Question

Find all values of x for which the series $\sum _{k=1}^{\infty }\frac{x^{2k}}{k^2}$converges.

Short answer

Series$\sum _{k=1}^{\infty }\frac{x^{2k}}{k^2}$converses when$x\in (-1,1).$

Step-by-step solution

Step 1. Given information.

The given series is$\sum _{k=1}^{\infty }\frac{x^{2k}}{k^2}.$

Step 2. Value of x.

Use the ratio test and Find the value of $\rho =\underset{k\to \infty }{\lim }\frac{a_{k+1}}{a_k}$

role="math" localid="1649196887689" $$\begin{gathered} \rho =\underset{k\to \infty }{\lim }\frac{\left(\frac{x^{2(k+1)}}{{\left(k+1\right)}^2}\right)}{\left(\frac{x^{2k}}{k^2}\right)} \\ =\underset{k\to \infty }{\lim }\frac{x^{2k}·x^2·k^2}{{\left(k+1\right)}^2·x^{2k}} \\ =x^2·\underset{k\to \infty }{\lim }{\left(\frac{k}{k+1}\right)}^2 \\ =x^2 \end{gathered}$$

According to the ratio test, series will converge when $\rho <1.$

role="math" localid="1649197012614" $$\begin{gathered} x^2<1 \\ -1<x<1 \end{gathered}$$

So series converses whenrole="math" localid="1649197120658" $x\in (-1,1)$