real analysis - $sum limits_{n=1}^{infty}n(frac{2}{3})^n$ Evalute Sum

How can you compute the limit of
$\sum \limits_{n=1}^{\infty} n(2/3)^n$

Evidently it is equal to 6 by wolfram alpha but how could you compute such a sum analytically?

Answer

$$\begin{align*} \sum_{n=1}^\infty n(2/3)^n &= \sum_{m=1}^\infty \sum_{n=m}^\infty (2/3)^n \\ &= \sum_{m=1}^\infty \frac{(2/3)^m}{1-2/3} \\ &= \frac{2/3}{(1-2/3)^2} = 6. \end{align*}$$