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

Question

How to go about summing $\sum_{k=0}^{\infty} \frac{k^2}{2^k}$.

Answer 1

Write $c^k$ as the forward difference of $c^k/(c-1)$ with $c=1/2.$ Find the sum from $1$ to $n$ of the given expression by using the summation by parts formula twice. Then take the limit as $n$ goes to infinity.

Answer 2

Start with $\sum_{k=0}^\infty x^k=\frac{1}{1-x}$ for $|x|\lt 1$

$\frac{d}{dx}\frac{1}{1-x}=\sum_{k=0}^\infty kx^{k-1}=\frac{1}{(1-x)^2}$

$\frac{d}{dx}\frac{x}{(1-x)^2}=\sum_{k=0}^\infty k^2x^{k-1}=\frac{1+x}{(1-x)^3}$

Therefore $\sum_{k=0}^\infty k^2x^k=\frac{x(1+x)}{(1-x)^3}$

The question is the value of this last expression when $x=\frac{1}{2}$. Ans $=6$.

Answer 3

$$\sum_{k=0}^{\infty} x^k=\frac{1}{1-x}, |x|<1.$$ Differentate w.r. $x$, then $$\sum_{k=0}^{\infty} k x^k =\frac{x}{(1-x)^2}.$$ Differentiate again w.r. t. $x$, we get By putting $x=1/2$, we get $$\sum_{k=0}^{\infty} k^2 x^k= \frac{x(1+x)}{(1-x)^3}.$$ $$\sum_{k=0}^{\infty} \frac{k^2}{2^k}=6$$

Answer 4

Since $$\sum_{n=1}^{\infty}\frac{n}{2^n}=\sum_{n=0}^{\infty}\frac{n+1}{2^{n+1}}=\sum_{n=1}^{\infty}\frac{n}{2^{n+1}}+\sum_{n=0}^\infty\frac{1}{2^{n+1}}=\frac{1}{2}\sum_{n=1}^{\infty}\frac{n}{2^{n}}+1$$

implies that

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

we can use the same trick to see that

\begin{align}\sum_{n=1}^{\infty}\frac{n^2}{2^n}=\sum_{n=0}^{\infty}\frac{(n+1)^2}{2^{n+1}}&=\frac{1}{2}\sum_{n=1}^{\infty}\frac{n^2}{2^{n}}+\sum_{n=1}^{\infty}\frac{n}{2^{n}}+\sum_{n=0}^\infty\frac{1}{2^{n+1}}\\&= \frac{1}{2}\sum_{n=1}^{\infty}\frac{n^2}{2^{n}}+3 \end{align}

therefore $$\sum_{n=0}^{\infty}\frac{n^2}{2^{n}}=\sum_{n=1}^{\infty}\frac{n^2}{2^{n}}=6$$