Infinte series Sum of $\sum_{k=1}^{\infty}\frac {k^n}{2^k}$

Question

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.

Answer 1

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 $\operatorname{Li}_{-n}(\frac{1}{2})$.

Answer 2

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 $P$.