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.

Solutions Collecting From Web of "Closed-forms of infinite series with factorial in the denominator"

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 $e^{ax}$

$$ $$

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:
$$f_1(n)=\frac{1}{3}\left(1+\omega^2\cdot\omega^n+\omega\cdot\omega^{2n}\right)=\mathbb{1}_{n\equiv 1\!\pmod{3}}(n),$$
$$f_2(n)=\frac{1}{3}\left(1+\omega\cdot\omega^n+\omega^2\cdot\omega^{2n}\right)=\mathbb{1}_{n\equiv 2\!\pmod{3}}(n).$$

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With $\ds{\ell = 0,1,2}$:

\begin{align}
{\cal I}_{\ell}&\equiv\sum_{n = 0}^{\infty}{1 \over \pars{3n + \ell}!}=
\sum_{n = 0}^{\infty}\sum_{k=0}^{\infty}{\delta_{k,3n + \ell} \over k!}
=\sum_{n,k = 0}^{\infty}{1 \over k!}
\oint_{\atop{\atop\verts{z}\ =\ a\ >\ 1}}{1 \over z^{3n + \ell – k + 1}}
\,{\dd z \over 2\pi\ic}
\\[3mm]&=\oint_{\atop{\atop\verts{z}\ =\ a\ >\ 1}}{1 \over z^{\ell + 1}}
\bracks{\sum_{n = 0}^{\infty}\pars{1 \over z^{3}}^{n}}
\bracks{\sum_{k = 0}^{\infty}{z^{k} \over k!}}\,{\dd z \over 2\pi\ic}
\\[3mm]&=\oint_{\atop{\atop\verts{z}\ =\ a\ >\ 1}}{1 \over z^{\ell + 1}}
\,{1 \over 1 – 1/z^{3}}\,\expo{z}\,{\dd z \over 2\pi\ic}
=\oint_{\atop{\atop\verts{z}\ =\ a\ >\ 1}}
{z^{2 – \ell} \over z^{3} – 1}\,\expo{z}\,{\dd z \over 2\pi\ic}
\\[3mm]&=\sum_{m = -1}^{1}{z_{m}^{2 – \ell}\expo{z_{m}} \over 3z_{m}^{2}}\qquad
\mbox{where}\qquad z_{m} \equiv \exp\pars{2m\pi\ic \over 3}\,,\quad m = -1,0,1
\end{align}

Then,
\begin{align}
{\cal I}_{\ell} &= {1 \over 3}
\sum_{m = -1}^{1}z_{m}^{-\ell}\expo{z_{m}}
={1 \over 3}\,\expo{} + {2 \over 3}\,\Re\pars{z_{1}^{-\ell}\expo{z_{1}}}
={1 \over 3}\,\expo{}
+{2 \over 3}\,\Re\pars{\expo{-2\ell\pi\ic/3}\exp\pars{\expo{2\pi\ic/3}}}
\\[3mm]&={1 \over 3}\,\expo{}
+{2 \over 3}\,
\Re\pars{\expo{-2\ell\pi\ic/3}\exp\pars{-\,\half + {\root{3} \over 2}\,\ic}}
\\[3mm]&={1 \over 3}\,\expo{}
+{2 \over 3\root{\expo{}}}\,
\Re\exp\pars{\bracks{{\root{3} \over 2} – {2\pi \over 3}\,\ell}\ic}
\end{align}

$$
{\cal I}_{\ell}\equiv\sum_{n = 0}^{\infty}{1 \over \pars{3n + \ell}!}
={1 \over 3}\,\expo{}
+{2 \over 3\root{\expo{}}}\,\cos\pars{{\root{3} \over 2} – {2\pi \over 3}\,\ell}
\,,\qquad\ell = 0,1,2
$$

$$\begin{array}{rclcl}
{\cal I}_{0}&=&\color{#66f}{\large\sum_{n = 0}^{\infty}{1 \over \pars{3n}!}}
&=&{1 \over 3}\,\expo{}
+{2 \over 3\root{\expo{}}}\,\cos\pars{{\root{3} \over 2}}
\\[5mm]
{\cal I}_{1}&=&\color{#66f}{\large\sum_{n = 0}^{\infty}{1 \over \pars{3n + 1}!}}
&=&{1 \over 3}\,\expo{}
+{2 \over 3\root{\expo{}}}\
\overbrace{\cos\pars{{\root{3} \over 2} – {2\pi \over 3}}}
^{\ds{\color{#c00000}{\sin\pars{{\root{3} \over 2} – {\pi \over 6}}}}}
\\[5mm]
{\cal I}_{2}&=&\color{#66f}{\large\sum_{n = 0}^{\infty}{1 \over \pars{3n + 2}!}}
&=&{1 \over 3}\,\expo{}
+{2 \over 3\root{\expo{}}}\
\underbrace{\cos\pars{{\root{3} \over 2} – {4\pi \over 3}}}
_{\ds{\color{#c00000}{-\sin\pars{{\root{3} \over 2} + {\pi \over 6}}}}}
\end{array}
$$

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.