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Find where m is an integer: $$S = 1+\sum_{n=1}^{\infty}\frac{n!}{(mn)!}$$

When we rewrite this sum as one plus a function of $x$ and multiply the $n^{th}$ term of the series by $x^{mn-1}$, the resulting function solves for the differential equation $$\frac{d^{m-1}f(x)}{dx^{m-1}} = \frac1m + \frac1{xf(x)}$$ and hence we can replace $f(x)$ with $s(x)$ and set $s(0) = 1$. Then use that to find $s(1)$ which is $S$.

I'm looking for an alternative method because I'm having trouble dealing with this differential equation.

jg mr chapb
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    There's no nice closed form. Per Mathematica, when $m=4$, we obtain 1/24 HypergeometricPFQ[{1}, {5/4, 3/2, 7/4}, 1/256] which Mathematica doesn't know how to simplify. – Patrick Stevens Sep 05 '17 at 08:36

2 Answers2

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Considering $$S_m = \sum_{n=1}^{\infty}\frac{n!}{(mn)!}$$ beside the case $m=2$ for which $$S_2=\frac{1}{2} \sqrt[4]{e} \sqrt{\pi } \text{erf}\left(\frac{1}{2}\right)$$ as Patrick Stevens commented, we are left with hypergeometric functions which apparently do not simplify at all.

They show quite nice patterns $$S_3=\frac{1}{3!} \, _1F_2\left(1;\frac{4}{3},\frac{5}{3};\frac{1}{3 ^3}\right)$$ $$S_4=\frac{1}{4!} \, _1F_3\left(1;\frac{5}{4},\frac{6}{4},\frac{7}{4};\frac{1}{4^4}\right)$$ $$S_5=\frac{1}{5!} \, _1F_4\left(1;\frac{6}{5},\frac{7}{5},\frac{8}{5},\frac{9}{5};\frac{1}{5^5}\right)$$ $$S_6=\frac{1}{6!} \, _1F_5\left(1;\frac{7}{6},\frac{8}{6},\frac{9}{6},\frac{10}{6},\frac{11}{6};\frac{ 1}{6^6}\right)$$ $$S_7=\frac{1}{7!}\, _1F_6\left(1;\frac{8}{7},\frac{9}{7},\frac{10}{7},\frac{11}{7},\frac{12}{7},\frac{13}{7};\frac{1}{7^7}\right)$$

In fact, it seems that the asymptotics is just $S_m\sim\frac{1}{m!}$.

Claude Leibovici
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Using integrals, one gets (based on Prove the following series $\sum\limits_{s=0}^\infty \frac{1}{(sn)!}$):

$\displaystyle S_m = \sum\limits_{n=0}^\infty \frac{n!}{(mn)!} = \sum\limits_{n=0}^\infty \frac{1}{(mn)!} \int\limits_0^\infty \frac{x^n}{e^x} dx =\int\limits_0^\infty \frac{1}{e^x} \left( \sum\limits_{n=0}^\infty \frac{x^n}{(mn)!}\right) dx$

$\hspace{7mm} \displaystyle = \frac{1}{m} \int\limits_0^\infty \left(\sum\limits_{r=0}^{m-1} e^{\sqrt[m]{x}e^{i2\pi r/m} -x}\right) dx = \frac{1}{m} \sum\limits_{r=0}^{m-1} \int\limits_0^\infty e^{\sqrt[m]{x}e^{i2\pi r/m} -x} dx $

user90369
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