Closed form for $\sum_{n=0}^\infty {1\over (xn)!}$

Let $k=1,2,3,\cdots$ be a fixed integer. One may consider the set of roots $$ R_k=\{\omega\in\mathbb C\mid \omega^k=1\} $$ then, for each integer $n$, one has $$\sum\limits_{\omega \in R_k}\omega^n = \begin{cases} k, & \text{if $n=0$ (mod $k$)} \\ 0, & \text{otherwise} \end{cases}$$ then $$ \sum_{\omega \in R_k} e^{\omega z}=\sum\limits_{\omega \in R_k} \sum\limits_{n\ge0}\frac{\omega^nz^n}{n!}=\sum\limits_{n\ge0}\frac{z^n}{n!}\cdot\sum_{\omega \in R_k}\omega^n=k\cdot\sum\limits_{n\ge0}\frac{z^{kn}}{(kn)!} $$ that is

$$ \sum_{n\ge0}\frac{z^{kn}}{(kn)!}=\frac1k\cdot\sum_{\omega \in R_k} e^{\omega z} $$

from which one deduces the considered cases with $z=1$.

In 1903, the Swedish mathematician Gösta Mittag-Leffler introduced the following special function: $$ E_\alpha(z)=\sum_{n=0}^{\infty}\frac{z^n}{\Gamma(\alpha n+1)}, $$ where $z$ is a complex number, $\Gamma$ is the gamma function and $\alpha \geq 0$. The Mittag-Leffler function is a direct generalization of the exponential function to which it reduces for $\alpha = 1$. You can find more details about the function in this survey article.

We can express your sum in terms of the Mittag-Leffler function as the following. For a positive integer $k$, we have

$$ \sum_{n=0}^\infty \frac{1}{(kn)!} = E_k(1). $$

Note that for a positive integer $n$, we have $\Gamma(n) = (n-1)!$

Assuming $x$ is a positive integer, $$ \sum_{n=0}^\infty\frac1{(xn)!}=(e^{1}+e^{\zeta}+e^{\zeta^2}+\dots+e^{\zeta^{x-1}})/x\tag1 $$ where $\zeta$ is a primitive $x^\text{th}$ root of unity, e.g. $\zeta=\exp(i2\pi/x)$.

To see this, rewrite the sum as $$ \sum_{n=0}^\infty \frac{[\,x\text{ divides n}\,]}{n!}\tag2 $$ where $[P]$ is the Iverson bracket, equal to $1$ if $P$ is true and $0$ otherwise.

I claim that $$ (1+(\zeta^n)+(\zeta^n)^2+\dots+(\zeta^n)^{x-1})/x=[x\text{ divides }n]\tag3 $$ Once this is proved, substituting (3) into (2), and splitting into several sums, we obtain several exponential series and obtain (1).

If $n$ is a multiple of $x$, then $\zeta^n=1$, so (3) holds because $$ 1+(\zeta^n)+(\zeta^n)^2+\dots+(\zeta^n)^{x-1}=1+1+\dots+1=x $$ On the other hand, if $n$ is not a multiple of $x$, then $\zeta^n\neq 1$, so we can use the geometric series formula: $$ 1+(\zeta^n)+(\zeta^n)^2+\dots+(\zeta^n)^{x-1}=\frac{(\zeta^n)^x-1}{\zeta^n-1}=\frac{(\zeta^x)^n-1}{\zeta^n-1}=\frac{1^n-1}{\zeta^n-1}=0 $$ This proves both cases of (3).