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

Start with the geometric series
$$s(x) = \sum_{k=0}^{\infty} x^k = \frac{1}{1-x}$$
Then
$$x \frac{d}{dx} s(x) = \sum_{k=1}^{\infty} k x^k = \frac{x}{(1-x)^2}$$
Your case has $x=1/2$.
