TITLE: Prove the following series $\sum\limits_{s=0}^\infty \frac{1}{(sn)!}$
QUESTION [9 upvotes]: 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.

REPLY [4 votes]: $\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

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.