How do you prove $\sum \frac {n}{2^n} = 2$? How do you prove $$\sum_{n=1}^{\infty} \frac {n}{2^n} = 2\ ?$$
My attempt: I have been trying to find geometric series that converge to 2 which can bind the given series on either side. But I am unable to find these. Is there a general technique to find the sum? This is a high school interview question and must be easy enough to solve in a few minutes.
Please give any hints for the first step towards a solution.

Here is a simple way to see this. It ignores technical aspects of rearranging infinite series. 
Write down the series as follows:
$$
\quad \frac{1}{2} + \frac{2}{2^2} + \frac{3}{2^3} + \frac{4}{2^4} + \dots \\
= \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4} + \dots \\
\qquad  + \frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4} + \dots \\
\qquad 
\qquad 
\quad + \frac{1}{2^3} + \frac{1}{2^4} + \dots \\
\qquad 
\qquad + \frac{1}{2^4} + \dots \\
\vdots
$$
Each row is a geometric series. The values of the rows are $1, \frac{1}{2}, \, \frac{1}{2^2}, \frac{1}{2^3}$, which is again a geometric series. Add this series and you'll get the answer $\boxed{2}$.   
