Sumfunction of $\sum_{n=0}^{\infty} \frac{z^{4n}}{(4n)!}$ and $\sum_{n=1}^{\infty} \frac{z^{-n}}{4n}$. Sumfunction of $\sum_{n=0}^{\infty} \frac{z^{4n}}{4n!}$ and   $\sum_{n=1}^{\infty} \frac{z^{-n}}{4n}$. 
The first one looks like the cosine hyperbolicus but the $4n!$ anoys me. I tried using $\cosh(z^2)$ or something in that form but that doesn't work, for the second problem i tried to find something in the form of log(1+z) but that did give me problems with the $(-1)^n$. usually if i try to find the sumfunction i go through this phase:
1) try to find sumfunctions which look alike
2) if one does not work, try to differentiate to find your solution! 
Any hints for the first sum function, the second function will eventually work i think.
 A: Since the second one is $\frac{1}{4} \sum_{k \geq 1} \frac{s^k}{k}$ for $s =\frac{1}{z}$, it's just Maclaurin series expansion for $\frac{1}{4} \log \frac{1}{1-s}$ for $|s|<1$.
A: The first one is $\frac12 (\cos z + \cosh z)$. (Every second term will cancel when you expand.)
A: \begin{align}
f(z)&=\sum_{n=0}^{\infty} \frac{z^{4n}}{(4n)!}\\
f'(z)&=\sum_{n=1}^{\infty} \frac{z^{4n-1}}{(4n-1)!}\\
f^{(2)}(z)&=\sum_{n=1}^{\infty} \frac{z^{4n-2}}{(4n-2)!}\\
f^{(3)}(z)&=\sum_{n=1}^{\infty} \frac{z^{4n-3}}{(4n-3)!}\\
f^{(4)}(z)&=\sum_{n=1}^{\infty} \frac{z^{4n-4}}{(4n-4)!}\\
&=\sum_{n=0}^{\infty} \frac{z^{4n}}{(4n)!}=f(x)\\
\end{align}
Now note that \begin{align}
f(0)&=1\\
f'(0)&=0\\
f^{(2)}(0)&=0\\
f^{(3)}(0)&=0\\
\end{align}
Therefore we proceed to solve the following boundary condition differential equation $$f^{(4)}-f(x)=0$$
Taking a Laplace transform we obtain
$$s^4F(s)-s^3f(0)-s^2f'(0)-sf^{(2)}(0)-f^{(3)}(0)-F(s)=0$$
or $$F(s)=\frac{s^3}{s^4-1}=\frac14\frac{1}{s-1}+\frac14\frac{1}{s+1}+\frac12\frac{s}{s^2+1}$$ thus taking an inverse Laplace transform we obtain 
$$f(z)=\frac{e^{-z}}{4}+\frac{e^{z}}{4}+\frac12 \cos z =\frac12(\cos z + \cosh z)$$
Note: Stefan also mentioned this above.
The second was treated already.
A: The first series can be approached with the discrete Fourier transform. Since
$$ f(n) = \frac{1}{4}\left(1^n+(-1)^n+i^n+(-i)^n\right)$$
is the indicator function of the integers being multiples of four, we have:
$$\sum_{n\geq 0}\frac{x^{4n}}{(4n)!}=\sum_{n\equiv 0\pmod{4}}\frac{x^n}{n!}=\frac{1}{4}\left(e^x+e^{-x}+e^{ix}+e^{-ix}\right) = \frac{1}{2}\left(\cos x+\cosh x\right).$$
The second series is just:
$$\frac{1}{4}\sum_{n\geq 1}\frac{(1/z)^n}{n}=-\frac{1}{4}\log\left(1-\frac{1}{z}\right).$$
A: For the second sum use this this general result:
$$
\sum_{n=1}^{+\infty}\frac{1}{nz^n},\;\;\;\; z\in\mathbb C\;.
$$
We immediately see that $|z|>1$, in order to have absolute convergence.
We recall first two results:
$\bullet\;\;$First:
$$
\log(1+z)=\sum_{n=1}^{+\infty}(-1)^{n+1}\frac{z^n}{n},\;\;\;\forall |z|<1
$$
$\bullet\;\;$Second:
$$
\prod_{n=0}^{+\infty}\left(1+z^{2^{n}}\right)=
\sum_{n=0}^{+\infty}z^{n}=\frac{1}{1-z},\;\;\;\forall |z|<1
$$
The last one can be proved, showing by induction that $\prod_{k=0}^{N}\left(1+z^{2^{k}}\right)=\sum_{k=0}^{2^{N+1}-1}z^{k}$.
Ok:
\begin{align*}
\sum_{n=1}^{+\infty}\frac{1}{nz^n}=&
\sum_{n=1}^{+\infty}\frac{1}{n}\left(\frac{1}{z}\right)^n\\
=&\underbrace{\sum_{k=0}^{+\infty}\frac{1}{2k+1}\left(\frac{1}{z}\right)^{2k+1}-
\sum_{k=1}^{+\infty}\frac{1}{2k}\left(\frac{1}{z}\right)^{2k}}_{\log\left(1+\frac{1}{z}\right)}+
2\sum_{k=1}^{+\infty}\frac{1}{2k}\left(\frac{1}{z}\right)^{2k}\\
=&\log\left(1+\frac{1}{z}\right)+
\sum_{k=1}^{+\infty}\frac{1}{k}\left(\frac{1}{z^2}\right)^{k}\\
=&\log\left(1+\frac{1}{z}\right)+
\log\left(1+\frac{1}{z^2}\right)+\cdots\\
=&\sum_{n=0}^{+\infty}\log\left(1+\frac{1}{z^{2^n}}\right)\\
=&\log\left(\prod_{n=0}^{+\infty}\left(1+\left(\frac{1}{z}\right)^{2^n}\right)\right)\\
=&\log\left(\frac{z}{z-1}\right)
\end{align*}
