91Ó°ÊÓ

The power series is$\underset{k=1}{\overset{∞}{∑}}\frac{(-1{)}^k}{k{2}^k}(x-2{)}^k$

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,

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

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$x∈(0,4)$

When $x=0$

$$\begin{gathered} {\left.\underset{k=1}{\overset{∞}{∑}}\frac{(-1{)}^k}{k{2}^k}(x-2{)}^k\right|}_{x=0}=\underset{k=1}{\overset{∞}{∑}}\frac{(-1{)}^k}{k{2}^k}(0-2{)}^k \\ =\underset{k=1}{\overset{∞}{∑}}\frac{(-1{)}^k}{k{2}^k}(-1{)}^k{2}^k \\ =\underset{k=1}{\overset{∞}{∑}}\frac{1}{k}(-1{)}^{2k} \\ =\underset{k=1}{\overset{∞}{∑}}\frac{1}{k} \end{gathered}$$

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$\underset{k=1}{\overset{∞}{∑}}\frac{(-1{)}^k}{k{2}^k}(x-2{)}^k$is$(0,4)$