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This question already has an answer here:

As shown in the title, how do I find the sum of:

$$\sum\limits_{k=1}^\infty{\frac{k}{2^{k+1}}}=1$$

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marked as duplicate by colormegone, lab bhattacharjee, angryavian, Jyrki Lahtonen Aug 11 '15 at 5:43

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HINT:

Note that for $|x|<1$, $f(x)=\sum_{k=1}^{\infty}x^{k}=\frac{x}{1-x}$ implies that

$$x^2f'(x) = \sum_{k=1}^{\infty}kx^{k+1}$$

Then, let $x=1/2$

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  • $\begingroup$ Why $|x| < 1$ and take $x=2$? $\endgroup$ – GAVD Aug 11 '15 at 5:41
  • $\begingroup$ @GAVD $x=1/2$ sorry $\endgroup$ – Mark Viola Aug 11 '15 at 5:42

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