number theory - why $sum_{h=0}^{infty}{dfrac{h}{2^h}} = 2$

There is a summation in analysis of an algorithm which is the following:

$$\sum_{h=0}^{\infty}{\dfrac{h}{2^h}} = \dfrac{1/2}{(1-1/2)^2} = 2$$

But I don't can't solve this. I would be appreciated if anyone can solve this.

thanks.

Answer

Not sure what you need to solve (there doesn't seem to be an unknown in your expression), but assuming you want to derive the result:

$\sum_{n=0}^{\infty}x^n=\frac{1}{1-x}$ for $|x|<1$. Differentiate both sides to get $\sum_{n=0}^{\infty}nx^{n-1}=\frac{1}{(1-x)^2}$. Multiply both sides by $x$ to finally get:

$\sum_{n=0}^{\infty}nx^n=\frac{x}{(1-x)^2}$