Prove the following series $\sum\limits_{s=0}^\infty \frac{1}{(sn)!}$

Question

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.

the rhs equals $\frac{1}{2n}\Re\sum_{r=0}^{n-1}e^{e^{i 2 r \pi/n }}$ — tired, Sep 27 '16 at 19:50
From an old A-level book from the UK, one of the few questions in the book I'm having trouble with. — user51515, Sep 27 '16 at 19:54
You can try ordering online from bookdepository, Further Pure Mathematics 2 and 3 (OCR), most of it is relatively easy. But there are a few miscellaneous questions that are quite hard. If you like difficult questions try searching STEP papers, these questions are specifically targetted at students looking to study mathematics in UK universities. http://www.physicsandmathstutor.com/step/ — user51515, Sep 27 '16 at 20:27
A partial result:
observing that

$$ \sum_{s=0}^{\infty}\frac{1}{(4s)!}=\frac{1}2\sum_{s=0}^{\infty}\left(\frac{(-1)^s}{(2s)!}+\frac{1}{(2s)!}\right)=\frac{\cosh(1)+\cos(1)}{2} $$

which equals the rhs of OPs equation after some algebra. By the same scheme you may now tackle $n=8,16,32$ etc — tired, Sep 27 '16 at 20:43
For $$ \sum_{s=0}^{\infty}\frac{(-1)^s}{(4s)!} $$
play around with the real part of the series

$$ \sum_{s=0}^{\infty}\frac{(-1)^{s/2}}{(2s)!} $$ — tired, Sep 27 '16 at 20:53
@tired, I see what you are doing, I'll attempt it further from here. — user51515, Sep 27 '16 at 20:54
For what it's worth, the general name for this process is series multisection and you can find more information about it under that name (e.g., https://en.wikipedia.org/wiki/Series_multisection ) - I first learned about this in Melzak's Companion to Concrete Mathematics. — Steven Stadnicki, Sep 27 '16 at 23:42

user90369 · Accepted Answer · 2016-09-28T08:27:35.097

9

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.

edited Sep 28 '16 at 08:27

answered Sep 27 '16 at 20:55

user90369

11,518

i was so blind...also a big (+1) – tired Sep 27 '16 at 20:56
@tired : Thanks! :-) – user90369 Sep 27 '16 at 20:57
Thank you, I forgot I can go prove from the RHS. – user51515 Sep 27 '16 at 20:57
@user51515 : You are welcome - it was a nice question. :-) – user90369 Sep 27 '16 at 20:59
1

Note that $$\sum\limits_{r=0}^{n-1} e^{i\frac{2\pi k}{n}r}=\begin{cases} n&,k=\ell n\\0&,k\ne \ell n\end{cases}$$for integer $\ell$. – Mark Viola Sep 28 '16 at 02:16
@Dr. MV : Thank you! (I have corrected - I have to be more precisely, I know :-)) – user90369 Sep 28 '16 at 08:33
You're welcome. My pleasure. -Mark – Mark Viola Sep 28 '16 at 13:20

Felix Marin · Answer 2 · 2016-09-27T22:53:36.327

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

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.

Thanks for your answer, this is also very helpful. – user51515 Sep 28 '16 at 03:09 — user51515, Sep 28 '16 at 03:09
@user51515 You're welcome. It's nice it was helpful to you. – Felix Marin Sep 28 '16 at 03:11 — Felix Marin, Sep 28 '16 at 03:11
@FelixMarin: Nice elaboration! (+1) – Markus Scheuer Sep 28 '16 at 07:02 — Markus Scheuer, Sep 28 '16 at 07:02

Mark Fischler · Answer 3 · 2016-09-27T20:29:45.663

0

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.

edited Sep 27 '16 at 20:29

answered Sep 27 '16 at 20:16

Mark Fischler

41,743

I don't think the question involves anything on this level. But thanks for the help. – user51515 Sep 27 '16 at 20:30

Prove the following series $\sum\limits_{s=0}^\infty \frac{1}{(sn)!}$

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