I'm trying to understand how to get to this equality step by step. I appreciate your help.
As I recall the series is $\displaystyle \sum\limits_{k = 1}^\infty {\frac{k}{{{2^k}}}} $.
Lets consider $\displaystyle y = \sum\limits_{k = 1}^\infty {x^k}=\frac{x}{1-x} $ where $|x|<1$
Now
$\displaystyle \frac{dy}{dx} = \sum\limits_{k = 1}^\infty {kx^{k-1}}=\frac{1}{(1-x)^2} $, thus $\displaystyle x\cdot \frac{dy}{dx} = \sum\limits_{k = 1}^\infty {kx^{k}}=\frac{x}{(1-x)^2} $
So use $x=\dfrac{1}{2}$ to get $\displaystyle \sum\limits_{k = 1}^\infty {\frac{k}{{{2^k}}}}=~? $
P.S. SEE HERE