Question

Find the interval of convergence of the power series

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

Short answer

The interval of convergence of the power series$\sum _{k=1}^{\infty }\frac{(-1{)}^k}{k{2}^k}(x-2{)}^k$is$(0,4)$

Step-by-step solution

Given information 

The power series is$\sum _{k=1}^{\infty }\frac{(-1{)}^k}{k{2}^k}(x-2{)}^k$

The ratio test for absolute convergence will be used to determine the convergence interval.

Let, the first assume $b_k=\frac{(-1{)}^k}{k{2}^k}(x-2{)}^k$

therefore $b_{k+1}=\frac{(-1{)}^{k+1}}{(k+1)2^{k+1}}(x-2{)}^{k+1}$

Implies that,

$$\begin{gathered} \underset{k\to \infty }{\lim }\left|\frac{b_{k+1}}{b_k}\right|=\underset{k\to \infty }{\lim }\left|\frac{{\displaystyle \frac{{(-1)}^{k+1}}{(k+{12}^{k+1}}}{(x-2)}^{k+1}}{{\displaystyle \frac{{(-1)}^k}{k{2}^k}}{(x-2)}^k}\right| \\ =\underset{k\to \infty }{\lim }\frac{1}{2}\frac{k}{{(k+1)}^k}\left|x-2\right| \end{gathered}$$

The limit is $\frac{1}{2}|x-2|$

The ratio test for absolute convergence will be used to determine the convergence interval when $\frac{1}{2}|x-2|<1$that is $-2<x-2<2$

As a result, we may write $-2<x-2andx-2<2$

Implies that

$x>0$and $x<4$

or$x\in (0,4)$

Now, because the intervals are limited, we examine the series' behavior at the ends.

When $x=0$

$$\begin{gathered} {\left.\sum _{k=1}^{\infty }\frac{(-1{)}^k}{k{2}^k}(x-2{)}^k\right|}_{x=0}=\sum _{k=1}^{\infty }\frac{(-1{)}^k}{k{2}^k}(0-2{)}^k \\ =\sum _{k=1}^{\infty }\frac{(-1{)}^k}{k{2}^k}(-1{)}^k{2}^k \\ =\sum _{k=1}^{\infty }\frac{1}{k}(-1{)}^{2k} \\ =\sum _{k=1}^{\infty }\frac{1}{k} \end{gathered}$$

The series contains the diverging alternating multiple.

So, when $x=4$

The alternating multiple is the outcome, and it converges.

The interval of convergence of the power series

The interval of convergence of the power series$\sum _{k=1}^{\infty }\frac{(-1{)}^k}{k{2}^k}(x-2{)}^k$is$(0,4)$