Answer
$\Sigma_{n=0}^\infty \dfrac{(\pi)^n x^n}{2^nn!}$
Work Step by Step
Since, we know that the Taylor Series for $e^x$ is defined as:
$e^x=\Sigma_{n=0}^\infty \dfrac{x^n}{n!}=1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+...$
Now, $e^{(\frac{\pi x}{2})}=\Sigma_{n=0}^\infty \dfrac{(\pi x/2)^n}{n!}=1+(\pi x/2)+\dfrac{(\pi x/2)^2}{2!}+\dfrac{(\pi x/2)^3}{3!}+\dfrac{(\pi x/2)^4}{4!}+...$
or, $\Sigma_{n=0}^\infty \dfrac{(\pi)^n x^n}{2^nn!}$
