A simple series $\sum_{i=1}^\infty \frac{i}{2^i} = 2$ [duplicate]

I don't do math a long time, so I completely don't remember how to prove that: $$ \sum_{i=1}^\infty \frac{i}{2^i} = 2 $$

Can anybody help me?

Solution 1:

$S = x + 2x^2 + 3x^3 + \ldots $

It can be written as

$ \Rightarrow S = (x + x^2 + x^3 + \ldots)+(x^2 + x^3 + \ldots)+(x^3 + \ldots)+\ldots $

$\Rightarrow S = (x + x^2 + x^3 + ...)+x(x + x^2 + ...)+x^2(x + ...) + \ldots $

$\Rightarrow S = ( 1+x+x^2+ .. )\times( x+x^2+.. )$

$\Rightarrow S = \frac{1}{1-x}\times\frac{x}{1-x}$

$\Rightarrow S = \frac{x}{(1-x)^2} $

Put $x =0.5$ you will get the answer

Solution 2:

The series $$\sum_{i=1}^\infty x^i=\frac{x}{1-x}$$ is the geometric series and we can differentiate it term by term since it's a power series so we have $$\frac{d}{dx}\left(\frac{x}{1-x}\right)=\frac{1}{(1-x)^2}=\sum_{i=1}^\infty ix^{i-1}$$ so multiplying by $x$ gives $$\sum_{i=1}^\infty i x^{i}=\frac{x}{(1-x)^2}$$ and set $x=\frac 1 2$.