How to find $\sum_{n=0}^{+\infty} \dfrac{1}{(kn)!}$? 
I want to calculate $$S_k = \sum_{n=0}^{+\infty} \dfrac{1}{(kn)!}$$ when $k\in\Bbb N ^*$.

I tried to find a recurrence equation for $k$, but I found nothing really interesting. 
I already know that for $k=1, S_1 = e$ and $S_2 = ch(1)$, but I don't know what to do for $k\geq 3$.

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.
