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*}
$$