A series with factorials

Evaluate the series

$$\mathcal{S}=\sum_{n=0}^{\infty} \frac{1}{(3n)!}$$

Solution

We give three solutions. The third one is not of the poster though. 

1st solution: Let us apply a dicrete Fourier transform. Let $\omega$ be a root of unity, namely $\omega=e^{2\pi i /3}$. Define

$$f(n)=\frac{1}{3} \left ( 1+ \omega^n +\omega^{2n} \right)$$

Then it is clear that $f(n)=1_{n \equiv 0 \pmod 3}(n)$.

\begin{align*}
\sum_{n=0}^{\infty} \frac{1}{(3n)!} &=\sum_{n=0}^{\infty} \frac{f(n)}{n!} \\
 &= \frac{1}{3}\sum_{n=0}^{\infty} \frac{1+\omega^n +\omega^{2n}}{n!}\\
 &= \frac{1}{3}\sum_{n=0}^{\infty} \frac{1}{n!} + \frac{1}{3}\left ( \sum_{n=0}^{\infty} \frac{\omega^n}{n!}+ \frac{\omega^{2n}}{n!} \right ) \\
 &= \frac{e}{3}+ \frac{1}{3}\left ( e^{\omega}+ e^{\omega^2} \right )\\
 &= \frac{e}{3} + \frac{2 \cos \frac{\sqrt{3}}{2}}{3\sqrt{e}}
\end{align*}

2nd solution: Consider the function

$$F(x)=\sum_{n=0}^{\infty} \frac{x^{3n}}{(3n)!}$$

The Laplace transformation of the function

\begin{align*}
F(s) &=\int_{0}^{\infty} e^{-sx } f(x) \, {\rm d}x \\
 &=\int_{0}^{\infty} e^{-sx} \sum_{n=0}^{\infty} \frac{x^{3n}}{(3n)!} \, {\rm d}x \\
 &=\sum_{n=0}^{\infty} \frac{1}{(3n)!} \int_{0}^{\infty} e^{-sx} x^{3n} \, {\rm d}x \\
 &= \sum_{n=0}^{\infty}\frac{1}{(3n)!} \frac{\Gamma(3n+1)}{s^{3n+1}}  \\
 &= \sum_{n=0}^{\infty} \frac{(3n)!}{(3n)!} \cdot \frac{1}{s^{3n+1}} \\
 &= \sum_{n=0}^{\infty} \frac{1}{s^{3n+1}} \\
 &= \frac{s^2}{s^3-1}
\end{align*}

Now take the inverse Laplace. You can invoke either a contour integration (Bromwich contour) or partial decomposition. We shall evaluate it using contour integration.

\begin{align*}
\mathcal{L}^{-1} \left ( \frac{s^2}{s^3-1} \right ) &=\frac{1}{2\pi i} \int_{-a -i \infty}^{a+i \infty} \frac{s^2 e^{sx}}{s^3-1} \, {\rm d}s \\
 &= \sum \text{residues}\\
 &= \frac{e^x}{3}+ \frac{1}{3} e^{-\left ( \frac{1}{2}- \frac{i\sqrt{3}}{2} \right )x} +\frac{1}{3} e^{-\left ( \frac{1}{2}+ \frac{i\sqrt{3}}{2} \right )x}
\end{align*}

Thus

\begin{align*}
F(x) &=\frac{e^x}{3}+ \frac{1}{3} e^{-\left ( \frac{1}{2}- \frac{i\sqrt{3}}{2} \right )x} +\frac{1}{3} e^{-\left ( \frac{1}{2}+ \frac{i\sqrt{3}}{2} \right )x} \\
 &=\frac{e^x}{3} + \frac{2}{3}e^{-x/2} \cos \left ( \frac{\sqrt{3}x}{2} \right )
\end{align*}

For $x=1$ we get that

$$F(1)= \sum_{n=0}^{\infty} \frac{1}{(3n)!}= \frac{e}{3} + \frac{2 \cos \frac{\sqrt{3}}{2}}{3\sqrt{e}}$$

as before.

3rd solution: 

\begin{align*}
\sum_{n=0}^{\infty} \frac{1}{(3n)!} &=\sum_{n=0}^{\infty} \sum_{k=0}^{\infty} \frac{\delta_{k, 3n}}{k!} \\
 &=\frac{1}{2\pi i } \sum_{n=0}^{\infty} \sum_{k=0}^{\infty}\; \oint \limits_{\left | z \right |=R>1} z^{3n-k+1} \, {\rm d}z\\
 &= \frac{1}{2\pi i }\oint \limits_{ \left | z \right |=R>1} \frac{1}{z} \sum_{n=0}^{\infty} \frac{1}{z^{3n}}  \sum_{k=0}^{\infty} \frac{z^k}{k!} \, {\rm d}z\\
 &= \frac{1}{2\pi i} \oint \limits_{\left | z \right |=R>1} \frac{e^z}{z \left ( 1-\frac{1}{z^3} \right )}\, {\rm d}z\\
 &= \frac{1}{2\pi i }  \oint \limits_{\left | z \right |=R>1} \frac{e^z z^2}{z^3-1} \, {\rm d}z \\
 &=\sum_{m=-1}^{1}\frac{\zeta_m^2 e^{\zeta_m}}{3\zeta_m^2}
\end{align*}

where $\delta_{n, k}$ is delta Kronecker and $\zeta_m= e^{2\pi m i /3}$. Simplifying we get  the result.