I was trying to find to which function the next series converges. $$ \sum_{n=1}^{\infty} \ln(n)z^n $$ If we take the polylogarithm function $Li_s(z)$ defined as $$ Li_s(s)=\sum_{n=1}^{\infty} \frac{z^n}{n^s} $$ Then it is easily seen that $$ \sum_{n=1}^{\infty} \ln(n)z^n = - \left( \frac{\partial}{\partial s}Li_s(z)\right)_{s=0} $$
Now, my question is how to calculate $ \frac{\partial}{\partial s}Li_s(z)$, using an integral representation for $Li$, such as $$ Li_s(z)=\frac{1}{\Gamma(s)}\int_{0}^{\infty} \frac{zt^{s-1}}{e^t-z} dt $$
Is there any nice solution to this? All my attempts are unclear about it, especially because of the derivative of $\Gamma(s)$.
You might write your series as $$ \sum_{n=1}^\infty \int_0^1 \dfrac{(n-1) z^n}{(n-1)u + 1}\; du = \int_{0}^{1}\!{\frac {{z}^{2}}{u+1} {\mbox{$_2$F$_1$}\left(2,{\frac {u+1}{u}};\,{\frac {2\,u+1}{u}};\,z\right)}} \, du $$ but I don't think you'll get a closed form for the integral.