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Find the sum of $$\sum_{n=1}^{\infty} \frac{n^2}{2^n}$$ I know I need to manipulate the power series $\sum_{n=0}^{\infty}x^n$ with $x = \frac{1}{2}$, but I'm not sure how. Would differentiating it twice help?

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2 Answers 2

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Hint

Considering $$S=\sum_{i=1}^\infty n^2 x^n=\sum_{i=1}^\infty (n^2-n+n) x^n=\sum_{i=1}^\infty n(n-1) x^n+\sum_{i=1}^\infty n x^n$$ $$S=x^2\sum_{i=1}^\infty n(n-1) x^{n-2}+x\sum_{i=1}^\infty n x^{n-1}$$ $$S=x^2\left( \sum_{i=1}^\infty x^{n}\right)''+x\left( \sum_{i=1}^\infty x^{n}\right)'$$

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HINT:

If $S(x)=\sum_{n=1}^\infty x^n=\frac{1}{1-x}$ for $|x|<1$, then

$$x(xS'(x))'=\sum_{n=1}^\infty n^2x^n$$

Then, let $x=\frac12$.

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  • $\begingroup$ Please let me know how I can improve my answer. I really want to give you the best answer I can. -Mark $\endgroup$
    – Mark Viola
    Dec 2, 2016 at 14:33

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