1. Question
Could anyone recommend a useful method for approaching the following series?
$$\sum_{k=2}^{\infty}\zeta'(k)x^{k-1}$$ Where $\zeta(z)$ is the Riemann Zeta function.
I've seen that there are many series involving this function, but I haven't found any that involve its derivative, the analogue to this series without the derivative is
$$\sum_{k=2}^{\infty}\zeta(k)x^{k-1}=-\gamma-\psi(1-x)$$
which has a closed form solution, I was curious to know if there was a closed form for the version with the derivative too
Any suggestions are welcome
Update
I arrived at this formulation: $$\sum_{n=2}^{\infty}\zeta'\left(n\right)x^{n-1}=-\sum_{n=2}^{\infty}\sum_{k=1}^{\infty}\frac{\ln\left(n\right)}{k^{n}}x^{n-1}=-\sum_{k=1}^{\infty}\sum_{n=2}^{\infty}\frac{\ln\left(n\right)}{k^{n}}x^{n-1}=\frac{1}{x}\sum_{k=1}^{\infty}\text{Li}_{0}^{\left(1,0\right)}\left(\frac{x}{k}\right)$$
Where:
$$\text{Li}_{0}^{\left(1,0\right)}\left(z\right):=\left.\frac{\partial}{\partial \nu}\text{Li}_{\nu}\left(z\right)\right|_{\nu=0}=-\sum_{k=1}^{\infty}\ln\left(k\right)z^{k}$$
So
$$\sum_{n=2}^{\infty}\zeta'\left(n\right)x^{n-1}=\frac{1}{x}\frac{\partial}{\partial \nu}\left.\left[\sum_{k=1}^{\infty}\text{Li}_{\nu}\left(\frac{x}{k}\right)\right]\right|_{\nu=0}$$
This is not a full answer as it does not really provide a "closed form" representation of the series, though I highly suspect no such closed form exists. Rather, here is short discussion on some different forms of this and a related series that could be of some interest to the asker.
Firstly, we have the following easy to prove identity \begin{equation} \boxed{\sum_{k=2}^\infty \zeta'(k)x^k=\sum_{n=2}^\infty\frac{\log(n)x^2}{n(n-x)}} \end{equation} for $|x|<2$ which provides the analytic continuation of the first sum to the whole complex plane except for integers $x\geq 2$.
To see this, write
\begin{equation} \begin{split} \sum_{k=2}^\infty \zeta'(k)x^k&=\sum_{k=2}^\infty\left[\sum_{n=2}^\infty\frac{\log(n)}{n^k}\right]x^k\\ &=\sum_{n=2}^\infty\log(n)\sum_{k=2}^\infty\left(\frac{x}{n}\right)^k\\ &=\sum_{n=2}^\infty\frac{\log(n)x^2}{n(n-x)}\\ \end{split} \end{equation}
The related series representing the even part of the series discussed above admits the following representation \begin{equation} \boxed{\sum_{n=1}^\infty\zeta'(2n)y^{2n}\!=\!\frac1{2}[\gamma+\log(2\pi)](1-\pi y\cot(\pi y))+\pi^2y^2\csc(\pi y)\!\!\int_0^1\!\!\!\sin(\pi y(2x-1))\log\Gamma(x)dx} \end{equation} for $|y|<2$ (for $y\in\{-1,0,1\}$, the right-hand expression can be interpreted as a limit). Note that the integral converges for any complex $y$, so this identity again provides an analytic continuation of the sum to the whole complex plane except for integers $y$ with $|y|\geq 2$.
We begin with the following well known Fourier series
\begin{equation} \begin{split} \log\Gamma(x)&=\frac12\log(2\pi)+\sum_{k=1}^\infty \frac{1}{2k}\cos(2k\pi x)+\sum_{k=1}^\infty \frac{1}{\pi k}[\gamma+\log(2k\pi)]\sin(2k\pi x)\\ B_{2n-1}(x)&=2(2n-1)!(-1)^{2n-1}\sum_{k=1}^\infty \frac{\sin(2k\pi x)}{(2k\pi)^{2n-1}}\\ \end{split} \end{equation} for $0<x<1$ and $n\geq 1$, where $B_m(x)$ is the $m$-th Bernoulli polynomial. Now, by using the above series and appealing to the orthogonality relationships between sin and cos, we may show that. \begin{equation} \int_0^1 B_{2n-1}(x)\log\Gamma(x)dx=2(-1)^n\frac{(2n-1)!}{(2\pi)^{2n}}[(\gamma+\log(2\pi))\zeta(2n)-\zeta'(2n)] \end{equation} for $n\geq 1$. Finally, using the generating function identity
$$\frac{t \sinh((x-1/2)t)}{2\sinh(t/2)}=\sum_{n=1}^\infty B_{2n-1}(x)\frac{t^{2n-1}}{(2n-1)!}$$
we have that \begin{equation} \begin{split} \int_0^1 \frac{t\sinh((x-1/2)t)}{2\sinh(t/2)}\log\Gamma(x)dx&=2\sum_{n=1}^\infty (-1)^n[(\gamma+\log(2\pi))\zeta(2n)-\zeta'(2n)]\frac{t^{2n-1}}{(2\pi)^{2n}}\\ &=[\gamma+\log(2\pi)]\left(\frac1{t}+\frac{i}{2}\cot\left(\frac{t}{2i}\right)\right)-\frac2{t}\sum_{n=1}^\infty\zeta'(2n)\left(\frac{t}{2\pi i}\right)^{2n}\\ \end{split} \end{equation} Performing the change of variable $t\mapsto 2\pi iy$, we have that \begin{equation} \int_0^1 \frac{\pi i y \sin(\pi y(2x-1))}{\sin(\pi y)}\log\Gamma(x)dx=[\gamma+\log(2\pi)]\left(-\frac{i}{2\pi y}+\frac{i}{2}\cot(\pi y)\right)+\frac{i}{\pi y}\sum_{n=1}^\infty\zeta'(2n)y^{2n} \end{equation} taking the imaginary part of both sides, and rearranging, we have that \begin{equation} \sum_{n=1}^\infty\zeta'(2n)y^{2n}\!=\!\frac1{2}[\gamma+\log(2\pi)](1-\pi y\cot(\pi y))+\pi^2y^2\csc(\pi y)\!\!\int_0^1\!\!\!\sin(\pi y(2x-1))\log\Gamma(x)dx \end{equation} as desired.