Series to functions Find the function represented by the following series, and find the interval of convergence of the series. (Not all these series are power series.) sum_{k=1}^inftyfrac{x^{2k}}{4^k}

We know that $\sum _{k=0}^{\mathrm{\infty }}{x}^k=\frac{1}{1-x}$
Given function is $\sum _{k=0}^{\mathrm{\infty }}\frac{x^{2k}}{4^k}$
It can be written as $\sum _{k=0}^{\mathrm{\infty }}\frac{x^{2k}}{4^k}=\sum _{k=0}^{\mathrm{\infty }}(\frac{x^2}{4}{)}^k$
Comparing this with equation, we may write
$\sum _{k=0}^{\mathrm{\infty }}(\frac{x^2}{4}{)}^k=\frac{1}{1-\frac{x^2}{4}}$
$=\frac{4}{4-x^2}$ Now, for the function to be convergent,
$|\frac{x^2}{4}|<1$
$\Rightarrow |x^2|<4$
So the solution is -2
Therefore, the given series can be represented as $\frac{4}{4-x^2}$ and the interval of convergence is $(-2,2)$