jmc

We can rewrite this as two separate series $$\sum _{n=1}^{\infty }\left(\frac{2n}{3^n}-\frac{1}{2^n}\right)=2\sum _{n=1}^{\infty }\frac{n}{3^n}-\sum _{n=1}^{\infty }\frac{1}{2^n}.$$The first, $S={\sum }_{n=1}^{\infty }\frac{n}{3^n}=\frac{1}{3}+\frac{2}{9}+\frac{3}{27}+\cdots$ is an arithmetico-geometric series. Multipling by 3, the inverse of the common ratio, gives us $$3S=\sum _{n=1}^{\infty }\frac{n}{3^{n-1}}=1+\frac{2}{3}+\frac{3}{9}+\frac{4}{27}+\cdots$$Subtracting $S$ from $3S$ gives $$\begin{array}{rl}2S& =1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\cdots \\ & =\frac{1}{1-\frac{1}{3}}\\ & =\frac{3}{2}.\end{array}$$The second series is a geometric series so we have $$\sum _{n=1}^{\infty }\frac{1}{2^n}=\frac{\frac{1}{2}}{1-\frac{1}{2}}=1.$$Hence, $$2\sum _{n=1}^{\infty }\frac{n}{3^n}-\sum _{n=1}^{\infty }\frac{1}{2^n}=\frac{3}{2}-1=\boxed{\frac{1}{2}}.$$