Interesting Series

Calculus (Integrals, Series)
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Hamza Mahmood
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Interesting Series

#1

Post by Hamza Mahmood »

\[\sum_{{n=0}}^{\infty} \frac{1}{\left ( 3n \right )!}= ?\]
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Tolaso J Kos
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Re: Interesting Series

#2

Post by Tolaso J Kos »

Hi Hamza. Here is the first of the three solutions I am aware of.

We are applying a discrete Fourier transform. Let $\omega$ be a root of unity, namely $\omega=e^{2\pi i /3}$. Define

$$f(n)=\frac{1}{3} \left ( 1+ \omega^n +\omega^{2n} \right)$$

Then it is clear that $f(n)=1_{n \equiv 0 \pmod 3}(n)$.

\begin{align*}
\sum_{n=0}^{\infty} \frac{1}{(3n)!} &=\sum_{n=0}^{\infty} \frac{f(n)}{n!} \\
&= \frac{1}{3}\sum_{n=0}^{\infty} \frac{1+\omega^n +\omega^{2n}}{n!}\\
&= \frac{1}{3}\sum_{n=0}^{\infty} \frac{1}{n!} + \frac{1}{3}\left ( \sum_{n=0}^{\infty} \frac{\omega^n}{n!}+ \frac{\omega^{2n}}{n!} \right ) \\
&= \frac{e}{3}+ \frac{1}{3}\left ( e^{\omega}+ e^{\omega^2} \right )\\
&= \frac{e}{3} + \frac{2 \cos \frac{\sqrt{3}}{2}}{3\sqrt{e}}
\end{align*}

:)

More to come later.

P.S: I leave as an exercise for someone to verify that

$$\frac{2 \cos \frac{\sqrt{3}}{2}}{3\sqrt{e}} =\frac{1}{3}\left ( e^{\omega}+ e^{\omega^2} \right )$$
Imagination is much more important than knowledge.
Hamza Mahmood
Posts: 14
Joined: Fri Dec 04, 2015 4:54 pm

Re: Interesting Series

#3

Post by Hamza Mahmood »

Simply Beautiful Solution!!!
Thank you soooo much!
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Tolaso J Kos
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Re: Interesting Series

#4

Post by Tolaso J Kos »

Let us continue with the Laplace transform solution. There we go:

Consider the function

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

The Laplace transformation of the function

\begin{align*}
F(s) &=\int_{0}^{\infty} e^{-sx } f(x) \, {\rm d}x \\
&=\int_{0}^{\infty} e^{-sx} \sum_{n=0}^{\infty} \frac{x^{3n}}{(3n)!} \, {\rm d}x \\
&=\sum_{n=0}^{\infty} \frac{1}{(3n)!} \int_{0}^{\infty} e^{-sx} x^{3n} \, {\rm d}x \\
&= \sum_{n=0}^{\infty}\frac{1}{(3n)!} \frac{\Gamma(3n+1)}{s^{3n+1}} \\
&= \sum_{n=0}^{\infty} \frac{(3n)!}{(3n)!} \cdot \frac{1}{s^{3n+1}} \\
&= \sum_{n=0}^{\infty} \frac{1}{s^{3n+1}} \\
&= \frac{s^2}{s^3-1}
\end{align*}

Now take the inverse Laplace. You can invoke either a contour integration (Bromwich contour) or partial decomposition. We shall evaluate it using contour integration.

\begin{align*}
\mathcal{L}^{-1} \left ( \frac{s^2}{s^3-1} \right ) &=\frac{1}{2\pi i} \int_{-a -i \infty}^{a+i \infty} \frac{s^2 e^{sx}}{s^3-1} \, {\rm d}s \\
&= \sum \text{residues}\\
&= \frac{e^x}{3}+ \frac{1}{3} e^{-\left ( \frac{1}{2}- \frac{i\sqrt{3}}{2} \right )x} +\frac{1}{3} e^{-\left ( \frac{1}{2}+ \frac{i\sqrt{3}}{2} \right )x}
\end{align*}

Thus

\begin{align*}
F(x) &=\frac{e^x}{3}+ \frac{1}{3} e^{-\left ( \frac{1}{2}- \frac{i\sqrt{3}}{2} \right )x} +\frac{1}{3} e^{-\left ( \frac{1}{2}+ \frac{i\sqrt{3}}{2} \right )x} \\
&=\frac{e^x}{3} + \frac{2}{3}e^{-x/2} \cos \left ( \frac{\sqrt{3}x}{2} \right )
\end{align*}

For $x=1$ we get that

$$F(1)= \sum_{n=0}^{\infty} \frac{1}{(3n)!}= \frac{e}{3} + \frac{2 \cos \frac{\sqrt{3}}{2}}{3\sqrt{e}}$$

as before.
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Tolaso J Kos
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Re: Interesting Series

#5

Post by Tolaso J Kos »

Yet another solution which is not mine though. I have seen it many years ago.

\begin{align*}
\sum_{n=0}^{\infty} \frac{1}{(3n)!} &=\sum_{n=0}^{\infty} \sum_{k=0}^{\infty} \frac{\delta_{k, 3n}}{k!} \\
&=\frac{1}{2\pi i } \sum_{n=0}^{\infty} \sum_{k=0}^{\infty}\; \oint \limits_{\left | z \right |=R>1} z^{3n-k+1} \, {\rm d}z\\
&= \frac{1}{2\pi i }\oint \limits_{ \left | z \right |=R>1} \frac{1}{z} \sum_{n=0}^{\infty} \frac{1}{z^{3n}} \sum_{k=0}^{\infty} \frac{z^k}{k!} \, {\rm d}z\\
&= \frac{1}{2\pi i} \oint \limits_{\left | z \right |=R>1} \frac{e^z}{z \left ( 1-\frac{1}{z^3} \right )}\, {\rm d}z\\
&= \frac{1}{2\pi i } \oint \limits_{\left | z \right |=R>1} \frac{e^z z^2}{z^3-1} \, {\rm d}z \\
&=\sum_{m=-1}^{1}\frac{\zeta_m^2 e^{\zeta_m}}{3\zeta_m^2}
\end{align*}

where $\delta_{n, k}$ is Delta Kronecker and $\zeta_m = e^{2\pi m i /3}$. Simplifying we get the result.
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Re: Interesting Series

#6

Post by Tolaso J Kos »

And here is the generalization. Suppose that $\ell \geq 0$. Following the above approach (that with contours) we have:

\begin{align*}
\sum_{n=0}^{\infty}\frac{1}{\left ( 3n+\ell \right )!} &=\sum_{n=0}^{\infty} \sum_{k=0}^{\infty} \frac{\delta_{k, 3n+\ell}}{k!} \\
&=\frac{1}{2\pi i }\sum_{n=0}^{\infty} \sum_{k=0}^{\infty}\frac{1}{k!} \; \oint \limits_{\left | z \right |=R>1} \frac{{\rm d}z}{z^{3n+\ell-k+1}} \\
&= \frac{1}{2\pi i }\oint \limits_{\left | z \right |=R>1} \frac{1}{z^{\ell+1}} \sum_{n=0}^{\infty} \frac{1}{z^{3n}} \sum_{k=0}^{\infty} \frac{z^k}{k!}\\
&= \frac{1}{2\pi i } \oint \limits_{\left | z \right |=R>1} \frac{e^z z^{2-\ell}}{z^3-1} \, {\rm d}z\\
&= \sum_{m=-1}^{1}\frac{\zeta_m^{2-\ell} e^{\zeta_m}}{3\zeta^2_m}\\
&= \frac{e}{3} + \frac{2}{3\sqrt{e}} \mathfrak{Re} \left [ \exp \left ( \frac{\sqrt{3}}{2} - \frac{2\pi}{3} \ell \right )i \right ] \\
&= \frac{e}{3} + \frac{2}{3\sqrt{e}} \cos \left ( \frac{\sqrt{3}}{2} - \frac{2\pi \ell}{3} \right )
\end{align*}
Imagination is much more important than knowledge.
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