jmc

Let the sum be $S$. This series looks almost geometric, but not quite. We can turn it into a geometric series as follows: $$\begin{array}{rl}S& =\frac{1}{2^1}+\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+\cdots \\ \frac{1}{2}S& =\frac{1}{2^2}+\frac{2}{2^3}+\frac{3}{2^4}+\cdots \end{array}$$We subtract the second from the first to obtain $$\frac{1}{2}S=\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\cdots$$Now, we do have a geometric series, so we can find $\frac{1}{2}S=\frac{\frac{1}{2}}{1-\frac{1}{2}}=1$, and $S=\boxed{2}$.