A simple series $\sum_{i=1}^\infty \frac{i}{2^i} = 2$ 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?
 A: $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 
A: 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 $x=\frac 1 2$.
A: The standard procedure to solve this kind of series is to define a power series like
$$f(x)=\sum_{i=1}^\infty\frac{i}{2^i}x^i$$
so what we want to figure out is $f(1)$.
Then we manipulate the terms using differentiation. Since 
$$x\frac{d}{dx}(x^i)=xix^{i-1}=ix^i$$
you have
$$f(x)=\sum_{i=1}^\infty\frac{1}{2^i}x\frac{d}{dx}(x^i)$$
And by linearity of $x\frac{d}{dx}$,
$$f(x)=x\frac{d}{dx}\left(\sum_{i=1}^\infty\frac{1}{2^i}x^i\right)$$
Now using a property of exponents,
$$f(x)=x\frac{d}{dx}\left(\sum_{i=1}^\infty\left(\frac{x}{2}\right)^i\right)$$
We managed to get a power series out of this, which is great because
$$\sum_{i=1}^\infty\left(\frac{x}{2}\right)^i=\frac{1}{1-\frac{x}{2}}-1=\frac{2}{2-x}-1$$
So
$$f(x)=x\frac{d}{dx}\left(\frac{2}{2-x}-1\right)$$
$$f(x)=-\left(-x\frac{2}{(2-x)^2}\right)=\frac{2x}{(2-x)^2}$$
Finally, we evaluate at $x=1$,
$$f(1)=\frac{2}{(2-1)^2}=2$$
