Chapter 10, Problem 12RE

(a)

To determine

The radius of convergence of the series ${\displaystyle \sum _{n=0}^{\infty }\frac{{\left(-1\right)}^n{\left(x-1\right)}^2{}^{n+1}}{2n+1}}$

Answer to Problem 12RE

The radius of convergence for is 1. $R=1$

Explanation of Solution

Given information:

The given series is,

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

Formula used:

Ratio test is used.

Calculation:

Check for absolute convergence using ratio test. $\underset{n\to \infty }{\lim }\frac{\left|a_{n+1}\right|}{\left|a_n\right|}=\underset{n\to \infty }{\lim }\frac{\left|{\left(-1\right)}^{n+1}{\left(x-1\right)}^{2n+3}\right|}{2n+3}.\frac{2n+1}{\left|\left(-1\right){\left(x-1\right)}^{2n-1}\right|}$

  $\begin{array}{l}=\underset{n\to \infty }{\lim }\frac{\left(2n+1\right)}{\left(2n+3\right)}\left|{\left(x-1\right)}^2\right|\\ =\left|{\left(x-1\right)}^2\right|\end{array}$

Hence,

  $\left|{\left(x-1\right)}^2\right|<1$

Therefore,

  $\left|x-1\right|<1$

That gives

  $0<x<\mathrm{2..............}\left(1\right)$

Checking convergence at end points At $x=0$

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

  ${\displaystyle \sum _{n=0}^{\infty }\frac{{\left(-1\right)}^n}{2n+1}}\text{converges conditionally by the alternating series test}\mathrm{........}\left(2\right)$

Since,

From equation (1) $\left|x-1\right|<1\text{if}\left|x-1\right|<1$ then the series converges absolutely and hence convergence. Therefore, the radius of convergence for is 1. $R=1$

Conclusion:

The radius of convergence for is 1. $R=1$

(b)

To determine

The interval of convergence of the series ${\displaystyle \sum _{n=0}^{\infty }\frac{{\left(-1\right)}^n{\left(x-1\right)}^2{}^{n+1}}{2n+1}}$

Answer to Problem 12RE

The series always converges for interval $\left[0,2\right]$

Explanation of Solution

Given information:

The given series is,

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

Formula used:

Ratio test is used.

Calculation:

Since,

  $\underset{n\to \infty }{\lim }\frac{\left|a_{n+1}\right|}{\left|a_n\right|}=\left|\left(x-1\right)\right|$

From equation (1) and (2) as shown below,

  $0\le x\le 2$

Therefore,

The series always converges for interval $\left[0,2\right]$

Conclusion:

The series always converges for interval $\left[0,2\right]$

(c)

To determine

The series ${\displaystyle \sum _{n=0}^{\infty }\frac{{\left(-1\right)}^n{\left(x-1\right)}^2{}^{n+1}}{2n+1}}$ converges absolutely.

Answer to Problem 12RE

The series always converges for interval $\left[0,2\right]$

Explanation of Solution

Given information:

The given series is,

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

Formula used:

Ratio test is used.

Calculation:

Since,

  $\underset{n\to \infty }{\lim }\frac{\left|a_{n+1}\right|}{\left|a_n\right|}=\left|\left(x-1\right)\right|$

From equation (1) and (2) as shown below,

  $0\le x\le 2$

Therefore,

The series always converges for interval $\left[0,2\right]$

Conclusion:

The series always converges for interval $\left[0,2\right]$

(d)

To determine

The series ${\displaystyle \sum _{n=0}^{\infty }\frac{{\left(-1\right)}^n{\left(x-1\right)}^2{}^{n+1}}{2n+1}}$ converges conditionally.

Answer to Problem 12RE

The points of conditional convergence are at x = 0 and at x = 2.

Explanation of Solution

Given information:

The given series is,

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

Formula used:

Ratio test is used.

Calculation:

The points of conditional convergence are at x = 0 and at x = 2. Since, the rearrangement of the terms affects the convergence of the series.

Conclusion:

The points of conditional convergence are at x = 0 and at x = 2.