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