jmc

We have an arithmetico-geometric series with common ratio $\frac{1}{3}$. Let the sum be $S$. When we multiply by $\frac{1}{3}$ we get $$\frac{S}{3}=\frac{1}{3}+\frac{3}{9}+\frac{5}{27}+\frac{7}{81}+\cdots$$Subtracting this from the original series gives us $$\begin{array}{rl}\frac{2}{3}S& =1+\frac{2}{3}+\frac{2}{9}+\frac{2}{27}+\frac{2}{81}+\cdots \\ & =1+\frac{\frac{2}{3}}{1-\frac{1}{3}}=1+1=2.\end{array}$$Then $S=\boxed{3}$.