$\sum \limits_{n=1}^{\infty}n(\frac{2}{3})^n$ Evalute Sum 
Possible Duplicate:
How can I evaluate $\sum_{n=1}^\infty \frac{2n}{3^{n+1}}$ 

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?

Just a curioursity.
Using Fourier series one comes to:
$$\sum_{n=1}^\infty n\left( \frac{2}{3}\right)^n\ \cos nx = 6\ \frac{13 \cos x-12}{(13-12\cos x)^2}$$
hence:
$$\sum_{n=1}^\infty n\left( \frac{2}{3}\right)^n = \sum_{n=1}^\infty n\left( \frac{2}{3}\right)^n\ \cos nx \Bigg|_{x=0} =6\ \frac{13 \cos x-12}{(13-12\cos x)^2}\Bigg|_{x=0} =6\; .$$
