Does this Sum follow a general trend? 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?
 A: For $k>2$, I think that you can only face hypergeometric functions.
To show you the patterns
$$f(x,3)=\frac{x^3}{3!}  \,
   _0F_5\left(;\frac{4}{6},\frac{5}{6},\frac{7}{6},\frac{8}{6},\frac{9}{6};\frac{x^
   6}{6^6}\right)$$
$$f(x,4)=\frac{x^4}{4!}  \,
   _0F_7\left(;\frac{5}{8},\frac{6}{8},\frac{7}{8},\frac{9}{8},\frac{10}{8},\frac{11
   }{8},\frac{12}{8};\frac{x^8}{8^8}\right)$$
$$f(x,5)=\frac{x^5 }{5!} \,
   _0F_9\left(;\frac{6}{10},\frac{7}{10},\frac{8}{10},\frac{9}{10},\frac{11}{10},
\frac{12}{10},\frac{13}{10},\frac{14}{10},\frac{15}{10};\frac{x^{10}}{10^{10}}\right)$$
$$f(x,6)=\frac{x^6}{6!}  \,
   _0F_{11}\left(;\frac{7}{12},\frac{8}{12},\frac{9}{12},\frac{10}{12},\frac{11}{12},
\frac{13}{12},\frac{14}{12},\frac{15}{12},\frac{16}{12},\frac{17}{12},\frac{18}{12};\frac{
   x^{12}}{12^{12}}\right)$$
$$\color{blue}{f(x,k)=\frac{x^k}{k!}\,
   _0F_{2k-1}\left(;\frac{k+1}{2k},\frac{k+2}{2k},\cdots, \frac{k+2k}{2k};\frac{
   x^{2k}}{(2k)^{2k}}\right)}$$
$$f(1,k)=\frac 1 {k!}$$
A: 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.
