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

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

