Writing sigma notation $\sum^n_{i=1} \frac {i}{2^i}$ in closed form What would be a way to find the closed form of $\frac {1}{2} + \frac {2}{4}+\frac {3}{8}+\cdots+\frac {n}{2^n}=\sum^n_{i=1} \frac {i}{2^i}=s$
I've looked at $\frac {s}{2}=\frac {1}{4} + \frac {2}{8}+\frac {3}{16}+\cdots+\frac {n}{2^{n+1}}$
And then $s-\frac {s}{2}=\frac {s}{2}=\frac {1}{2} + \frac {1}{4}+\frac {1}{8}+\cdots+\frac {1}{2^n}-\frac {n}{2^{n+1}}$
Any idea where to go from here?
Please, if you are sure of where I'm trying to go with this, just ask.

As given in the comments, looking at the derivative of $\frac{1}{1-x}$ gives the answer, but you have mentioned you have been instructed to solve it using this method.
Notice that you almost reached a solution! You have 
$$ \frac{s}{2} = \frac{1}{2}+\cdots+\frac{1}{2^n} - \frac{n}{2^{n+1}}$$
The first part of the right hand side is a sum of a geometric sequence. Use this!
