Find the value for $\Sigma_1^{\infty} \frac{n}{2^n}$ I am messing around with series and trying to figure out how to calculate the sum. I know that $\Sigma_0^{\infty} x^n$ for $-1<x<1$ converges to $\dfrac{1}{1-x}$ = $1 + x + x^2 + x^3 +...+ x^n$. Term by term differentiation yields $\dfrac{1}{(1-x)^2}=1+2x+3x^2+4x^3+...+nx^{n-1}$. I found that if I let $x=\frac{1}{2}$ I get $1+1+\frac{3}{4}+\frac{4}{8}+...+ = \Sigma_0^{\infty}\dfrac{n}{2^{n-1}}$
This is not quite what I want but I am having trouble figuring out how to fix it.

$1/2 + 2/4 + 3/8 + 4/16 + 5/32 + 6/64 + 7/128 + …$
$= (1/2 + 1/4 + 1/8 + 1/16 + 1/32 + …) + (1/4 + 1/8 + /1/16 + 1/32 + …) + (1/8 + 1/16 + 1/32 + …) + … $
$= (1/2 + 1/4 + 1/8 + 1/16 + 1/32 + …) + 1/2.(1/2 + 1/4 + 1/8 + 1/16 + 1/32 + …) + $
$1/4.(1/2 + 1/4 + 1/8 + 1/16 + 1/32 + …) + 1/8.(1/2 + 1/4 + 1/8 + 1/16 + 1/32 + …) + …$
$= (1/2 + 1/4 + 1/8 + 1/16 + 1/32 + …) .(1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + …)$
$= 1 . 2$
$= 2.$
