Prove the following series $\sum\limits_{s=0}^\infty \frac{1}{(sn)!}$ Prove that, $$\sum\limits_{s=0}^\infty \frac{1}{(sn)!}=\frac{1}{n}\sum\limits_{r=0}^{n-1}\exp\left(\cos\frac{2r\pi}{n}\right)\cos\left(\sin\frac{2r\pi}{n}\right)$$ 
I don't have a real idea on how to start approaching this question, some hints and suggestions would be helpful. 
 A: It's $\sum\limits_{r=0}^{n-1} e^{i\frac{2\pi k}{n}r}=\frac{e^{i\frac{2\pi k}{n}n}-1}{e^{i\frac{2\pi k}{n}}-1}$ with $=0$ for $k\neq l\cdot n$ and $=n$ for $k=l\cdot n$, $l\in\mathbb{Z}$.
It follows $\sum\limits_{r=0}^{n-1} e^{e^{i\frac{2\pi}{n}r}}=\sum\limits_{k=0}^\infty\frac{1}{k!}\sum\limits_{r=0}^{n-1} e^{i\frac{2\pi k}{n}r}=n\sum\limits_{k=0}^\infty\frac{1}{(nk)!}$ and therefore the claim.
A: $\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

The roots of $\ds{z^{n} - 1 = 0}$ are given by
  $\ds{\braces{\exp\pars{{2\pi r \over n}\,\ic}\ \mid\ r = 0,1,\ldots,n - 1}}$. Note that
  $\ds{{1 \over n}\sum_{r = 0}^{n - 1}\exp\pars{{2\pi rs \over n}\,\ic}}$ is equal to $\ds{1}$ whenever $\ds{n \mid s}$ and it vanishes out otherwise.

\begin{align}
\color{#f00}{\sum_{s = 0}^{\infty}{1 \over \pars{sn}!}} & =
\sum_{s = 0}^{\infty}{1 \over s!}\,
\bracks{{1 \over n}\sum_{r = 0}^{n - 1}\exp\pars{{2\pi rs \over n}\,\ic}} =
{1 \over n}\sum_{r = 0}^{n - 1}\sum_{s = 0}^{\infty}{1 \over s!}\,
\bracks{\exp\pars{{2\pi r \over n}\,\ic}}^{s}
\\[5mm] & =
{1 \over n}\sum_{r = 0}^{n - 1}\exp\pars{\exp\pars{{2\pi r \over n}\,\ic}} =
{1 \over n}\sum_{r = 0}^{n - 1}\exp\pars{\cos\pars{2\pi r \over n}}
\exp\pars{\ic\sin\pars{2\pi r \over n}}
\\[5mm] & =
{1 \over n}\sum_{r = 0}^{n - 1}\exp\pars{\cos\pars{2\pi r \over n}}\bracks{%
\cos\pars{\sin\pars{2\pi r \over n}} + 
\sin\pars{\sin\pars{2\pi r \over n}}\ic}\label{1}\tag{1}
\end{align}

\begin{align}
\color{#f00}{\sum_{s = 0}^{\infty}{1 \over \pars{sn}!}} & =
\color{#f00}{{1 \over n}\sum_{r = 0}^{n - 1}\exp\pars{\cos\pars{2\pi r \over n}}
\cos\pars{\sin\pars{2\pi r \over n}}}
\end{align}


It's clear that the imaginary part of \eqref{1} vanishes out.

A: The expression on the left is the hypergeometric function with no upper and $n-1$ lower parameters of values 
$$
\frac1n, \frac2n \cdots \frac{n-1}{n}
$$
at value $z = n^{-n}$.
I don't know that his is very helpful:  The expression on the left has sorts of functions that are one-lower-index hypergeometrics (e.g., sin and cos) but it is not a simple convolution or anything.
