In my personal works I came across a sum that looks like $$f(x,k)= \sum_{n=0}^\infty \frac{x^{k(2n+1)}}{(k(2n+1))!}. $$ We have: $$f(x,1)= \sinh(x)$$ and $$f(x,2)= \frac{1}{2} (\cosh(x) + \cos(x)).$$
Rr-writing these in their exponential forms we get $$f(x,1)= \frac{e^x}{2} + \frac{e^{-x}}{2}$$ And again, $$f(x,2) = \frac{e^x}{4} + \frac{e^{-x}}{4} - \frac{e^{ix}}{4} - \frac{e^{-ix}}{4}$$
My question is, is there a general trend that can be found for any natural number, $k$?
How would one evaluate $f(x,3)$ or $f(x,4)$, for example?
As Somos commented, one can also use the closed formula for multi-section series: One gets \begin{align} \sum_{n=0}^\infty \frac{z^{2kn+k}}{(2kn+k)!} &= \frac{1}{2k} \sum_{j=0}^{2k-1} (-1)^j \exp( e^{i \pi j/k} z) \\ &= \frac{1}{k} \sum_{j=0}^{k-1} (-1)^j \frac{1}{2}(\exp( e^{i \pi j/k} z) + (-1)^k \exp( - e^{i \pi j/k} z)). \end{align} Depending on $k$, the last term in the sum can be rewritten as $\cosh(...)$, resp. $\sinh(...)$ as follows: If $k$ is even, we get $$\sum_{n=0}^\infty \frac{z^{2kn+k}}{(2kn+k)!} = \frac{1}{k} \sum_{j=0}^{k-1} (-1)^j \cosh( e^{i \pi j/k} z).$$ If $k$ is odd, we get $$\sum_{n=0}^\infty \frac{z^{2kn+k}}{(2kn+k)!} = \frac{1}{k} \sum_{j=0}^{k-1} (-1)^j \sinh( e^{i \pi j/k} z).$$ This identity is a generalization of the pattern you have found.