<s>[INST] Closed-forms of infinite series with factorial in the denominator How to evaluate the closed-forms of series

\begin{equation}
1)\,\, \sum_{n=0}^\infty\frac{1}{(3n)!}\qquad\left|\qquad2)\,\, \sum_{n=0}^\infty\frac{1}{(3n+1)!}\qquad\right|\qquad3)\,\, \sum_{n=0}^\infty\frac{1}{(3n+2)!}\\
\end{equation}

Of course Wolfram Alpha can give us the closed-forms
\begin{align}
\sum_{n=0}^\infty\frac{1}{(3n)!}&=\frac{e}{3}+\frac{2\cos\left(\frac{\sqrt{3}}{2}\right)}{3\sqrt{e}}\\
\sum_{n=0}^\infty\frac{1}{(3n+1)!}&=\frac{e}{3}+\frac{2\sin\left(\frac{\sqrt{3}}{2}-\frac{\pi}{6}\right)}{3\sqrt{e}}\\
\sum_{n=0}^\infty\frac{1}{(3n+2)!}&=\frac{e}{3}-\frac{2\sin\left(\frac{\sqrt{3}}{2}+\frac{\pi}{6}\right)}{3\sqrt{e}}
\end{align}
but how to get those closed-forms by hand? I can only notice that
\begin{equation}
\sum_{n=0}^\infty\frac{1}{n!}=\sum_{n=0}^\infty\frac{1}{(3n)!}+\sum_{n=0}^\infty\frac{1}{(3n+1)!}+\sum_{n=0}^\infty\frac{1}{(3n+2)!}=e
\end{equation}
Could anyone here please help me? Any help would be greatly appreciated. Thank you.
PS: Please don't work backward.
 [/INST] 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.
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