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How to find this sum :

$\sum_{n=0}^\infty \dfrac{n^2}{2^n}$

$\sum_{n=0}^\infty \dfrac{n^2}{2^n}=\dfrac{1}{2}+\dfrac{4}{4}+\dfrac{9}{8}+\dfrac{16}{16}+\dfrac{25}{32}+\dfrac{36}{64}+\dfrac{49}{128}+\dots$

Now $\sum_{n=0}^\infty \dfrac{n}{2^n}\leqslant \sum_{n=0}^\infty \dfrac{n^2}{2^n}$

And I know that $\sum_{n=0}^\infty \dfrac{n}{2^n}=2$.

But how to find this sum ? I am confused.Please give some hints.

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    No you are wrong @MathMajor – 2017-01-19
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    Maple is evidently incorrect; c.f. the answer of Andre Nicolas here: http://math.stackexchange.com/questions/757263/how-to-find-answer-to-the-sum-of-series-sum-n-1-infty-fracn2n?rq=1 – 2017-01-19
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    Start with $1+x+x^2+\cdots=1/(1-x)$. Differentiate. Multiply by $x$. Differentiate again. Plug in $x=1/2$. All the steps valid, because we are well within the radius of convergence. Surprised if this is not a duplicate. – 2017-01-19
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    @JyrkiLahtonen;what is meant by "well within radius ...";will you please explain – 2017-01-19
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    It means that we are in an interval where the series and its derivatives converge uniformly, and termwise differentiation is thus justified (by the usual results on power series). – 2017-01-19

3 Answers 3

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Hint: For $|x|<1$ \begin{align*} \sum_{n=0}^{\infty }n^{2}x^{n}&=\sum_{n=0}^{\infty }n\left ( n+1 \right )x^{n}-\sum_{n=0}^{\infty }nx^{n} \\ &=x\sum_{n=0}^{\infty }n\left ( n+1 \right )x^{n-1}-x\sum_{n=0}^{\infty }nx^{n-1}\\ &=x\left ( \sum_{n=0}^{\infty }x^{n+1} \right )''-x\left ( \sum_{n=0}^{\infty }x^{n} \right )'\\ &=-\frac{x\left ( x+1 \right )}{\left ( x-1 \right )^{3}} \end{align*} then let $x=\dfrac{1}{2}$ you will get the answer.

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    how is the last line obtained i.e. the derivative ?? – 2017-01-19
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    @BenStokes Yes! Solve the inner series and take the derivative of its result. – 2017-01-19
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    yes i got your point;One more question :why is $|x|<1$ needed? – 2017-01-19
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    @Ben Stokes If $|x|\geq 1$, both of the series are divergent. – 2017-01-19
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Interesting... let's look at it.

$S=\frac{1}{2}+\frac{4}{4}+\frac{9}{8}+...$

$\frac{S}{2}=\frac{1}{4}+\frac{4}{8}+\frac{9}{16}+...$

$\frac{S}{2}=\frac{1}{2}+\frac{3}{4}+\frac{5}{8}+...$

$\frac{S}{4}=\frac{1}{4}+\frac{3}{8}+\frac{5}{16}+...$

$\frac{S}{4}=\frac{1}{2}+\frac{2}{4}+\frac{2}{8}+...=\frac{1}{2}+1=\frac{3}{2}$

So $S=6.$

EDIT: According to the comments this is wrong. I'll check over it (?)

EDIT2: Apparently that solution (which is for $\Sigma_{n=0}^{\infty}\frac{n}{2^n}$) gives 2 for the original summand. I can't read, oops. :\

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    Between the 2nd and 3rd lines, I personally lost you.... – 2017-01-19
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    Subtract the first term of the 2nd line from the second term from the 1st line. Then the second term of the 2nd line from the third term in the 1st line. And so on. The same thing happens on lines 4-5. – 2017-01-19
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    Ah, okay. Your solution might actually be fine then! And in fact I believe you *do* obtain $S/4 = \frac{3}{2}$ as you claimed, since you get $\frac{1}{2} + \frac{2}{4} + \frac{4}{8}$ ultimately. – 2017-01-19
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    Really nice solution +1 – 2017-01-19
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If $-1< x < 1$, we have, by differentiation and adding :

$$\begin{align} \sum_{n=0}^\infty x^n & = \frac 1{1-x} \implies & \\ \sum_{n=1}^\infty nx^{n-1} & = \frac 1{(1-x)^2} \implies \\ \sum_{n=1}^\infty nx^n & = \frac x{(1-x)^2} \implies & \small \\ \sum_{n=1}^\infty n^2x^{n-1} & = \frac {1+x}{(1-x)^3} \implies & \small \\ \sum_{n=1}^\infty n^2x^n & = \frac {x(1+x)}{(1-x)^3} \end{align}$$ Put $x=\frac{1}{2}$. We are done.

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    And this is because $-1 < x < 1$ is the [radius of convergence](https://en.wikipedia.org/wiki/Radius_of_convergence) of both the series $\sum_n x^n$ and $\sum_n nx^{n-1}$. – 2017-01-19