Solving $\sum_{x=1}^{\infty}{x^n}(\frac12)^x$

Question

For finding the $n$-th moment of $$f(x)=\begin{cases} 2^{-x}& x \in \mathbb{N} \\ 0 & \text{otherwise} \end{cases}$$ about $0$ , I need to solve $$\sum_{x=1}^{\infty}\frac{x^n}{2^x}$$ which I have no clue how it can be solved and where I should start. If it can't be solved for all $n$, is there any solutions for $n=1,2,3,\ldots$?

Answer 1

This is the definition of polylogarithm functions$$\sum_{x=1}^{\infty}{x^n}\left(\frac{1}{2}\right)^x=\text{Li}_{-n}\left(\frac{1}{2}\right)$$

Starting at $n=0$, the first values are $$\{1,2,6,26,150,1082,9366,94586,1091670,14174522,204495126\}$$ They correspond to sequence $A000629$ at $OEIS$ (have a look here).

A very good approximation is given by

$$\text{Li}_{-n}\left(\frac{1}{2}\right)\sim \frac{n!}{\left[\log(2)\right]^{n+1}}$$ (refering to Benoit Cloitre's comment in the linked page).