Evaluate $\sum\limits_{n=1}^{\infty}\frac{n^2}{2^n}$

Question

A user just posted a similar question (What is the summation of $\sum_{k=0}^{\infty}\frac{k}{2^k}$?) with $n$ instead of $n^2$, and this question got some original answers, so I would like to propose the question:

Find the sum $$\sum\limits_{n=1}^{\infty}\frac{n^2}{2^n}.$$

I am really interested to see what original answers are given to this question. There are number of different approaches, some I havent seen or though of before. Thus although the question may be unoriginal, some of the solutions are not. Thus it is inappropriate to close this question.

Answer 1

We know that $$\sum_{n\geq 0} r^n = \frac{1}{1-r}$$ Now, differentiate both sides: $$\sum_{n\geq 0} \left(r^{n}\right) = \left(\frac{1}{1-r}\right)'$$ $$\sum_{n\geq 0} nr^{n-1} = \frac{1}{(1-r)^2}$$ Differentiating again: $$\sum_{n\geq 0} (nr^{n-1})' = \left(\frac{1}{(1-r)^2}\right)'$$ $$\sum_{n\geq 0} n(n-1)r^{n-2} = \frac{2}{(1-r)^3}$$ $$\sum_{n\geq 0} n^2r^{n-2}-\sum_{n\geq 0} nr^{n-2} = \frac{2}{(1-r)^3}$$ $$\sum_{n\geq 0} n^2r^{n-2} =\sum_{n\geq 0} nr^{n-2}+ \frac{2}{(1-r)^3}$$ Now multiple both sides by $r^2$: $$\sum_{n\geq 0} n^2r^{n} =\sum_{n\geq 0} nr^{n}+ \frac{2r^2}{(1-r)^3}$$ $$\sum_{n\geq 0} n^2r^{n} =\frac{r}{(1-r)^2}+ \frac{2r^2}{(1-r)^3}$$

Answer 2

\begin{align*} \sum\limits_{n=1}^{N}\frac{(n+1)^2}{2^n}-\sum\limits_{n=1}^{N}\frac{n^2}{2^n}&=\sum\limits_{n=1}^{N}\frac{2n+1}{2^n}\\ 2\sum\limits_{n=1}^{N}\frac{(n+1)^2}{2^{n+1}}-\sum\limits_{n=1}^{N}\frac{n^2}{2^n}&=2\sum\limits_{n=1}^{N}\frac{n}{2^n}+\sum\limits_{n=1}^{N}\frac{1}{2^n}\\ 2\sum\limits_{n=2}^{N}\frac{n^2}{2^n}-\sum\limits_{n=1}^{N}\frac{n^2}{2^n}&=2\sum\limits_{n=1}^{N}\frac{n}{2^n}+\sum\limits_{n=1}^{N}\frac{1}{2^n}\\ 2\left(\sum\limits_{n=1}^{N}\frac{n^2}{2^n}-\frac{1}{2}+\frac{(N+1)^2}{2^{N+1}}\right)-\sum\limits_{n=1}^{N}\frac{n^2}{2^n}&=2\sum\limits_{n=1}^{N}\frac{n}{2^n}+\sum\limits_{n=1}^{N}\frac{1}{2^n}\\ \sum\limits_{n=1}^{N}\frac{n^2}{2^n}&=2\sum\limits_{n=1}^{N}\frac{n}{2^n}+\sum\limits_{n=1}^{N}\frac{1}{2^n}+1-\frac{(N+1)^2}{2^N}\\ \sum\limits_{n=1}^{\infty}\frac{n^2}{2^n}&=2\sum\limits_{n=1}^{\infty}\frac{n}{2^n}+\sum\limits_{n=1}^{\infty}\frac{1}{2^n}+1\\ &=2(2)+(1)+1\\ &=6 \end{align*}

Answer 3

Using a probabilistic interpretation as the second moment $E(X^2)$ of a random varaible $X$ with a geometric distribution $\text{Geo}(p=1/2$) (beginning at $1$ !), it suffices to know (https://en.wikipedia.org/wiki/Geometric_distribution) its mean and its variance, resp. :

$$E(X)=\tfrac{1}{p} \ \ \text{and} \ \ E(X^2)-E(X)^2=\tfrac{1-p}{p^2}$$

to be able to conclude that:

$$E(X^2)=(\tfrac{1}{p})^2+\tfrac{1-p}{p^2}=\tfrac{2-p}{p^2} \ \ \text{with} \ \ p=\tfrac12,$$ finally giving the sum : $6.$

Answer 4

Following my telescoping sum strategy from the similar question mentioned in your question -

$$\sum_{k=0}^m \left( \frac{(k+1)^2}{2^{k+1}} - \frac{k^2}{2^k} \right) = \frac{(m+1)^2}{2^{m+1}}$$

$$ \Longrightarrow \sum_{k=0}^m \frac{-k^2+2k+1}{2^{k+1}} = \frac{(m+1)^2}{2^{k+1}}$$

$$ \Longrightarrow \frac{1}{2} \sum_{k=0}^m \frac{k^2}{2^k} = \sum_{k=0}^m \frac{k}{2^k} + \frac{1}{2} \sum_{k=0}^m \frac{1}{2^k} - \frac{(m+1)^2}{2^{m+1}}$$

And using the partial summation for the first term of the RHS (found in my answer to the similar question), we have

$$\sum_{k=0}^m \frac{k^2}{2^k} = 2 \left( 2- \frac{m+2}{2^m} \right) + \left( 2-\frac{1}{2^m} \right) - \frac{(m+1)^2}{2^m}$$

and taking the limit as $m$ approaches infinity yields

$$\sum_{k=0}^{\infty} \frac{k^2}{2^k} = 6.$$

Answer 5

By stars and bars we have $$ \frac{1}{(1-x)^{k+1}}=\sum_{n\geq 0}\binom{n+k}{k} x^n $$ for any $x\in(-1,1)$, in particular $$ \sum_{n\geq 0}\binom{n+k}{k}\frac{1}{2^n} = 2^{k+1} $$ for any $k\in\mathbb{N}$. Since $n^2 = 2\binom{n+2}{2}-3\binom{n+1}{1}+\binom{n+0}{0}$ we have $$ \sum_{n\geq 0}\frac{n^2}{2^n} = 2\cdot 2^3 - 3\cdot 2^2 + 2^1 = \color{red}{6}. $$