How can we show that $ \sum_{n=1}^{\infty} \frac{n}{2^n} = 2 $? How can we prove the following?
$$ \sum_{n=1}^{\infty} \frac{n}{2^n} = 2 $$
It would be great to see multiple ways, or hints, about how this can be proven. 
I know this is a power series $\sum_{n=0}^\infty c_n \, (x-a)^n$ where $c_n = n$ and $x = \frac{1}{2}$ and $a = 0$, but otherwise I am stuck.
 A: Let 
$$
I=\sum_{i=1}^\infty\frac{i}{2^i}.
$$
Then
$$
2I=\sum_{i=1}^\infty\frac{i}{2^{i-1}}=1+\sum_{i=2}^\infty\frac{i}{2^{i-1}}
=1+\sum_{i=1}^\infty\frac{i+1}{2^{i}}=1+I+\sum_{n=1}^\infty\frac{1}{2^i}=2+I.
$$
Thus $I=2$.
Note. If $2$ in $\frac{i}{2^i}$ is replaced by $a>1$, then $I$ becomes $\frac{2}{a-1}$.
A: Consider $f(x)=\sum\limits_{i=0}^\infty x^i=\frac{1}{1-x}$ for $|x|<1$. Then
$$f'(x)=\sum\limits_{i=0}^\infty i x^{i-1}=\sum\limits_{i=1}^\infty i x^{i-1}=\frac1{(1-x)^2}$$ for $|x|<1$. Plugging in $x=\frac12$, we have
$$\frac12f'(1/2)=\sum_{i=1}^\infty\frac i{2^i}=\frac1{2 \left(1-\frac12\right)^2}=2$$
So $$\sum\limits_{i=1}^\infty\frac i{2^i}=2$$
A: As a change of pace from the usual derivative, we can show that the result is an immediate consequence of the summagration-by-parts formula for two sequences:
$$\sum_{k=m}^{n}f_{k}\left(g_{k+1}-g_{k}\right)=\left[f_{n+1}g_{n+1}-f_{m}g_{m}\right]-\sum_{k=m}^{n}g_{k+1}\left(f_{k+1}-f_{k}\right).$$
Letting $f_{k}=k$ and $g_{k}=-2^{1-k}$, we have:
$$\sum_{k=m}^{n}k\,2^{-k}=\left[-(n+1)2^{1-n-1}+m\,2^{1-m}\right]+\sum_{k=m}^{n}2^{1-k-1}\left(k+1-k\right).$$
It should be fairly clear where to go from there.
