Sum of series $\sum n(\frac23)^n$

Question

$$\sum^{\infty}_{n=0} n\left(\frac23\right)^n$$

I've no idea what to do with it. If there is no n before fraction, it would be simply sum of convergent geometric series.

Answer 1

Recall that for $\vert x \vert < 1$, we have $$\sum_{n=0}^{\infty} x^n = \dfrac1{1-x}$$ Differentiating both sides, we get that $$\sum_{n=0}^{\infty} nx^{n-1} = \dfrac1{(1-x)^2}$$ Hence, $$\sum_{n=0}^{\infty} nx^{n} = \dfrac{x}{(1-x)^2}$$ Take $x=2/3$ to get what you want.