Compute Power Series Convergence to a function 
Consider the next power series
  $$
\sum_{n=1}^{\infty} \ln (n) z^n
$$
  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?
 A: Here is an approach.


*

*Assume $|z|<1$. 
From $0\leq\ln n<n$, $\:n=1,2,3,\ldots$, we get
$$
\left|\sum_{n=1}^{\infty} \ln n\: z^n\right|\leq  \sum_{n=1}^{\infty} \ln n\: |z|^n \leq \sum_{n=1}^{\infty} n |z|^n=\frac{|z|}{(1-|z|)^2}<\infty
$$

*Assume $|z|\geq1$. 
Then 
$$
\lim\limits_{z \to \infty}  \left|\ln n \:z^n\right| = \infty \neq 0.
$$
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  

$$
\sum_{n=1}^{\infty} \ln n\: z^n =\frac{-z\log (1-z)-z^2 \gamma (z)}{1-z} \tag1
$$

where $\gamma(\cdot)$ denotes a special function called the generalized-Euler-constant function  which has been studied by Jonathan Sondow and Petros Hadjicostas, amongst others. You will find the relation $(1)$ in this paper (p. $9$). 
A: Here is another answer. Clearly, the convergence radius is given by 
$$
R=\lim_{n \to \infty} \frac{\ln(n)}{\ln(n+1)} =\lim_{n \to \infty} \frac{n+1}{n}= 1
$$
Consider now the polylogaritmic function given by
$$
\text{Li}_s(z):=\sum_{n=1}^{\infty} \frac{z^n}{n^s}.
$$
Then is easily seen that for $z \in \left\{ z \in \mathbb{C} : |z|<1 \right\}$ we have
$$
\sum_{n=1}^{\infty} \ln(n) z^n = - \left(\frac{\partial}{\partial s} \text{Li}_s(z) \right)_{s=0}
$$
