jmc

Let the sum be $S$. This series looks almost geometric, but not quite. We can turn it into a geometric series as follows: $$\begin{array}{rl}S& =\frac{1}{3^1}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+\cdots \\ \frac{1}{3}S& =\frac{0}{3^1}+\frac{1}{3^2}+\frac{2}{3^3}+\frac{3}{3^4}+\cdots \\ \frac{2}{3}S=S-\frac{1}{3}S& =\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\cdots \end{array}$$Now, we do have a geometric series, so we can find $\frac{2}{3}S=\frac{\frac{1}{3}}{1-\frac{1}{3}}=\frac{1}{2}$, and $S=\boxed{\frac{3}{4}}$.