# Showing that $\sum\limits_{l=0}^\infty \frac{1}{(3l+2)!}=\frac{1}{3}(e-\frac{2}{\sqrt{e}} \cos(\frac{\pi}{3}-\frac{\sqrt{3}}{2}))$

It is known that $\frac{1}{2\pi i} \int_C \frac{e^z}{z^3-1}dz=\sum\limits_{l=0}^\infty \frac{1}{(3l+2)!}$.

Now I will like to evaluate $\frac{1}{2\pi i} \int_C \frac{e^z}{z^3-1}dz$ using Cauchy's Integral Formula, to show that $$\sum\limits_{l=0}^\infty \frac{1}{(3l+2)!}=\frac{1}{3}(e-\frac{2}{\sqrt{e}} \cos(\frac{\pi}{3}-\frac{\sqrt{3}}{2}))$$

There is a hint where we can assume the identity

$$\prod\limits_{j=1,j\neq k}^m =(e^{2\pi ij/m}-e^{2\pi ik/m})=\frac{m}{e^{2\pi ik/m}}$$

I managed the following:

$$\frac{1}{2\pi i} \int_C \frac{e^z}{z^3-1}dz=(\frac{e^z}{z+1})'|_{z=1}=\frac{e}{4}$$

by Cauchy's Integral Formula for higher derivatives, which is clearly not the desired solution.

Where did I go wrong in my working? I feel like I may have a conceptual error of some sort..

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

With $\ds{r \in \braces{1,\expo{\pm\ 2\pi\ic/3}}}$:

\begin{align} {1 \over 2\pi\ic}\oint_{\verts{z}\ =\ 1^{\large +}} {\expo{z} \over z^{3} - 1}\,\dd z & = \sum_{r}{\expo{r} \over 3r^{2}} = {1 \over 3}\sum_{r}{r\expo{r} \over r^{3}} = {1 \over 3}\sum_{r}r\expo{r} \\[5mm] & = {1 \over 3}\expo{} + {2 \over 3}\,\Re\bracks{\expo{2\pi\ic/3} \exp\pars{\expo{2\pi\ic/3}}} \\[5mm] &= {1 \over 3}\expo{} + {2 \over 3}\,\Re \exp\pars{{2\pi \over 3}\,\ic + \cos\pars{2\pi \over 3} + \ic \sin\pars{2\pi \over 3}} \\[5mm] &= {1 \over 3}\expo{} + {2 \over 3}\,\Re \exp\pars{- {1 \over 2} + \bracks{{2\pi \over 3} + {\root{3} \over 2}}\,\ic} \\[5mm] &= \bbx{{1 \over 3}\expo{} + {2 \over 3}\expo{-1/2} \cos\pars{{2\pi \over 3} + {\root{3} \over 2}}} \end{align}

Note that $\ds{\cos\pars{{2\pi \over 3} + {\root{3} \over 2}} = -\cos\pars{{\pi \over 3} - {\root{3} \over 2}}}$.

• Thank you for your answer. It is straightforward and concise! Just to confirm - when you equate $\frac{1}{2\pi i}\int_{C}\frac{e^z}{z^3-1}dz=\sum\limits_r \frac{e^r}{3r^2}$, do you mean to sum up all the residues at $r$, for $r \in \{1, e^{\pm 2\pi i/3} \}$? – Stoner May 3 '18 at 5:23
• @Stoner Yes. The residue at $r$ is given by $\displaystyle\lim_{z \to r}{\left(z - r\right)\mathrm{e}^{z} \over z^{3} - 1} = {\mathrm{e}^{r} \over 3r^{2}} = {1 \over 3}\,r\,\mathrm{e}^{r}$ because $\displaystyle r^{3} = 1$. Thanks. – Felix Marin May 3 '18 at 5:39
• Thank you very much for the clarification. :) – Stoner May 3 '18 at 5:45
• @Stoner ${\bullet\quad\bullet \brace \smile}$ – Felix Marin May 3 '18 at 5:47