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.
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 $