I recently came across the following function
$$\sum_{k=1}^\infty(\log(k))^n\frac{z^k}{k}$$
I found it while dealing with the polylogarithm function, $Li_n (z)$ (Notice that if instead of $(\log(k))^n$ we had $k^n$ then the above expression would become $Li_{1-n}(z)$. Still these functions are quite different.)
I was wondering if this function is known, and if there are good numerical approximations to estimate it?
Thank you in advance for your help.
Your sum can be expressed in term of derivatives of a polylogarithm with respect to order. For all $|z|<1$, we have: $$\begin{align} (-1)^n\frac{ \partial^n}{\partial s^n} \left.\operatorname{Li}_s(z)\right|_{s=1} &= (-1)^n \frac{\partial^n}{\partial s^n} \left.\left(\sum_{k=1}^\infty \frac{z^k}{k^s}\right)\right|_{s=1} \\ &= (-1)^n \sum_{k=1}^\infty z^k \frac{\partial^n}{\partial s^n}\left.\left(\frac{1}{k^s}\right)\right|_{s=1} \\ &= (-1)^n \sum_{k=1}^\infty z^k\left.\left(\frac{(-1)^n\log^n(k)}{k^s}\right)\right|_{s=1} \\ &= \sum_{k=1}^\infty \frac{z^k \log^n(k)}{k}. \end{align}$$