Closed form of $\sum_{n=0}^{\infty} \frac{n^k}{2^n}$

Question

I'm trying to find a closed-form for $$\sum_{n=0}^{\infty} \frac{n^k}{2^n}$$ on natural $k$.

First I look at $\sum_{n=0}^{\infty}\frac{1}{2^n}=2$ and $\sum_{n=0}^{\infty}\frac{n}{2^n}=2$. After that, I realized that I needed calculus to get further, so I took the generating functions approach with that.

I have used the generating function $\sum_{n=0}^{\infty}nx^n=\frac{x}{(1-x)^2}=f(x)$. Differentiating yields $\sum_{n=0}^{\infty}n^2x^{n-1}=f'(x)$ and so $\sum_{n=0}^{\infty}n^2x^n=xf'(x)$. Differentiating over and over again gives:

$$\sum_{n=0}^{\infty}n^kx^n=x^{k-1}f^{(k-1)}(x)$$ $$\sum_{n=0}^{\infty}\frac{n^k}{2^n}=\frac{f^{(k-1)}\left(\frac12\right)}{2^{k-1}}.$$

From here, I was wondering if I could get more of a "closed-form" expression by finding a closed form for the term $f^{(k-1)}\left(\frac12\right)$ (remember that $f(x)=\frac{x}{(1-x)^2}$). I differentiated it manually a few times but couldn't find a pattern. If there is a nice form, is there any easier way to find it?

Thanks in advance.

Answer 1

This is the Hurwitz-Lerch transcendent Phi function:

$$\Phi \left(\frac{1}{2},-k,0\right)$$

Answer 2

Let \begin{eqnarray*} f_k &=& \sum_{n=0}^{\infty} \frac{n^k}{2^n} \\ F(x) &=& \sum_{k=0}^{\infty} f_k \frac{x^k}{k!}. \end{eqnarray*} Inverting the order of the plums gives (use the exponetial sum & geometric sum formulea) \begin{eqnarray*} F(x)= \frac{1}{1-e^x/2}. \end{eqnarray*} Half of this is the exponetial generating function of the Ordered Bell numbers see https://oeis.org/A000670