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

How to evaluate the closed-forms of series

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

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 $$\sum_{n=0}^\infty\frac{1}{n!}=\sum_{n=0}^\infty\frac{1}{(3n)!}+\sum_{n=0}^\infty\frac{1}{(3n+1)!}+\sum_{n=0}^\infty\frac{1}{(3n+2)!}=e$$ Could anyone here please help me? Any help would be greatly appreciated. Thank you.

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I believe this link might prove itself helpful. –  Lucian Aug 30 at 18:54
Thank you Mr. @Lucian. BTW, where did you get those formulas? –  Anastasiya-Romanova Aug 30 at 18:58

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 $e^{ax}$



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It seems hard to me. I haven't learnt Laplace transform yet. +1 anyway. Thank you. –  Anastasiya-Romanova Aug 30 at 19:25
@V-Moy: It is not hard to learn. You are welcome. –  Mhenni Benghorbal Aug 30 at 19:27
@V-Moy: See what I added. –  Mhenni Benghorbal Aug 30 at 19:30
OK, thanks. I'll learn it. –  Anastasiya-Romanova Aug 30 at 19:39
I think I understand the concept of Laplace transform. Let $$f(x)=\sum_{n=0}^\infty \frac{x^m}{m!}$$ where $m=an+b$. Multiplying both sides by $e^{-px}$, then integrating \begin{align} \int_0^\infty f(x)e^{-px}\,dx&=\sum_{n=0}^\infty \int_0^\infty\frac{x^m}{m!}e^{-px}\,dx\\ F(p)&=\sum_{n=0}^\infty\frac{1}{p^{m+1}}\\ &=\sum_{n=0}^\infty\frac{1}{p^{an+b+1}}\\ &=\frac{p^{a-b-1}}{p^{a}-1}\\ \end{align} then to get $f(x)$ we use the inverse of $F(p)$, right? –  Anastasiya-Romanova Aug 30 at 20:26

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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Oh Lord, the previous answer is using Laplace transform and this one is using Fourier transform. Does 'transform' stand for transformation? If so, how many transformation are there in math? BTW, +1 Mr. D'Aurizio. –  Anastasiya-Romanova Aug 30 at 19:45


\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}

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

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

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Although I don't understand your approach but +1 Mr. Marin. Thanks for providing me the general formula for my OP. ≧◠◡◠≦✌ –  Anastasiya-Romanova Aug 31 at 13:22

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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