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 $$\sum_{n=0}^\infty (n+1)x^n= (\frac{1}{1-x})^2$$
 A: Integrate the general term of the series
$$\int (n+1) x^n \, dx=x^{n+1}+C$$
We know that
$$\sum _{n=0}^{\infty } x^{n+1}=\frac{1}{1-x}-1;\;|x|<1$$
Now differentiate and get the result
$$\sum _{n=0}^{\infty }(n+1)x^n=\frac{1}{(1-x)^2};\;|x|<1$$
A: As I wrote in my comment, the result follows immediately by termwise differentiation of the identity $\sum\limits_{n\geq 0}x^n=\dfrac 1{1-x}$ for $|x|\lt 1$
If you have to do it algebraically, note that,
$$\left(1+x+x^2+\ldots\right)\left(1+x+x^2+\ldots\right)=1+2x+3x^2+\ldots$$
A: 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: $$F(x) = \sum_{n=0}^\infty (n+1)x^n = 1+x + 2x^2 + 3x^3 + ...$$
$$xF(x) = \sum_{n=0}^\infty (n+1) x^{n+1} = x+ x^2 +2x^3  + ...$$
$$F(x) - xF(x) =1+ x + x^2 + x^3 + x^4... = \sum_{n=0}^\infty x^n = \dfrac{1}{1 - x}.$$
$$F(x)=\sum_{n=0}^\infty(n+1) x^n=\dfrac{1}{(1 - x)^2}.$$
