Is there an expression using the main constants of mathematics as result of the following infinite sum:

Question

$$\sum_{k=0}^\infty {{\pi^{k\over 2}}\over {\Gamma({k\over 2} +1)}}$$

I've found, that

$\sum_{k=0}^\infty {{\pi^{k\over 2}}\over {\Gamma({k\over 2} +1)}}$ = $e^{\pi} + 2\sum_{k=0}^\infty {{({4\pi})^{k}k!}\over {(2k+1)!}}$

But does that help?

Answer 1

You can have a closed form in terms of the $\rm {erf}$ function

$$ \operatorname{erf}(x) = \frac{2}{\sqrt{\pi}}\int_{0}^x e^{-t^2}\,\mathrm dt, $$

as

$$ {{\rm e}^{\pi }} \left( 1+{\rm erf} \left( \sqrt { \pi } \right) \right) .$$