TITLE: Behavior of a sum on the boundary of convergence/divergence
QUESTION [5 upvotes]: I am seeking the behavior of the sum
$$
\sum_{n = 1}^{\infty}\frac{\ln\left(n\right)}{n}\,z^{n}
$$
as $z \to 1^{-}$.  I know that at $z = 1$, it diverges.  So, ideally, I would like to know how it behaves as a function of $z$ in a left neighborhood of $1$.  I thought that in some way it may be related to the Polylogarithm function, but I can't seem to figure it out.
Thanks in advance for any help.

REPLY [3 votes]: We have $\mathcal{L}(\log x)=-\frac{\gamma+\log(s)}{s}$, hence $\mathcal{L}^{-1}\left(\frac{\log s}{s}\right) =-\left(\log(x)+\gamma\right)$. Additionally
$$ \mathcal{L}\left(\frac{\log x}{x}\right) = \frac{\pi^2}{12}+\frac{\left(\gamma+\log x\right)^2}{2},$$
hence a pretty good approximation is
$$ \sum_{n\geq 1}\frac{\log n}{n}\,x^n \sim \frac{(\gamma+\log(1-x))^2}{2}\quad \text{for }x\to 1^- $$
by just rephrasing Antonio Vargas' approach in terms of $\mathcal{L}/\mathcal{L}^{-1}$ and Hardy Littlewood's tauberian theorem.