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