Sum of the series $k^2/2^k$

Question

I am trying to calculate the following series but I don't know where to start.

$\sum_{k=1}^{\infty} \frac{k^2}{2^k}$

I appreciate any advice I can get:)

Answer 1

I'll derive a solution, similar to a previous post which you can find here.

For instance, you can consider the geometric series $\sum_{k\geq 0}r^k=(1-r)^{-1}$. Start by taking the first derivative of the series. Since the derivative is pushed into the sum, you will have $$\sum_{k\geq 1}kr^{k-1}=\frac{1}{(1-r)^2}.$$ Notice the index change since $r$'s power cannot be smaller than 0. Now, multiply on both sides by $r$ to obtain $$r\sum_{k\geq 1}kr^{k-1}=\sum_{k\geq 1}kr^k=\frac{r}{(1-r)^2}.$$ Now repeat the same process, so take the derivative of the above expression to obtain $$\begin{align} \sum_{k\geq 1}k^2r^{k-1}&=\left(\frac{r}{(1-r)^2}\right)^{'}\\ &=\frac{(1-r)^2-r[(2(-1)(1-r)]}{(1-r)^4}\\ &=\frac{(1-r)^2+2r(1-r)}{(1-r)^4}\\ &=\frac{1+r}{(1-r)^3},\end{align}$$ so that if you multiply by $r$ again on both sides, you get $$\sum_{k\geq 1}k^2r^k=\frac{r(1+r)}{(1-r)^3}.$$ If you now let $r=1/2$, you get $$\sum_{k\geq 1}2^{-k}k^2=\frac{3/4}{1/8}=6.$$ Hope this helps.