calculus - Why $sum_{k=1}^{infty} frac{k}{2^k} = 2$?

Can you please explain why $$\sum_{k=1}^{\infty} \dfrac{k}{2^k} = \dfrac{1}{2} +\dfrac{ 2}{4} + \dfrac{3}{8}+ \dfrac{4}{16} +\dfrac{5}{32} + \dots = 2$$

I know $1 + 2 + 3 + ... + n = \dfrac{n(n+1)}{2}$

Answer

$$\begin{gather*} |x|<1:\quad f(x)=\sum_{n=1}^{\infty} x^n=\frac{x}{1-x} \\ xf'(x)=\sum_{n=1}^{\infty} nx^n=\frac{x}{(1-x)^2} \end{gather*}$$

Let $x=\frac{1}{2}$