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$e$ and $\pi$ both have plenty of different sum representations that show up with a simple google search but its harder to find for the multiplication of $e*\pi$. Does anyone know a sum that equals the desired result? What would be the proper method of combining known sums for $e$ and $\pi$ into the proper result such as foiling using ceasero summation? Thanks for the help

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  • $\begingroup$ You could take the Cauchy product of two absolutely convergent series for $e$ and $\pi$. $\endgroup$ – Hayden Oct 20 '16 at 2:17
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Probably the simplest is the following. Since $1-\frac{1}{e\pi}<1 $ we have $$\sum_{k\geq0}\left(1-\frac{1}{e\pi}\right)^{k}=\frac{1}{1-\left(1-\frac{1}{e\pi}\right)}=\color{red}{e\pi}.$$

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If you accept the trigonometric function $\sin$ then the $\exp$ function should be accepted too (no ?) : $$e\,\pi=\sum_{k=0}^\infty (-1)^k\,\frac{\sin\left(\left(k+\frac 12\right)\,\exp(1)\right)}{2\;\left(k+\frac 12\right)^2}$$ (and yes $\,e=\exp(1)\,$ could be replaced by $x$ in a certain range...)


A neat infinite product by Melzak (from the ratio of the $n$-dimensional volumes of the sphere and its largest inscribed cylinder : Mekzak $1961$ "Infinite Products for $\pi e$ and $\pi/e$") : $$e\,\pi=6\prod_{n=1}^\infty \left(\frac n{n+1}\right)^{2n}\left(\frac{2n+3}{2n+1}\right)^{2n+1}$$ Combining two infinite products may seem easy but the problem remains...

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