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Find the value of the series $\sum\limits_{n=1}^ \infty \dfrac{n}{2^n}$

The series on expanding is coming as $\dfrac{1}{2}+\dfrac{2}{2^2}+..$

I tried using the form of $(1+x)^n=1+nx+\dfrac{n(n-1)}{2}x^2+..$ and then differentiating it but still it is not coming .What shall I do with this?

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marked as duplicate by Antonio Vargas, Justpassingby, Community Dec 14 '15 at 12:45

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  • $\begingroup$ This might help $\endgroup$ – glip-glop Dec 14 '15 at 12:36
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    $\begingroup$ Looks like the derivative of a geometric series to me $\endgroup$ – tired Dec 14 '15 at 12:37
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    $\begingroup$ See this for other ideas. $\endgroup$ – David Mitra Dec 14 '15 at 12:38
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    $\begingroup$ Just differentiate $\frac{1}{2(1-x)}=\frac12\sum x^n$ and set $x=\frac12$. $\endgroup$ – TonyK Dec 14 '15 at 12:39
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HINT:

$$\sum_{n=1}^{\infty}\frac{n}{2^n}=\lim_{m\to\infty}\sum_{n=1}^{m}\frac{n}{2^n}=\lim_{m\to\infty}\frac{-m+2^{m+1}-2}{2^m}=$$ $$\lim_{m\to\infty}\frac{-2^{1-m}+2-2^{-m}m}{1}=\frac{0+2-0}{1}=\frac{2}{1}=2$$

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