# How do I find the sum of $\sum\limits_{k=1}^\infty{\frac{k}{2^{k+1}}}=1$? [duplicate]

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

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

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

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