Answer
$\dfrac{4}{4+x^2}$
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 $\dfrac{x^2}{4}$ in the above series to obtain:
$\Sigma_{k=0}^{\infty} (-1)^k \dfrac{x^{2k}}{4^k}= (1+\dfrac{x^2}{4})^{-1}=\dfrac{4}{4+x^2}$
Therefore, the function represented by the power series is:
$ (1+\dfrac{x^2}{4})^{-1}=\dfrac{4}{4+x^2}$
