So I wanted to calculate/prove that $$\frac{1}{(1-q)^2} = \displaystyle\sum_{n=0}^{\infty}(n+1)q^n,$$ $q\in \mathbb{C},$ $\mid q \mid < 1$ using the Cauchy-Product.
So this is my solution and I'm asking for verification.
\begin{align*}\frac{1}{(1-q)^2} &= \frac{1}{(1-q)} * \frac{1}{(1-q)}\\ &= \left(\displaystyle\sum_{n=0}^{\infty}q^n\right)*\left(\displaystyle\sum_{n=0}^{\infty}q^n\right)\\ &= \displaystyle\sum_{l=0}^{\infty} \displaystyle\sum_{k=0}^{l} q^k * q^{l-k}\\ &= \displaystyle\sum_{l=0}^{\infty} (q^0 * q^{l-0} + q^{1} * q^{l-1} + \cdots + q^{l-2} * q^{l-(l-2)} + q^{l-1} * q^{l-(l-1)} +q^{l} * q^{l-l)})\\ &=\displaystyle\sum_{l=0}^{\infty} (q^{0+l-0} + q^{1+l-1} + ... + q^{l-2+l-(l-2)} + q^{l-1+l-(l-1)} + q^{l+(l-l)})\\ &=\displaystyle\sum_{l=0}^{\infty} (q^{l} + q^{l} + ... + q^{l} + q^{l} + q^{l}) \\ &=\displaystyle\sum_{l=0}^{\infty}(l+1)*q^l \quad(\text{ since } q^l\text{ is summed up } l+1 \text{ times}) \end{align*}
Is this correct?
Correct but you can shorten it.
Since $q^{k}*q^{l-k}=q^{k+l-k}=q^{l},$ so after the 2nd equality you have $$\frac{1}{(1-q)^2}=\sum_{l=0}^{\infty}\sum_{k=0}^l q^l=\sum_{l=0}^{\infty}(l+1)q^l.$$