Chapter 10.5, Problem 47E

a.

To determine

To find: the interval of convergence of the series.

Answer to Problem 47E

Interval of convergence is $\left(\frac{1}{2},\frac{3}{2}\right)$ .

Explanation of Solution

Given:

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

Formula used:

Power series about $x=a$ , is given by ${\displaystyle \sum _{n=0}^{\infty }{c}_n{(x-a)}^n}$

If $\underset{n\to \infty }{\lim }\left|\frac{u_{n+1}}{u_n}\right|<1$ , then the series converges. And if $\underset{n\to \infty }{\lim }\left|\frac{u_{n+1}}{u_n}\right|>1$ , then the series diverges.

For convergent series the limit will result in $\frac{|x-a|}{R}<1$ , where R is the radius of convergence and $|x-a|<R$ will give the interval of convergence.

Calculation:

Applying limit we get,

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

As $\underset{n\to \infty }{\lim }\left|\frac{u_{n+1}}{u_n}\right|<1$ , thus the series converges.

The radius of convergence is $R=\frac{1}{2}$ .

Interval of convergence will be: $|x-1|<\frac{1}{2}$

  $\begin{array}{l}-\frac{1}{2}<x-1<\frac{1}{2}\\ -\frac{1}{2}+1<x<\frac{1}{2}+1\\ \frac{1}{2}<x<\frac{3}{2}\end{array}$

The interval of convergence is $\left(\frac{1}{2},\frac{3}{2}\right)$ .

b.

To determine

To find: Values of x for which the series converges absolutely.

Answer to Problem 47E

The series converges absolutely for interval $\left(\frac{1}{2},\frac{3}{2}\right)$ .

Explanation of Solution

Given:

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

Formula used:

Power series about $x=a$ , is given by ${\displaystyle \sum _{n=0}^{\infty }{c}_n{(x-a)}^n}$

For convergent series the limit will result in $\frac{|x-a|}{R}<1$ , where R is the radius of convergence and $|x-a|<R$ will give the interval of convergence.

The series converges absolutely for the values of x in the interval of convergence.

Calculation:

From part a, the interval of convergence is $\left(\frac{1}{2},\frac{3}{2}\right)$ .

Thus the series converges absolutely for values of x belonging in the interval $\left(\frac{1}{2},\frac{3}{2}\right)$ .

c.

To determine

To find: Values of x for which the series converges conditionally.

Answer to Problem 47E

The series does not converge conditionally at any point.

Explanation of Solution

Given:

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

Formula used:

Power series about $x=a$ , is given by ${\displaystyle \sum _{n=0}^{\infty }{c}_n{(x-a)}^n}$

For convergent series the limit will result in $\frac{|x-a|}{R}<1$ , where R is the radius of convergence and $|x-a|<R$ will give the interval of convergence.

The series converges at end points then the series converges conditionally at that points.

Calculation:

Test the end points of the interval of convergence,

At $x=\frac{1}{2}$

  $\begin{array}{l}{\displaystyle \sum _{n=0}^{\infty }{(-2)}^n(n+1){(\frac{1}{2}-1)}^n}\\ {\displaystyle \sum _{n=0}^{\infty }{(-2)}^n(n+1){(-\frac{1}{2})}^n}\\ {\displaystyle \sum _{n=0}^{\infty }{(-2\times -\frac{1}{2})}^n(n+1)}\\ {\displaystyle \sum _{n=0}^{\infty }(n+1)}\end{array}$

The series diverges.

At $x=\frac{3}{2}$

  $\begin{array}{l}{\displaystyle \sum _{n=0}^{\infty }{(-2)}^n(n+1){(\frac{3}{2}-1)}^n}\\ {\displaystyle \sum _{n=0}^{\infty }{(-2)}^n(n+1){(\frac{1}{2})}^n}\\ {\displaystyle \sum _{n=0}^{\infty }{(-2\times \frac{1}{2})}^n(n+1)}\\ {\displaystyle \sum _{n=0}^{\infty }{(-1)}^n(n+1)}\end{array}$

The series diverges.

Therefore, the series does not converge conditionally anywhere.