Tasty summation!

$\sum_{n=1}^{\infty} \dfrac {2n}{3^{n+1}}$ $\frac{2}{3}\sum_{n=1}^{\infty} \dfrac {n}{3^{n}}$ $\sum_{n=1}^{\infty} \dfrac {n}{3^{n}}\rightarrow S_1=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}...~~(1)$ $~~~~~~~~~~~\rightarrow \frac{S_1}{3}=\frac{1}{3^2}+\frac{2}{3^3}+\frac{3}{3^4}+\frac{4}{3^5}...~~(2)$ $~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\rightarrow \frac{2S_1}{3}=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}...~~(1)-(2)$ $\rightarrow \frac{2S_1}{3}=\frac{\frac{1}{3}}{1-\frac{1}{3}}=\frac{1}{2}$ $\rightarrow S_1=\frac{3}{4}$ $\frac{2}{3}\sum_{n=1}^{\infty} \dfrac {n}{3^{n}}=\frac{2}{3}S_1=\frac{2}{3}\times\frac{3}{4}=$ $\sum_{n=1}^{\infty} \dfrac {2n}{3^{n+1}}=\Large\color{#3D99F6}{\boxed{\color{forestgreen}{\boxed{\frac{1}{2}}}}}$