Find the value of this series

Question

what is the value of this series $$\sum_{n=1}^\infty \frac{n^2}{2^n} = \frac{1}{2}+\frac{4}{4}+\frac{9}{8}+\frac{16}{16}+\frac{25}{32}+\cdots$$

I really tried, but I couldn't, help guys?

Answer 1

Since $n^2 = 2\binom{n}{2}+\binom{n}{1}$ we have: $$ S=\sum_{n\geq 1}\frac{n^2}{2^n}=\left.\left(2\frac{x^2}{(1-x)^3}+\frac{x}{(1-x)^2}\right)\right|_{x=\frac{1}{2}}=\color{red}{6}.$$ As an alternative to the negative binomial series, we may also use: $$ S = 2S-S = \sum_{n\geq 1}\frac{n^2}{2^{n-1}}-\sum_{n\geq 1}\frac{n^2}{2^n}=1+\sum_{n\geq 1}\frac{(n+1)^2-n^2}{2^n}=2+2\sum_{n\geq 1}\frac{n}{2^n}$$ so that, in the same way: $$ \sum_{n\geq 1}\frac{n}{2^n} = T = 2T-T = \sum_{n\geq 1}\frac{n}{2^{n-1}}-\sum_{n\geq 1}\frac{n}{2^n} = 1+\sum_{n\geq 1}\frac{1}{2^n}=2 $$ and $S=2+2\cdot 2=\color{red}{6}$.

Answer 2

For $x$ such that $|x|<1$ we have $$f(x)=\sum_{n=0}^\infty x^n=\frac1{1-x}.$$ The derivative of $f$ is $$f'(x)=\sum_{n=0}^\infty nx^{n-1}=\frac{1}{\left(1-x\right)^2},$$ such that $$xf'(x)=\sum_{n=1}^\infty nx^n=\frac x{\left(1-x\right)^2}.$$ A second derivative gives $$xf''(x)+f'(x)=\sum_{n=1}^\infty n^2x^{n-1}=\frac1{\left(1-x\right)^2}+\frac{2x}{\left(1-x\right)^3}$$ so you deduce that $$\sum_{n=0}^\infty n^2x^n=\frac{x+x^2}{\left(1-x\right)^3}.$$ With $x=1/2$, this gives $\sum_n n^22^{-n}=6$.

EDIT: Another equivalent solution

Write the series for $-1<x<1$ $$f(x)=\sum_{n=0}^\infty x^n=\sum_{n=0}^\infty \mathrm e^{n\ln x}=\frac1{1-x}=\frac1{1-\mathrm e^{\ln x}}.$$ Thus, the series we look for is $$\frac{\mathrm d^2f}{\mathrm d(\ln x)^2}=\sum_{n=0}^\infty n^2\mathrm e^{n\ln x}=\frac{\mathrm d}{\mathrm d\ln x}\left(\frac{\mathrm e^{\ln x}}{\left(1-\mathrm e^{\ln x}\right)^2}\right)=\frac{2\mathrm e^{2\ln x}}{\left(1-\mathrm e^{\ln x}\right)^3}+\frac{\mathrm e^{\ln x}}{\left(1-\mathrm e^{\ln x}\right)^2}=\frac{x+x^2}{\left(1-x\right)^3}.$$ The result is obtained setting $x=1/2$ and we get $$\sum_{n=0}^\infty \frac{n^2}{2^n}=6.$$

Answer 3

Hint:

$$\sum_{n=1}^{\infty} \frac{n^2}{2^n} = \sum_{i=1}^{\infty} (2i - 1) \sum_{j=i}^{\infty} \frac{1}{2^j}.$$

Start by using the geometric series formula on $\displaystyle \sum_{j=1}^{\infty} \frac{1}{2^j}$ to simplify the double series into a singular series. Then you will have a series that looks like $\displaystyle \sum_{i=1}^{\infty} \frac{i}{2^i}$. Just as I broke your initial series with quadratic term $n^2$ into a double series with linear term $i$, you can break this series with linear term $i$ into a double series with constant term $c$.

In a clearer form:

\begin{align} &\frac{1}{2} + &\frac{4}{4} + &\frac{9}{8} + &\frac{16}{16} + \dots =\\ &\frac{1}{2} + &\frac{1}{4} + &\frac{1}{8} + &\frac{1}{16} + \dots +\\ & &\frac{3}{4} + &\frac{3}{8} + &\frac{3}{16} + \dots + \\ & & & \frac{5}{8} + &\frac{5}{16} + \dots + \\ & & & &\frac{7}{16} + \dots + \\ \end{align}

Notice that each of the sums are geometric series, which can be evaluated easily.

Sorry about the weird formatting, I don't know where these spaces are coming from.

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Answer 4

Assuming as @Soke does, your series is $$\sum_{n=1}^\infty \frac{n^2}{2^n}$$ we can use the geometric series in order to find a closed form solution. The trick is to take derivatives and multiply by $x$.

Note that $$f(x) = \sum_{n=1}^\infty x^n = \frac{x}{1-x}$$ yields a derivative of $$\sum_{n=1}^\infty n x^{n-1} = \frac{d}{dx} \frac{x}{1-x} = \frac{1}{(1-x)^2}.$$ We multiply it by $x$ to get: $$\sum_{n=1}^\infty nx^n = \frac{x}{(1-x)^2}.$$

If we take another derivative we find $$\sum_{n=1}^\infty n^2 x^{n-1} = \frac{d}{dx} \frac{x}{(1-x)^2} = \frac{(1-x)^2+2(1-x)x}{(1-x)^4}= \frac{1+x}{(1-x)^3}.$$

Finally we multiply by $x$ again: $$\sum_{n=1}^\infty n^2 x^n = \frac{x(1+x)}{(1-x)^3}$$ and the answer you are looking for corresponds to $x=1/2$.

Answer 5

Here's a bit of a twist on the second approach in Jack D'Aurizio's answer. It's main virtue (if any) is that it avoids explicitly evaluating the auxiliary sum $\sum{n\over2^n}$, getting it to drop out instead.

It's convenient to start the sum at $n=0$ instead of $n=1$. Borrowing Jack's notation, we have

$$S=\sum_{n=0}^\infty{n^2\over2^n},\quad T=\sum_{n=0}^\infty{n\over2^n},\quad\text{and}\quad2=\sum_{n=0}^\infty{1\over2^n}$$

Then

$$\begin{align} S-2&=\sum_{n=0}^\infty{n^2-1\over2^n}\\ &=\sum_{n=0}^\infty{(n+1)(n-1)\over2^n}\\ &=\sum_{m=1}^\infty{m(m-2)\over2^{m-1}}\\ &=2\sum_{m=1}^\infty{m^2\over2^m}-4\sum_{m=1}^\infty{m\over2^m}\\ &=2S-4T \end{align}$$

so

$$S=4T-2$$

which will turn out to be better written as

$$4S=16T-8$$

Likewise (and here's where the twist comes in), we have

$$\begin{align} S-8&=\sum_{n=0}^\infty{n^2-4\over2^n}\\ &=\sum_{n=0}^\infty{(n+2)(n-2)\over2^n}\\ &=\sum_{m=2}^\infty{m(m-4)\over2^{m-2}}\\ &=4\sum_{m=2}^\infty{m^2\over2^m}-16\sum_{m=2}^\infty{m\over2^m}\\ &=4(S-{1\over2})-16(T-{1\over2})\\ &=4S-16T+6 \end{align}$$

so

$$3S=16T-14$$

Subtracting this equation from the cleverly written $4S=16T-8$ leaves

$$S=6$$

As promised, we haven't bothered computing $T$ (although it's clear its value is easily obtained).