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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.

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  • $\begingroup$ Have you tried plugging in approximants of your function to WolframAlpha or some other computer algebra system? $\endgroup$ – Jose Arnaldo Bebita-Dris May 28 '15 at 9:27
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Your sum can be expressed in term of derivatives of a polylogarithm with respect to order. For all $|z|<1$, we have: $$\begin{align} (-1)^n\frac{ \partial^n}{\partial s^n} \left.\operatorname{Li}_s(z)\right|_{s=1} &= (-1)^n \frac{\partial^n}{\partial s^n} \left.\left(\sum_{k=1}^\infty \frac{z^k}{k^s}\right)\right|_{s=1} \\ &= (-1)^n \sum_{k=1}^\infty z^k \frac{\partial^n}{\partial s^n}\left.\left(\frac{1}{k^s}\right)\right|_{s=1} \\ &= (-1)^n \sum_{k=1}^\infty z^k\left.\left(\frac{(-1)^n\log^n(k)}{k^s}\right)\right|_{s=1} \\ &= \sum_{k=1}^\infty \frac{z^k \log^n(k)}{k}. \end{align}$$

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