# Compute Power Series Convergence to a function Consider the next power series sum_(n=1)^(oo) ln (n) z^n Find the convergence radius and a the function f to which the series converges. I have easily found that R=1 is the convergence radius, however I can not find the function. I was trying to found an elemental function with this power series expantion, but I have failed. Anyone knows such function and how to prove the convergence?

Compute Power Series Convergence to a function
Consider the next power series
$\sum _{n=1}^{\mathrm{\infty }}\mathrm{ln}\left(n\right){z}^{n}$
Find the convergence radius and a the function f to which the series converges.
I have easily found that $R=1$ is the convergence radius, however I can not find the function. I was trying to found an elemental function with this power series expantion, but I have failed. Anyone knows such function and how to prove the convergence?
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blogpolisft
Here is an approach.
Assume $|z|<1$
From $0\le \mathrm{ln}n, $\phantom{\rule{mediummathspace}{0ex}}n=1,2,3,\dots$, we get
$|\sum _{n=1}^{\mathrm{\infty }}\mathrm{ln}n\phantom{\rule{mediummathspace}{0ex}}{z}^{n}|\le \sum _{n=1}^{\mathrm{\infty }}\mathrm{ln}n\phantom{\rule{mediummathspace}{0ex}}|z{|}^{n}\le \sum _{n=1}^{\mathrm{\infty }}n|z{|}^{n}=\frac{|z|}{\left(1-|z|{\right)}^{2}}<\mathrm{\infty }$
Assume $|z|\ge 1$
Then
$\underset{z\to \mathrm{\infty }}{lim}|\mathrm{ln}n\phantom{\rule{mediummathspace}{0ex}}{z}^{n}|=\mathrm{\infty }\ne 0.$
This proves that our power series admits a radius of convergence equal to 1.
Let $z$ be a complex number such that $|z|<1$, then
$\begin{array}{}\text{(1)}& \sum _{n=1}^{\mathrm{\infty }}\mathrm{ln}n\phantom{\rule{mediummathspace}{0ex}}{z}^{n}=\frac{-z\mathrm{log}\left(1-z\right)-{z}^{2}\gamma \left(z\right)}{1-z}\end{array}$
where $\gamma \left(\cdot \right)$ denotes a special function called the generalized-Euler-constant function which has been studied by Jonathan Sondow and Petros Hadjicostas, amongst others.
###### Did you like this example?
Iris Vaughn
$R=\underset{n\to \mathrm{\infty }}{lim}\frac{\mathrm{ln}\left(n\right)}{\mathrm{ln}\left(n+1\right)}=\underset{n\to \mathrm{\infty }}{lim}\frac{n+1}{n}=1$
Consider now the polylogaritmic function given by
${\text{Li}}_{s}\left(z\right):=\sum _{n=1}^{\mathrm{\infty }}\frac{{z}^{n}}{{n}^{s}}.$
Then is easily seen that for $z\in \left\{z\in \mathbb{C}:|z|<1\right\}$ we have
$\sum _{n=1}^{\mathrm{\infty }}\mathrm{ln}\left(n\right){z}^{n}=-{\left(\frac{\mathrm{\partial }}{\mathrm{\partial }s}{\text{Li}}_{s}\left(z\right)\right)}_{s=0}$