Chapter 10.5, Problem 61E

To determine

To find : the value of $x$ for which the series converges.

Answer to Problem 61E

The value of $x$ for all original series converges are: $-\frac{1}{2}\le x<\frac{1}{2}$

Explanation of Solution

Given information:

Given expression is : ${\displaystyle \sum }_{k=0}^{\infty }\frac{2^k{x}^k}{\ln (k+2)}$ .

Calculation:

The given series is

  ${\displaystyle \sum }_{k=0}^{\infty }\frac{2^k{x}^k}{\ln (k+2)}$

The series of absolute values is

  ${\displaystyle \sum }_{k=0}^{\infty }\left|a_k\right|=\frac{2^k|x{|}^k}{\ln (k+2)}$

The ratio Test Let $\sum a_n$ be a series with positive terms, and with

  $\underset{n\to \infty }{\lim }\frac{a_{m+1}}{a_n}=L$

Then, the series converges if $L<1,$ the series diverges if $L>1$ and the test is inconclusive if

  $L=1$

Using the ratio test, we check for absolute convergence as follows:

  $\begin{array}{cc}\underset{n\to \infty }{\lim }\left|\frac{a_{k+1}}{a_k}\right|& =\underset{n\to \infty }{\lim }\frac{2^{k+1}|x{|}^{k+1}}{\ln (k+3)}\times \frac{\ln (k+2)}{2^k|x{|}^k}\\ & =2\left|x\right|\end{array}$

The series converges absolutely for

  $2\left|x\right|<1$

  $\left|x\right|<\frac{1}{2}$

  $-\frac{1}{2}<x<\frac{1}{2}$

And diverges for

  $2\left|x\right|>1$

  $\left|x\right|>\frac{1}{2}$

And when $x=-\frac{1}{2},$ the series is

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

This converges by the Altemating Series Test.

And when $x=\frac{1}{2},$ the series is

  ${\displaystyle \sum }_{k=0}^{\infty }\frac{2^k{x}^k}{\ln (k+2)}={\displaystyle \sum }_{k=0}^{\infty }\frac{{(1)}^k}{\ln (k+2)}$

This diverges by the Direct Comparison Test, since $\frac{1}{\ln (k+2)}>\frac{1}{k}$ for all $k\ge 2,$ and ${\displaystyle \sum }_{k=1}^{\infty }\frac{1}{k}$ diverges (harmonic).

The original series converges for $-\frac{1}{2}\le x<\frac{1}{2}$