Processing math: 100%
Skip to main content
Library homepage
 

Text Color

Text Size

 

Margin Size

 

Font Type

Enable Dyslexic Font
Mathematics LibreTexts

5.1E: Exercises

( \newcommand{\kernel}{\mathrm{null}\,}\)

Exercise 5.1E.1

In exercises 1 - 4, state whether each statement is true, or give an example to show that it is false.

1) If n=1anxn converges, then anxn0 as n.

Answer:
True. If a series converges then its terms tend to zero.

2) n=1anxn converges at x=0 for any real numbers an.

3) Given any sequence an, there is always some R>0, possibly very small, such that n=1anxn converges on (R,R).

Answer:
False. It would imply that anxn0 for |x|<R. If an=nn, then anxn=(nx)n does not tend to zero for any x0.

4) If n=1anxn has radius of convergence R>0 and if |bn||an| for all n, then the radius of convergence of n=1bnxn is greater than or equal to R.

 

Exercise 5.1E.2

5) Suppose that n=0an(x3)n converges at x=6. At which of the following points must the series also converge? Use the fact that if an(xc)n converges at x, then it converges at any point closer to c than x.

a. x=1

b. x=2

c. x=3

d. x=0

e. x=5.99

f. x=0.000001

Answer:
It must converge on (0,6] and hence at: a. x=1; b. x=2; c. x=3; d. x=0; e. x=5.99; and f. x=0.000001.

6) Suppose that n=0an(x+1)n converges at x=2. At which of the following points must the series also converge? Use the fact that if an(xc)n converges at x, then it converges at any point closer to c than x.

a. x=2

b. x=1

c. x=3

d. x=0

e. x=0.99

f. x=0.000001

Answer

Add texts here. Do not delete this text first.

Exercise 5.1E.3

In the following exercises, suppose that |an+1an|1 as n. Find the radius of convergence for each series.

7) n=0an2nxn

Answer:
|an+12n+1xn+1an2nxn|=2|x||an+1an|2|x| so R=12

8) n=0anxn2n

9) n=0anπnxnen

Answer:
|an+1(πe)n+1xn+1an(πe)nxn|=π|x|e|an+1an|π|x|e so R=eπ

10) n=0an(1)nxn10n

11) n=0an(1)nx2n

Answer:
|an+1(1)n+1x2n+2an(1)nx2n|=|x2||an+1an||x2| so R=1

12) n=0an(4)nx2n

Answer

Add texts here. Do not delete this text first.

Exercise 5.1E.4

In exercises 13 - 22, find the radius of convergence R and interval of convergence for anxn with the given coefficients an.

13) n=1(2x)nn

Answer:
an=2nn so an+1xan2x. so R=12. When x=12 the series is harmonic and diverges. When x=12 the series is alternating harmonic and converges. The interval of convergence is I=[12,12).

14) n=1(1)nxnn

15) n=1nxn2n

Answer:
an=n2n so an+1xanx2 so R=2. When x=±2 the series diverges by the divergence test. The interval of convergence is I=(2,2).

16) n=1nxnen

17) n=1n2xn2n

Answer:
an=n22n so R=2. When x=±2 the series diverges by the divergence test. The interval of convergence is I=(2,2).

18) k=1kexkek

19) k=1πkxkkπ

Answer:
ak=πkkπ so R=1π. When x=±1π the series is an absolutely convergent p-series. The interval of convergence is I=[1π,1π].

20) n=1xnn!

21) n=110nxnn!

Answer:
an=10nn!,an+1xan=10xn+10<1 so the series converges for all x by the ratio test and I=(,).

22) n=1(1)nxnln(2n)

Answer

Add texts here. Do not delete this text first.

Exercise 5.1E.5

In exercises 23 - 28, find the radius of convergence of each series.

23) k=1(k!)2xk(2k)!

Answer:
ak=(k!)2(2k)! so ak+1ak=(k+1)2(2k+2)(2k+1)14 so R=4

24) n=1(2n)!xnn2n

25) k=1k!135(2k1)xk

Answer:
ak=k!135(2k1) so ak+1ak=k+12k+112 so R=2

26) k=12462k(2k)!xk

27) n=1xn(2nn) where (nk)=n!k!(nk)!

Answer:
an=1(2nn) so an+1an=((n+1)!)2(2n+2)!2n!(n!)2=(n+1)2(2n+2)(2n+1)14 so R=4

28) n=1sin2nxn

Answer

Add texts here. Do not delete this text first.

Exercise 5.1E.6

In exercises 29 - 32, use the ratio test to determine the radius of convergence of each series.

29) n=1(n!)3(3n)!xn

Answer:
an+1an=(n+1)3(3n+3)(3n+2)(3n+1)127 so R=27

30) n=123n(n!)3(3n)!xn

31) n=1n!nnxn

Answer:
an=n!nn so an+1an=(n+1)!n!nn(n+1)n+1=(nn+1)n1e so R=e

32) n=1(2n)!n2nxn

Answer

Add texts here. Do not delete this text first.

Exercise 5.1E.7

In the following exercises, given that 11x=n=0xn with convergence in (1,1), find the power series for each function with the given center a, and identify its interval of convergence.

33) f(x)=1x;a=1 (Hint: 1x=11(1x))

Answer:
f(x)=n=0(1x)n on I=(0,2)

34) f(x)=11x2;a=0

35) f(x)=x1x2;a=0

Answer:
n=0x2n+1 on I=(1,1)

36) f(x)=11+x2;a=0

37) f(x)=x21+x2;a=0

Answer:
n=0(1)nx2n+2 on I=(1,1)

38) f(x)=12x;a=1

39) f(x)=112x;a=0.

Answer:
n=02nxn on (12,12)

40) f(x)=114x2;a=0

41) f(x)=x214x2;a=0

Answer:
n=04nx2n+2 on (12,12)

42) f(x)=x254x+x2;a=2

Answer

Add texts here. Do not delete this text first.

Exercise 5.1E.8

Use the result of exercise 43 to find the radius of convergence of the given series in the subsequent exercises (44 - 47).

43) Explain why, if |an|1/nr>0, then |anxn|1/n|x|r<1 whenever |x|<1r and, therefore, the radius of convergence of n=1anxn is R=1r.

Answer:
|anxn|1/n=|an|1/n|x||x|r as n and |x|r<1 when |x|<1r. Therefore, n=1anxn converges when |x|<1r by the nth root test.

44) n=1xnnn

45) k=1(k12k+3)kxk

Answer:
ak=(k12k+3)k so (ak)1/k12<1 so R=2

46) k=1(2k21k2+3)kxk

47) n=1an=(n1/n1)nxn

Answer:
an=(n1/n1)n so (an)1/n0 so R=
Answer

Add texts here. Do not delete this text first.

Exercise 5.1E.9

48) Suppose that p(x)=n=0anxn such that an=0 if n is even. Explain why p(x)=p(x).

49) Suppose that p(x)=n=0anxn such that an=0 if n is odd. Explain why p(x)=p(x).

Answer:
We can rewrite p(x)=n=0a2n+1x2n+1 and p(x)=p(x) since x2n+1=(x)2n+1.

50) Suppose that p(x)=n=0anxn converges on (1,1]. Find the interval of convergence of p(Ax).

51) Suppose that p(x)=n=0anxn converges on (1,1]. Find the interval of convergence of p(2x1).

Answer:
If x[0,1], then y=2x1[1,1] so p(2x1)=p(y)=n=0anyn converges.
Answer

Add texts here. Do not delete this text first.

Exercise 5.1E.10

In the following exercises, suppose that p(x)=n=0anxn satisfies limnan+1an=1 where an0 for each n. State whether each series converges on the full interval (1,1), or if there is not enough information to draw a conclusion. Use the comparison test when appropriate.

52) n=0anx2n

53) n=0a2nx2n

Answer:
Converges on (1,1) by the ratio test

54) n=0a2nxn (Hint:x=±x2)

55) n=0an2xn2 (Hint: Let bk=ak if k=n2 for some n, otherwise bk=0.)

Answer:
Consider the series bkxk where bk=ak if k=n2 and bk=0 otherwise. Then bkak and so the series converges on (1,1) by the comparison test.
Answer

Add texts here. Do not delete this text first.

Exercise 5.1E.11

56) Suppose that p(x) is a polynomial of degree N. Find the radius and interval of convergence of n=1p(n)xn.

57) [T] Plot the graphs of 11x and of the partial sums SN=Nn=0xn for n=10,20,30 on the interval [0.99,0.99]. Comment on the approximation of 11x by SN near x=1 and near x=1 as N increases.

Answer:

The approximation is more accurate near x=1. The partial sums follow 11x more closely as N increases but are never accurate near x=1 since the series diverges there.

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

58) [T] Plot the graphs of ln(1x) and of the partial sums SN=Nn=1xnn for n=10,50,100 on the interval [0.99,0.99]. 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 Sn=Nn=1xnn2 for n=10,50,100 on the interval [0.99,0.99]. Comment on the behavior of the sums near x=1 and near x=1 as N increases.

Answer:

The approximation appears to stabilize quickly near both x=±1.

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

60) [T] Plot the graphs of the partial sums SN=Nn=1(sinn)xn for n=10,50,100 on the interval [0.99,0.99]. 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 SN=Nn=0(1)nx2n+1(2n+1)! for n=3,5,10 on the interval [2π,2π]. Comment on how these plots approximate sinx as N increases.

Answer:

The polynomial curves have roots close to those of sinx up to their degree and then the polynomials diverge from sinx.

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.

62) [T] Plot the graphs of the partial sums SN=Nn=0(1)nx2n(2n)! for n=3,5,10 on the interval [2π,2π]. Comment on how these plots approximate cosx as N increases.

 

Answer

Add texts here. Do not delete this text first.

 

Contributors

Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org.


5.1E: Exercises is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.

Support Center

How can we help?