# Closed-forms of infinite series with factorial in the denominator

How to evaluate the closed-forms of series

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

\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

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

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
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
\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.