Proof complex series

Question

I have to prove this: $\displaystyle\sum_{n=1}^\infty n\alpha^n = \displaystyle\frac{\alpha}{(1-\alpha)^2}$ if $|\alpha | < 1$

I think this is a geometric series, and i have to solve it with a limit but don't know how to raise it

Answer 1

Recall that $$1 + \alpha + \alpha^2 + \cdots = \frac{1}{1 - \alpha}, \quad |\alpha| < 1.$$ Now, complex power series can be differentiated term-wise (see this for justification). Thus, differentiating both sides of the above equation we find that $$1 + 2\alpha + 3\alpha^2 + \cdots = \frac{1}{(1 - \alpha)^2}, \quad |\alpha| < 1.$$ If we multiply both sides by $\alpha$, we obtain: $$\alpha + 2\alpha^2 + 3\alpha^3 + \cdots = \frac{\alpha}{(1 - \alpha)^2}, \quad |\alpha| < 1.$$