A infinite series sum $\ \sum _{0}^{\infty} \dfrac{r}{2^r} =? $

Question

$\ \sum _{0}^{\infty} \dfrac{r}{2^r} =? $

what I was able to find is a bound to this convergent series by using

for|x| <1 $\ \sum _{k=0}^{\infty} kx^k = \dfrac{x}{(1-x)^2}$

and substituting x=0.5

is there a series formula available to evaluate it ?

Answer 1

You can derive the formula by using the derivative of $x^k$.

$$\sum_{k=0}^\infty kx^k$$ $$=\sum_{k=0}^\infty x\frac{d}{dx}x^k$$ $$=x\frac{d}{dx}\sum_{k=0}^\infty x^k$$ $$=x\frac{d}{dx}\frac{1}{1-x}=x\frac{d}{dx}(1-x)^{-1}$$

You have to apply the chain rule here.

$$=x\frac{-1}{\left(1-x\right)^2}\left(-1\right)$$ $$=\frac{x}{\left(1-x\right)^2}$$

Answer 2

You have almost answered your own question.

$$\sum_{r=0}^\infty r (\frac{1}{2})^r = \frac{1/2}{(1-1/2)^2} = \frac{1/2}{1/4} = 2$$