Sum of $\sum \limits_{n=0}^{\infty} \frac{1}{(kn)!}$

If $(c_n)$ is any sequence with period $k$ (that is, $c_{n+k}=c_n$) then it's possible to evaluate $\sum c_n/n!$ using tricks involving $k$-th roots of unity.

Let $\omega=e^{2\pi i/k}$. Consider the $k$ sequences

$s_0:1,1,1\dots$

$s_1: 1, \omega,\omega^2,\omega^3,\dots$

$s_2: 1, \omega^2,\omega^4,\omega^6,\dots$

$s_3: 1, \omega^3,\omega^6,\omega^9,\dots$

...

$s_{k-1}: 1, \omega^{k-1},\omega^{2(k-1)},\dots$.

Using tricks analogous to finding Fourier coefficients you can find $a_0,\dots a_{k-1}$ so that $$(c_n)=a_0(s_0)+\dots+a_k(s_{k-1}).$$

Hence $$\sum\frac{c_n}{n!}=\sum_{j=0}^{k-1}a_j\sum_n\frac{\omega^{jn}}{n!} =\sum_{j=0}^{k-1}a_je^{\omega^j}.$$

If you do that for your sequence you get $$\sum_{n=0}^\infty\frac{1}{(kn)!}=\frac1k\sum_{j=0}^{k-1}e^{\omega^j}.$$

Edit: Thomas Andrews makes a comment that I should have included: Since the original sum is real, it follows that $$\sum_{n=0}^\infty\frac{1}{(kn)!}=\frac{1}{k}\sum_{j=0}^{k-1}e^{\cos 2\pi j/k}\cos(\sin2\pi j/k).$$

Edit: If one is familiar with "abstract harmonic analysis", here just harmonic analysis on compact abelian groups, one sees that those "tricks analogous to finding Fourier coefficients" are in fact finding Fourier coefficients for a certain function on the group $\Bbb Z/k\Bbb Z$.

This is a different approach to the idea in David Ullrich's answer.

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As long as $\frac nk\not\in\mathbb{Z}$, $$ \begin{align} \sum_{j=0}^{k-1}e^{2\pi ij\frac nk} &=\frac{e^{2\pi in}-1}{e^{2\pi i\frac nk}-1}\\ &=0 \end{align} $$ if $\frac nk\in\mathbb{Z}$, then $$ \begin{align} \sum_{j=0}^{k-1}e^{2\pi ij\frac nk} &=\sum_{j=0}^{k-1}1\\ &=k \end{align} $$ Thus, $$ \begin{align} \sum_{n=0}^\infty\frac1{(kn)!} &=\sum_{n=0}^\infty\overbrace{\left(\frac1k\sum_{j=0}^{k-1}e^{2\pi ij\frac nk}\right)}^{1\iff\frac nk\in\mathbb{Z}}\frac1{n!}\\ &=\frac1k\sum_{j=0}^{k-1}\sum_{n=0}^\infty\frac{\left(e^{2\pi ij/k}\right)^n}{n!}\\ &=\frac1k\sum_{j=0}^{k-1}e^{\large e^{2\pi ij/k}}\\ &=\frac1k\sum_{j=0}^{k-1}e^{\cos(2\pi j/k)+i\sin(2\pi j/k)}\\ &=\frac1k\sum_{j=0}^{k-1}e^{\cos(2\pi j/k)}\left(\cos(\sin(2\pi j/k))+i\sin(\sin(2\pi j/k))\right)\\ &=\frac1k\sum_{j=0}^{k-1}e^{\cos(2\pi j/k)}\cos(\sin(2\pi j/k)) \end{align} $$ The last step follows since the imaginary parts of the $j$ and $k-j$ terms cancel.