Answer
$-\ln (1-x^2)$
Work Step by Step
The power series of $(1-x)$ can be written as:
$-x-\dfrac{x^2}{2}-\dfrac{x^3}{3}+......=-\Sigma_{k=1}^{\infty} \dfrac{x^k}{k}$
Multiply the above equation by $-1$ and replace $x$ by $x^2$ in the above series to obtain:
$(x^2+\dfrac{(x^2)^2}{2}+\dfrac{(x^2)^3}{3}+......=\Sigma_{k=0}^{\infty} \dfrac{x^{2k}}{k}$
Therefore, the function represented by the power series is:
$\Sigma_{k=0}^{\infty} \dfrac{x^{2k}}{k}=-\ln (1-x^2)$
