Closed form for the series $\sum_{n=1}^{\infty}‎\frac{\pi^n}{n!n^p}z^n$.

Question

I know that the following series $$\sum_{n=1}^{\infty}‎\frac{\pi^n}{n!n^p}z^n ;\;z\in\mathbb{C},$$ is convergent by applying the ratio test. Now I wanted to know that Does the series have any closed form. Anyone can help me? Thanks. Do attention that $p‎>‎1$.

Answer 1

For $p=0$, it is $e^{\pi z}-1$, since $$ f_0(z) := \sum_{n=1}^\infty \frac{z^n}{n!} = e^{z}-1 $$ For $p=1$, take $f_0/z$ and integrate $$ f_1(z) := \int_0^z \frac{f_0(t)}{t}\;dt = \sum_{n=1}^\infty \frac{z^n}{n! n} =\mathrm{Ei}(z) - \gamma - \log 2 $$ an "exponential integral" function, and yours is $f_1(\pi z)$.

For $p=2$ repeat, $$ f_2(z) = \int_0^z\frac{f_1(t)}{t}\;dt =\mathrm{Ei}\left( z \right)-\log \left( z \right)+z-{e^{z}}- \gamma\,z+1 $$ and yours is $f_2(\pi z)$. Continue $$ f_3(z) = \int_0^z\frac{f_2(t)}{t}\;dt $$ still more complicated expression involving $\mathrm{Ei}(z)$.

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As Claude noted, these may also be written in terms of generalized hypergeometric functions (but this is little more than the definition):

$$ f_p(z) = \sum_{n=1}^\infty \frac{z^n}{n!\,n^p} = z\;{}_{p+1}F_{p+1}\! \left(\begin{align}1,1,\cdots,1\\2,2,\cdots,2\end{align};z\right) $$