Chapter 9.5, Problem 61E

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To determine

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

Answer to Problem 61E

The values of $x$ for which the series converges is $-\frac{1}{2}\le x<\frac{1}{2}$ .

Explanation of Solution

Given information:

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

Calculation:

By applying the Ratio Test and looking at the interval’s endpoints, now identifying the convergence interval.

Taking the absolute value of the terms in the series and looking at the following limit, only using the Ratio Test for series with non-negative terms.

  $\begin{array}{c}\underset{k\to \infty }{\lim }\frac{\left|\frac{2^{k+1}{x}^{k+1}}{\ln \left(k+1+2\right)}\right|}{\frac{2^k{x}^k}{\ln \left(k+2\right)}}=\underset{k\to \infty }{\lim }\left(\frac{2^{k+1}{\left|x\right|}^{k+1}}{\ln \left(k+3\right)}\cdot \frac{\ln \left(k+2\right)}{2^k{\left|x\right|}^k}\right)\\ =\underset{k\to \infty }{\lim }\left(\frac{\ln \left(k+2\right)}{\ln \left(k+3\right)}\cdot 2\left|x\right|\right)\\ =2\left|x\right|\end{array}$

The series converges if $2\left|x\right|<1$ , therefore $-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.<x<\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ .

To test the convergence at the endpoints,

At left endpoints $x=-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ ,

  ${\displaystyle \sum {\left(-1\right)}^k\frac{1}{\ln \left(k+2\right)}}$

To determine whether the given series converges or not by using Alternating Series Test,

The series satisfies the conditions of the test since $u=\frac{1}{\ln \left(k+2\right)}$ is positive for every $k$ , decreasing as $k$ increases, and approaching zero as $k$ approaches infinity. As a result, the series converges (conditionally) at $x=\frac{1}{2}$ .

At right endpoints $x=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ ,

  ${\displaystyle \sum \frac{1}{\ln \left(k+2\right)}}$

To determine whether the given series converges or not by using Direct comparison Test,

Since $k\ge \ln \left(+2\right)$ for every $k\ge 2$ , to get $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$k$}\right.\le \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\ln \left(k+2\right)$}\right.$ .

  ${\displaystyle \sum _{k=1}^{\infty }\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$k$}\right.}\le {\displaystyle \sum _{k=1}^{\infty }\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\ln \left(k+2\right)$}\right.}$

Obtaining a divergent lower bound for the given series, the Direct Comparison Test indicates that the series diverges at $x=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ .

From the above steps, it is known that the series diverges if $-\frac{1}{2}\le x<\frac{1}{2}$ .

Therefore, the values of $x$ for which the series converges is $-\frac{1}{2}\le x<\frac{1}{2}$ .