Knowing that: $ \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots = 1 $
And hence by removing successive terms from the left:
$$ \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots = \frac{1}{2} $$ $$ \frac{1}{8} + \frac{1}{16} + \cdots = \frac{1}{4} $$ $$ \vdots $$
We have: $$\begin{eqnarray*} \sum_{i=1}^\infty \frac{i}{2^i} &=& \frac{1}{2} + \frac{2}{4} + \frac{3}{8} + \frac{4}{16} + \cdots \\ &=& (\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots) + (\frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots) + (\frac{1}{8} + \frac{1}{16} + \cdots) + \cdots \\ && \text{Substituting the identites from above:} \\ &=& 1 + \frac{1}{2} + \frac{1}{4} + \cdots \\ && \text{And substituting the first identity again:} \\ &=& 1 + 1 \\ &=& 2 \end{eqnarray*}$$