Interesting Series with Zeta Function I was trying to find another representation for the value of an integral when I found the following series: 
$$f (z)=\sum_{n \in \Bbb N} (-z)^{n-1}\frac {(2^n-1)}{2^n}\zeta (n+1) $$
For $|z|<1$ and $\zeta $ is the zeta function. 
I was wondering if this series could be expressed in a closed form as it looks similar to the series expansion of the polygamma function and whether the convergence region could be extended beyond the unit disk.
EDIT: it appears to be something of the form $$f (z)= \frac {\zeta (2)}{z}+\frac {1}{z}(\psi^{(0)}(1-\frac {1}{z})-\psi^{(0)}(1-\frac {1}{2z}))$$
 A: If you define:
$$g(z)=\sum_{n=1}^\infty z^{n+1}\zeta(n+1)$$
Then we have:
$$f(z)=\frac{g(-z)-2g(-\frac{z}{2})}{z^2}$$
So it suffices to study the simpler looking function $g$ if we want to understand $f$.
Now expanding $g$ and interchanging the order of summation gives us:
$$g(z)=\sum_{n=1}^\infty z^{n+1}\zeta(n+1)=\sum_{n=1}^\infty\sum_{m=1}^\infty (\frac{z}{m})^{n+1}=\sum_{m=1}^\infty \frac{(z/m)^2}{1-(z/m)}=z\left(\sum_{m=1}^\infty\frac{z}{m(m-z)}\right)\\=-z\left(\gamma -\gamma+\sum_{m=1}^\infty\frac{-z}{m(m-z)}\right)=-z\gamma-z\psi(1-z)$$
Which if you then back substitute this formula into our functional equation, you can express $f$ in terms of the digamma function.
A: Indeed this can be expressed in terms of the digamma function. With
$$
\psi(z+1)= -\gamma -\sum_{n=1}^\infty(-z)^n\zeta(n+1)\;,
$$
we have
\begin{align}
f(z)&=\sum_{n=1}^\infty(-z)^{n-1}\zeta(n+1)-\sum_{n=1}^\infty\frac{(-z)^{n-1}}{2^n}\zeta(n+1)\\
&=\frac{\psi(z+1)}z-\frac{\psi\left(\frac z2+1\right)}z\;.
\end{align}
A: You are perfectly correct thinking about  the similarity to the series expansion of the polygamma function.
In fact $$\sum_{n=1}^\infty (-z)^{n-1}\frac {(2^n-1)}{2^n}\zeta (n+1)=\frac{\psi (z+1)-\psi \left(\frac{z}{2}+1\right)}{z}$$
I was typing when joriki's answer came (I was using the same arguments). I wonder where I made a mistake since my result slightly differs from his.
