Closed form of $\displaystyle\sum_{n=1}^\infty x^n\ln(n)$ Is there a closed form of this : 
$$\sum_{n=1}^\infty x^n\ln(n),$$
 where $|x|<1$.
Thanks in advance.

It admits a very simple approximation using Euler-Maclaurin formula:
$$
I_k = \int_{1}^{\infty}x^k \log x dx = -\frac{x^{k+1}}{k+1}\Bigg|_{1}^{\infty}- \int_{1}^{\infty}\frac{x^{k}dx}{k+1} = O\bigg(\frac{1}{k^2}\bigg)
$$
This is since $\lim_{m \to 1} \lim _{n \to \infty} \frac{1}{2}(f(n)-f(m))=0$
