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

the sequence $$\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+.....$$
is an $arithmetico  geometric series$
the sum of an$AG$ series of the form
$$S_{\infty}=a+(a+d)r+(a+2d)r^2+(a+3d)r^3+.....\infty=\frac{a}{1-r}+\frac{dr}{(1-r)^2}$$converting the above series inti this form
$$\frac{1}{2}( \frac{1}{1} +\frac{2}{2}+ \frac{3}{4}+...... )$$
here$a=1$,$d=1$,$r=\frac{1}{2}$.
you get the answer $2$.
