How to find $\sum_{n=0}^{+\infty} \dfrac{1}{(kn)!}$?

This can be done via series multisection. If one has a power series $$f(x)=\sum_{n=0}^\infty a_n x^n$$ then one can pick out the terms with $k\mid n$ as follows. Let $\zeta=\exp(2\pi i/k)$ and consider $$\sum_{j=0}^{k-1}f(\zeta^j x)=\sum_{j=0}^{k-1}\sum_{n=0}^\infty a_n\zeta^{jn}x^n =\sum_{n=0}^\infty a_nx^n\sum_{j=0}^{k-1}\zeta^{jn}.$$ The inner sum is a GP and is zero, unless $k\mid n$, in which case it equals $k$. Therefore $$\sum_{j=0}^{k-1}f(\zeta^j x) =k\sum_{m=0}^\infty a_{km}x^{km}.$$ Taking $x=1$, $$\sum_{m=0}^\infty a_{km}=\frac1k\sum_{j=0}^{k-1}f(\zeta^j).$$

So, $$\sum_{m=0}^\infty \frac1{(km)!}=\frac1k\sum_{j=0}^{k-1}\exp(\zeta^j) =\frac1k\sum_{j=0}^{k-1}\exp(\cos(2\pi j/k))\cos(\sin(2\pi j/k))$$ on taking real parts.

This is closely related to series multisection.

Let $$f(z)=\sum^\infty_{n=0}\frac1{n!}z^n=e^z$$

Let $a_n=\frac1{n!}$.

Then, $$S_k=\sum^\infty_{m=0}a_{km}z^{km}$$ where $z=1$.

By the multisection formula, we immediately obtain $$S_k=\frac1k\sum^{k-1}_{p=0}\text{exp}\left(e^{\frac{2\pi ip}{k}}\right)$$

THOUGHTS:

How to evaluate $S_\infty$ from the formula for $S_k$ I derived above?

Follow-up:

$S_\infty$ is a Riemann sum! $$S_\infty=\int_0^\infty e^{e^{2\pi ix}}dx$$

RESPONSE TO COMMENT:

For reference: Wikipedia page