Show convergence of the following series. I honestly am not sure how to start. I think we will use convergence of geometric series. This section has to do with rearrangement of series.
Prove that if $$0 \leq x \lt 1$$ then $$\sum_{n=0}^\infty (n+1)x^n= (\frac{1}{1-x})^2$$

Hints-We have:
$$\sum_{n=0}^\infty x^n=\frac{1}{1-x}$$
and the convergence is absolute. Hence
$$\sum_{n=0}^\infty nx^{n-1}=\sum_{n=0}^\infty \frac{d}{dx}x^n=\frac{d}{dx}\sum_{n=0}^\infty x^n=\frac{d}{dx}\frac{1}{1-x}=\frac{1}{\left(1-x\right)^2}.$$
Added: $$F(x) = \sum_{n=0}^\infty (n+1)x^n = 1+x + 2x^2 + 3x^3 + ...$$
$$xF(x) = \sum_{n=0}^\infty (n+1) x^{n+1} = x+ x^2 +2x^3  + ...$$
$$F(x) - xF(x) =1+ x + x^2 + x^3 + x^4... = \sum_{n=0}^\infty x^n = \dfrac{1}{1 - x}.$$
$$F(x)=\sum_{n=0}^\infty(n+1) x^n=\dfrac{1}{(1 - x)^2}.$$
