Answer
$(1+\dfrac{x}{3})^{-1}=\dfrac{3}{3+x}$
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 $\dfrac{x}{3}$ in the above series to obtain:
$\Sigma_{k=0}^{\infty} (-1)^k \dfrac{x^k}{3^k}= (1+\dfrac{x}{3})^{-1}$
Therefore, the function represented by the power series is:
$(1+\dfrac{x}{3})^{-1}=\dfrac{3}{3+x}$
