Closed-forms of infinite series with factorial in the denominator

Another possible approach is to use the discrete Fourier transform. Let $\omega=\exp\frac{2\pi i}{3}$. Then: $$f(n)=\frac{1}{3}\left(1+\omega^n+\omega^{2n}\right)=\mathbb{1}_{n\equiv 0\!\pmod{3}}(n),$$ hence: $$\color{red}{\sum_{n=0}^{+\infty}\frac{1}{(3n)!}}=\sum_{n=0}^{+\infty}\frac{f(n)}{n!}=\frac{1}{3}\left(\exp(1)+\exp(\omega)+\exp(\omega^2)\right)=\color{red}{\frac{e}{3}+\frac{2}{3\sqrt{e}}\cos\frac{\sqrt{3}}{2}.}$$ The other two series can be computed with the same technique, by noticing that: $$f_1(n)=\frac{1}{3}\left(1+\omega^2\cdot\omega^n+\omega\cdot\omega^{2n}\right)=\mathbb{1}_{n\equiv 1\!\pmod{3}}(n),$$ $$f_2(n)=\frac{1}{3}\left(1+\omega\cdot\omega^n+\omega^2\cdot\omega^{2n}\right)=\mathbb{1}_{n\equiv 2\!\pmod{3}}(n).$$

Related techniques: (I). Here is an approach which enables you to tackle your problems. Let's consider the series

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

Taking the Laplace transform gives

$$ F(s) = \sum_{n=0}^{\infty}\frac{1}{s^{3n+1}} = \frac{s^2}{s^3-1}. $$

To finish the problem you need to find the inverse Laplace of $F(s)$. One technique is partial fraction

$$ F(s) = \frac{1}{3(s-1)} + \frac{1}{3(s+1/2-i\sqrt{3}/2)} + \frac{1}{3(s+1/2+i\sqrt{3}/2)} .$$

Notes:

1) Laplace transform is defined as

$$ F(s) = \int_{0}^{\infty}f(x)e^{-sx}dx. $$

2) Laplace transform of $x^m$ is

$$ \frac{\Gamma(m+1)}{s^{m+1}} $$

3) Laplace transform of $e^{ax}$ is

$$ \frac{1}{s-a}. $$

Or equivalently the inverse Laplace of $\frac{1}{s-a}$ is $e^{ax}$

$$ $$

This problem can be solved without advanced techniques. We have the Taylor series $$ e^x = \sum_{n\geq 0} \frac{x^n}{n!}. $$ As the questioner noted, plugging $x=1$ yields the equation $A+B+C=e$ connecting the three unknown sums. However, plugging in any cube root of unity also sheds light on the question because the numerators $x^n$ will repeat with period $3$. Let $\omega = e^{2\pi i/3}$; plugging $x = \omega$ yields $A + \omega B + \omega^2 C = e^{\omega}$, and plugging $x = \omega^2$ yields $A + \omega^2 B + \omega C = e^{\omega^2}$. We now have three equations that can be solved easily for the three unknowns $A$, $B$, and $C$. For instance, adding all three equations together yields $$ 3A = e + e^\omega + e^{\omega^2} = e + e^{-1/2}\left(\cos \frac{\sqrt{3}}{2} + i \sin \frac{\sqrt{3}}{2}\right) + e^{-1/2}\left(\cos \frac{-\sqrt{3}}{2} + i \sin \frac{-\sqrt{3}}{2}\right) $$ $$ = e + \frac{2}{\sqrt{e}} \cos \frac{\sqrt{3}}{2}, $$ in accordance with Wolfram Alpha.