Let $G(x)=\frac{1}{(1-x)^2}$. Prove that $G(x)=\sum_{n=0}^{\infty}(n+1)x^n$.

Question

Let $G(x)=\frac{1}{(1-x)^2}$. Prove that $$G(x)=\sum_{n=0}^{\infty}(n+1)x^n.$$
The solution given uses the Cauchy Product. It is shown below:
$$\begin{aligned} G(x)&=\left(\sum_{n=0}^{\infty}x^n\right)\left(\sum_{n=0}^{\infty}x^n\right)\\ &=\sum_{n=0}^{\infty}\sum_{k=0}^{n}x^n\\ &=\sum_{n=0}^{\infty}(n+1)x^n. \end{aligned}$$
My question is that I don't understand why the equations after the second equal symbol are reached. What I know is that the Cauchy Product is used here, but how it comes to these equations?

Answer 1

The Cauchy product for general series is: $$ \left(\sum_{n=0}^\infty a_n\right)\left(\sum_{n=0}^\infty b_n\right)=\sum_{n=0}^\infty \sum_{k=0}^n a_kb_{n-k} $$ In you case, $a_n=b_n=x^n$ for all $n$, so $a_kb_{n-k}=x^kx^{n-k}=x^n$. Making that replacement explains the second equality. The third equality follows since you $\sum_{k=0}^nx^n$ is the sum of $n+1$ copies of $x^n$, and therefore equal to $(n+1)x^n$.