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This question already has an answer here:

How to evaluate those infinite series? How are they called?

$$ \sum_{k=1}^\infty \frac{k}{2^k} \quad \text{and} \quad \sum_{k=1}^\infty \frac{k^2}{2^k} $$

I'm really sorry for asking, but I can't figure out how to google such stuff when I even don't know the names/categories.

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marked as duplicate by Travis, Clement C., Aditya Hase, Santosh Linkha, Namaste Dec 17 '14 at 15:13

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    $\begingroup$ "differentiated geometric series" $\endgroup$ – David Mitra Dec 17 '14 at 14:34
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    $\begingroup$ This may help you compute the sums. $\endgroup$ – David Mitra Dec 17 '14 at 14:35
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    $\begingroup$ @DavidMitra Thanks a lot, that's all I need to know ;-) $\endgroup$ – NoBackingDown Dec 17 '14 at 14:37
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$$\sum_{k=1}^{\infty}x^k=\frac{x}{1-x}\rightarrow\frac{d}{dx}\sum_{k=1}^{\infty}x^k=\sum_{k=1}^{\infty}kx^{k-1}=\frac{1}{(1-x)^2}\rightarrow \sum_{k=1}^{\infty}kx^{k}=\frac{x}{(1-x)^2}$$

Now let $x=\frac{1}{2}$, hence we have $\sum_{k=1}^{\infty}k({\frac{1}{2}})^{k}=\frac{0.5}{(1-0.5)^2}=2$

Now if we take the derivative, we will have $$\frac{d}{dx}\sum_{k=1}^{\infty}kx^{k}=\sum_{k=1}^{\infty}k^2x^{k-1}=\frac{d}{dx}(\frac{x}{(1-x)^2})=\frac{1+x}{(1-x)^3}\rightarrow \sum_{k=1}^{\infty}k^2x^{k}=\frac{x(1+x)}{(1-x)^3}$$ Now if you let $x=\frac{1}{2}$, then you have: $$\sum_{k=1}^{\infty}k^2(\frac{1}{2})^{k}=\frac{0.5(1+0.5)}{(1-0.5)^3}=\frac{1.5}{0.25}=6$$

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    $\begingroup$ Thank you AMIR for explicitly solving the problem. The strategy of iterated differentiation and replenishing of x's is worthwhile to memorize. $\endgroup$ – NoBackingDown Dec 17 '14 at 15:10

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