Answer
$\Sigma_{n=0}^\infty (\dfrac{1}{n!}+(-1)^n)x^n$
Work Step by Step
Consider the Maclaurin Series for $e^x$ and $\dfrac{1}{1+x}$ is defined as:
$e^x=1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+...$;
$\dfrac{1}{1+x}=(1-x+x^2-x^3+x^4+...)$
Then $e^x+\dfrac{1}{1+x}=[1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+...]+[1-x+x^2-x^3+x^4+...]$
$\implies (1+1)x^0+(1-1)x^1+(\dfrac{1}{2!}+1)x^2+...=\Sigma_{n=0}^\infty (\dfrac{1}{n!}+(-1)^n)x^n$
