Infinte series Sum of $\sum_{k=1}^{\infty}\frac {k^n}{2^k}$ Consider the infinite some over all "k" from 1 to infinity: $$\sum_{k=1}^{\infty}\frac {k^n}{2^k}$$ for a given "n" we get different results such as:


*

*n = 0; sum = 1

*n = 1; sum = 2

*n = 2; sum = 6

*n = 3; sum = 26

*n = 4; sum = 150

*n = 5; sum = 1082

*n = 6; sum = 9366


My question - is there a general formula? - I cannot find one.
 A: The polylogarithm $\operatorname{Li}_s(z):=\sum_{k\ge 1}\frac{z^k}{k^s}$ is so-called because $\operatorname{Li}_1(z)=-\ln(1-z)$ for $|z|<1$, and $\partial_{\ln z}\operatorname{Li}_s(z)=\operatorname{Li}_{s-1}(z)$. This series definition of $\operatorname{Li}_s(z)$ converges if $|z|<1$, or $|z|=1$ and $\Re s>1$. (We can improve on this with analytic continuation.) Your expression is $\operatorname{Li}_{-n}(\frac{1}{2})$.
A: Let 
$$S_0(x):=\sum_{k=1}^\infty x^k,$$
$$S_1(x):=x\frac{dS_0}{dx}=\sum_{k=1}^\infty kx^k,$$
$$\cdots$$
$$S_n(x):=x\frac{dS_{n-1}}{dx}=\sum_{k=1}^\infty k^nx^k.$$
We have
$$S_0(x)=\frac x{1-x}\to S_0\left(\frac12\right)=1$$
$$S_1(x)=\frac{x}{(1-x)^2}\to S_1\left(\frac12\right)=2$$
$$S_2(x)=\frac{x(x+1)}{(1-x)^3}\to S_2\left(\frac12\right)=6$$
and more generally
$$S_n(x)=\frac{P_n(x)}{(1-x)^n}.$$
Then
$$S_{n+1}(x)=\frac{P_{n+1}(x)}{(1-x)^{n+1}}=x\frac{P'_n(x)(1-x)^n+nP_n(x)(1-x)^{n-1}}{(1-x)^{2n}}=\frac{x(1-x)P'_n(x)+nxP_n(x)}{(1-x)^{n+1}}$$
gives a recurrence relation for $P$.
