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.

$$
\sum_{i=1}^n i x^i = \sum_{i=1}^n x \frac d {dx} x^i = x \frac d {dx} \sum_{i=1}^n x^i = x \frac d {dx}\  \frac{x - x^{n+1}}{1-x} = \cdots
$$
(Now evaluate the derivative and then substitute $\dfrac 1 2$ for $x$.)
