n=1}^{\infty }{\left(-1\right)}^{n}\frac{{x}^{n}}{\text{ln}\left(2n\right)}$

In the following exercises, find the radius of convergence of each series.

23. $\displaystyle\sum _{k=1}^{\infty }\frac{{\left(k\text{!}\right)}^{2}{x}^{k}}{\left(2k\right)\text{!}}$

Show Solution

${a}_{k}=\frac{{\left(k\text{!}\right)}^{2}}{\left(2k\right)\text{!}}$ so

$\frac{{a}_{k+1}}{{a}_{k}}=\frac{{\left(k+1\right)}^{2}}{\left(2k+2\right)\left(2k+1\right)}\to \frac{1}{4}$ so

$R=4$

24.

$\displaystyle\sum _{n=1}^{\infty }\frac{\left(2n\right)\text{!}{x}^{n}}{{n}^{2n}}$

25. $\displaystyle\sum _{k=1}^{\infty }\frac{k\text{!}}{1\cdot 3\cdot 5\text{$\cdots$ }\left(2k - 1\right)}{x}^{k}$

Show Solution

${a}_{k}=\frac{k\text{!}}{1\cdot 3\cdot 5\cdots\left(2k - 1\right)}$ so

$\frac{{a}_{k+1}}{{a}_{k}}=\frac{k+1}{2k+1}\to \frac{1}{2}$ so

$R=2$

26.

$\displaystyle\sum _{k=1}^{\infty }\frac{2\cdot 4\cdot 6\text{$\cdots$ }2k}{\left(2k\right)\text{!}}{x}^{k}$

27. $\displaystyle\sum _{n=1}^{\infty }\frac{{x}^{n}}{\left(\begin{array}{c}2n\\ n\end{array}\right)}$ where $\left(\begin{array}{c}n\\ k\end{array}\right)=\frac{n\text{!}}{k\text{!}\left(n-k\right)\text{!}}$

Show Solution

${a}_{n}=\frac{1}{\left(\begin{array}{c}2n\\ n\end{array}\right)}$ so

$\frac{{a}_{n+1}}{{a}_{n}}=\frac{{\left(\left(n+1\right)\text{!}\right)}^{2}}{\left(2n+2\right)\text{!}}\frac{2n\text{!}}{{\left(n\text{!}\right)}^{2}}=\frac{{\left(n+1\right)}^{2}}{\left(2n+2\right)\left(2n+1\right)}\to \frac{1}{4}$ so

$R=4$

28.

$\displaystyle\sum _{n=1}^{\infty }{\sin}^{2}n{x}^{n}$

In the following exercises, use the ratio test to determine the radius of convergence of each series.

29. $\displaystyle\sum _{n=1}^{\infty }\frac{{\left(n\text{!}\right)}^{3}}{\left(3n\right)\text{!}}{x}^{n}$

Show Solution

$\frac{{a}_{n+1}}{{a}_{n}}=\frac{{\left(n+1\right)}^{3}}{\left(3n+3\right)\left(3n+2\right)\left(3n+1\right)}\to \frac{1}{27}$ so

$R=27$

30.

$\displaystyle\sum _{n=1}^{\infty }\frac{{2}^{3n}{\left(n\text{!}\right)}^{3}}{\left(3n\right)\text{!}}{x}^{n}$

31. $\displaystyle\sum _{n=1}^{\infty }\frac{n\text{!}}{{n}^{n}}{x}^{n}$

Show Solution

${a}_{n}=\frac{n\text{!}}{{n}^{n}}$ so

$\frac{{a}_{n+1}}{{a}_{n}}=\frac{\left(n+1\right)\text{!}}{n\text{!}}\frac{{n}^{n}}{{\left(n+1\right)}^{n+1}}={\left(\frac{n}{n+1}\right)}^{n}\to \frac{1}{e}$ so

$R=e$

32.

$\displaystyle\sum _{n=1}^{\infty }\frac{\left(2n\right)\text{!}}{{n}^{2n}}{x}^{n}$

In the following exercises, given that $\frac{1}{1-x}=\displaystyle\sum _{n=0}^{\infty }{x}^{n}$ with convergence in $\left(-1,1\right)$ , find the power series for each function with the given center a, and identify its interval of convergence.

33. $f\left(x\right)=\frac{1}{x};a=1$ (Hint: $\frac{1}{x}=\frac{1}{1-\left(1-x\right)}$ )

Show Solution

$f\left(x\right)=\displaystyle\sum _{n=0}^{\infty }{\left(1-x\right)}^{n}$ on

$I=\left(0,2\right)$

34.

$f\left(x\right)=\frac{1}{1-{x}^{2}};a=0$

35. $f\left(x\right)=\frac{x}{1-{x}^{2}};a=0$

Show Solution

$\displaystyle\sum _{n=0}^{\infty }{x}^{2n+1}$ on

$I=\left(-1,1\right)$

36.

$f\left(x\right)=\frac{1}{1+{x}^{2}};a=0$

37. $f\left(x\right)=\frac{{x}^{2}}{1+{x}^{2}};a=0$

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$\displaystyle\sum _{n=0}^{\infty }{\left(-1\right)}^{n}{x}^{2n+2}$ on

$I=\left(-1,1\right)$

38.

$f\left(x\right)=\frac{1}{2-x};a=1$

39. $f\left(x\right)=\frac{1}{1 - 2x};a=0$ .

Show Solution

$\displaystyle\sum _{n=0}^{\infty }{2}^{n}{x}^{n}$ on

$\left(-\frac{1}{2},\frac{1}{2}\right)$

40.

$f\left(x\right)=\frac{1}{1 - 4{x}^{2}};a=0$

41. $f\left(x\right)=\frac{{x}^{2}}{1 - 4{x}^{2}};a=0$

Show Solution

$\displaystyle\sum _{n=0}^{\infty }{4}^{n}{x}^{2n+2}$ on

$\left(-\frac{1}{2},\frac{1}{2}\right)$

42.

$f\left(x\right)=\frac{{x}^{2}}{5 - 4x+{x}^{2}};a=2$

Use the next exercise to find the radius of convergence of the given series in the subsequent exercises.

43. Explain why, if ${|{a}_{n}|}^{\frac{1}{n}}\to r>0$ , then ${|{a}_{n}{x}^{n}|}^{\frac{1}{n}}\to |x|r<1$ whenever $|x|<\frac{1}{r}$ and, therefore, the radius of convergence of $\displaystyle\sum _{n=1}^{\infty }{a}_{n}{x}^{n}$ is $R=\frac{1}{r}$ .

Show Solution

${|{a}_{n}{x}^{n}|}^{\frac{1}{n}}={|{a}_{n}|}^{\frac{1}{n}}|x|\to |x|r$ as

$n\to \infty$ and

$|x|r<1$ when

$|x|<\frac{1}{r}$ . Therefore,

$\displaystyle\sum _{n=1}^{\infty }{a}_{n}{x}^{n}$ converges when

$|x|<\frac{1}{r}$ by the nth root test.

44.

$\displaystyle\sum _{n=1}^{\infty }\frac{{x}^{n}}{{n}^{n}}$

45. $\displaystyle\sum _{k=1}^{\infty }{\left(\frac{k - 1}{2k+3}\right)}^{k}{x}^{k}$

Show Solution

${a}_{k}={\left(\frac{k - 1}{2k+3}\right)}^{k}$ so

${\left({a}_{k}\right)}^{\frac{1}{k}}\to \frac{1}{2}<1$ so

$R=2$

46.

$\displaystyle\sum _{k=1}^{\infty }{\left(\frac{2{k}^{2}-1}{{k}^{2}+3}\right)}^{k}{x}^{k}$

47. $\displaystyle\sum _{n=1}^{\infty }{a}_{n}={\left({n}^{\frac{1}{n}}-1\right)}^{n}{x}^{n}$

Show Solution

${a}_{n}={\left({n}^{\frac{1}{n}}-1\right)}^{n}$ so

${\left({a}_{n}\right)}^{\frac{1}{n}}\to 0$ so

$R=\infty$

48. Suppose that

$p\left(x\right)=\displaystyle\sum _{n=0}^{\infty }{a}_{n}{x}^{n}$ such that

${a}_{n}=0$ if n is odd. Explain why

$p\left(x\right)=-p\left(\text{-}x\right)$ .

49. Suppose that $p\left(x\right)=\displaystyle\sum _{n=0}^{\infty }{a}_{n}{x}^{n}$ such that ${a}_{n}=0$ if n is even. Explain why $p\left(x\right)=p\left(\text{-}x\right)$ .

Show Solution

We can rewrite

$p\left(x\right)=\displaystyle\sum _{n=0}^{\infty }{a}_{2n+1}{x}^{2n+1}$ and

$p\left(x\right)=p\left(\text{-}x\right)$ since

${x}^{2n+1}=\text{-}{\left(\text{-}x\right)}^{2n+1}$ .

50. Suppose that

$p\left(x\right)=\displaystyle\sum _{n=0}^{\infty }{a}_{n}{x}^{n}$ converges on

$\left(-1,1\right]$ .
Find the interval of convergence of

$p\left(Ax\right)$ .

51. Suppose that $p\left(x\right)=\displaystyle\sum _{n=0}^{\infty }{a}_{n}{x}^{n}$ converges on $\left(-1,1\right]$ . Find the interval of convergence of $p\left(2x - 1\right)$ .

Show Solution

$x\in \left[0,1\right]$ , then

$y=2x - 1\in \left[-1,1\right]$ so

$p\left(2x - 1\right)=p\left(y\right)=\displaystyle\sum _{n=0}^{\infty }{a}_{n}{y}^{n}$ converges.

In the following exercises, suppose that $p\left(x\right)=\displaystyle\sum _{n=0}^{\infty }{a}_{n}{x}^{n}$ satisfies $\underset{n\to \infty }{\text{lim}}\frac{{a}_{n+1}}{{a}_{n}}=1$ where ${a}_{n}\ge 0$ for each n. State whether each series converges on the full interval $\left(-1,1\right)$ , or if there is not enough information to draw a conclusion. Use the comparison test when appropriate.

52.

$\displaystyle\sum _{n=0}^{\infty }{a}_{n}{x}^{2n}$

53. $\displaystyle\sum _{n=0}^{\infty }{a}_{2n}{x}^{2n}$

Show Solution

Converges on

$\left(-1,1\right)$ by the ratio test

54.

$\displaystyle\sum _{n=0}^{\infty }{a}_{2n}{x}^{n}\left(Hint\text{:}x=\pm \sqrt{{x}^{2}}\right)$

55. $\displaystyle\sum _{n=0}^{\infty }{a}_{{n}^{2}}{x}^{{n}^{2}}$ (Hint: Let ${b}_{k}={a}_{k}$ if $k={n}^{2}$ for some n, otherwise ${b}_{k}=0.$ )

Show Solution

Consider the series

$\displaystyle\sum {b}_{k}{x}^{k}$ where

${b}_{k}={a}_{k}$ if

$k={n}^{2}$ and

${b}_{k}=0$ otherwise. Then

${b}_{k}\le {a}_{k}$ and so the series converges on

$\left(-1,1\right)$ by the comparison test.

56. Suppose that

$p\left(x\right)$ is a polynomial of degree N. Find the radius and interval of convergence of

$\displaystyle\sum _{n=1}^{\infty }p\left(n\right){x}^{n}$ .

57. [T] Plot the graphs of $\frac{1}{1-x}$ and of the partial sums ${S}_{N}=\displaystyle\sum _{n=0}^{N}{x}^{n}$ for $n=10,20,30$ on the interval $\left[-0.99,0.99\right]$ . Comment on the approximation of $\frac{1}{1-x}$ by ${S}_{N}$ near $x=-1$ and near $x=1$ as N increases.

Show Solution

This figure is the graph of y = 1/(1-x), which is an increasing curve with vertical asymptote at 1.

The approximation is more accurate near $x=-1$ . The partial sums follow $\frac{1}{1-x}$ more closely as N increases but are never accurate near $x=1$ since the series diverges there.

58. [T] Plot the graphs of

$\text{-}\text{ln}\left(1-x\right)$ and of the partial sums

${S}_{N}=\displaystyle\sum _{n=1}^{N}\frac{{x}^{n}}{n}$ for

$n=10,50,100$ on the interval

$\left[-0.99,0.99\right]$ . Comment on the behavior of the sums near

$x=-1$ and near

$x=1$ as N increases.

59. [T] Plot the graphs of the partial sums ${S}_{n}=\displaystyle\sum _{n=1}^{N}\frac{{x}^{n}}{{n}^{2}}$ for $n=10,50,100$ on the interval $\left[-0.99,0.99\right]$ . Comment on the behavior of the sums near $x=-1$ and near $x=1$ as N increases.

Show Solution

This figure is the graph of y = -ln(1-x) which is an increasing curve passing through the origin.

The approximation appears to stabilize quickly near both $x=\pm 1$ .

60. [T] Plot the graphs of the partial sums

${S}_{N}=\displaystyle\sum _{n=1}^{N}\sin{n}{x}^{n}$ for

$n=10,50,100$ on the interval

$\left[-0.99,0.99\right]$ . Comment on the behavior of the sums near

$x=-1$ and near

$x=1$ as N increases.

61. [T] Plot the graphs of the partial sums ${S}_{N}=\displaystyle\sum _{n=0}^{N}{\left(-1\right)}^{n}\frac{{x}^{2n+1}}{\left(2n+1\right)\text{!}}$ for $n=3,5,10$ on the interval $\left[-2\pi ,2\pi \right]$ . Comment on how these plots approximate $\sin{x}$ as N increases.

Show Solution

This figure is the graph of the partial sums of (-1)^n times x^(2n+1) divided by (2n+1)! For n=3,5,10. The curves approximate the sine curve close to the origin and then separate as the curves move away from the origin.

The polynomial curves have roots close to those of $\sin{x}$ up to their degree and then the polynomials diverge from $\sin{x}$ .

62. [T] Plot the graphs of the partial sums

${S}_{N}=\displaystyle\sum _{n=0}^{N}{\left(-1\right)}^{n}\frac{{x}^{2n}}{\left(2n\right)\text{!}}$ for

$n=3,5,10$ on the interval

$\left[-2\pi ,2\pi \right]$ . Comment on how these plots approximate

$\cos{x}$ as N increases.

Power Series: Problem Sets

Problem Set: Power Series and Functions

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