Answer
$\dfrac{4x}{4+x}$
Work Step by Step
The power series of $(1+x)^{-1}$ can be written as:
$1-x+x^2-x^3+......=\Sigma_{k=0}^{\infty} (-1)^k x^k$
Replace $x$ by $\dfrac{x}{4}$ in the above series to obtain:
$\Sigma_{k=0}^{\infty} (-1)^k \dfrac{x^{2k+1}}{4^k}=x (1+\dfrac{x}{4})^{-1}$
Therefore, the function represented by the power series is:
$x (1+\dfrac{x}{4})^{-1}=\dfrac{4x}{4+x}$
