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Let $n \in \Bbb N$ be an integer. We want to compute $$\sum_{k=0}^\infty \frac{1}{(nk)!} $$

This clearly makes me think of the Taylor's series of $\exp$ and I know the answer is linked to $e$ but I think I would need to have $\sum_{k=0}^\infty \frac{1}{((n-j)k)!} $ for all others $j$ which makes the problem a lot harder.

I also thought of using Stirling's approximation above some rank but I would lose a lot of precision in the first terms.

The only remaining tool I can think of is $n!=\Gamma(n+1)$ but don't know how useful it can be.

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Kilkik
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1 Answers1

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The root of unity $\xi = e^{2\pi i / n}$ has the property $$ 1+\xi^{k}+\xi^{2 \, k}+\cdots \xi^{(n-1) \, k} = \begin{cases} n , & \text{ if } k \equiv 0 \text{ mod } n\\ 0 , & \text{otherwise} \end{cases} $$ and hence $$ \sum_{k=0}^\infty \frac{x^{n \, k}}{(n \, k)!} \, = \, \frac{1}{n}(e^{x}+e^{\xi \, x}+e^{\xi^2 \, x}+\cdots + e^{\xi^{n-1} \, x}) \, .$$ Setting $x = 1$ you get an expression for $\sum_{k=0}^\infty \frac{1}{(n \, k)!}$ as a sum of exponentials of roots of unity.

user1363745
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