<s>[INST] How can I rewrite $\frac{1}{(1-z)^4}$ as a power series? I'm supposed to rewrite $\frac{1}{(1-z)^4}$ as a power series, using the Cauchy-Product. It is given that $|z| < 1$.
So far I have
\begin{aligned}
\frac{1}{(1-z)^4} &= \left( \frac{1}{(1-z)^2} \right)^2\\
&= \left( \sum_{n=0}^{\infty} (n+1)z^n \right)^2\\
&=  \sum_{n=0}^{\infty} \sum_{k=0}^{n} (n-k+1)z^{n-k}kz^k  \\
&=  \sum_{n=0}^{\infty} \sum_{k=0}^{n} k(k+1)z^{n}  \\
\end{aligned}
First of all, I'm not sure whether every step was correctly so far. Additionally, I don't know how I can reduce that to a power series.
 [/INST] The third line should be
\begin{eqnarray*}
&=  \sum_{n=0}^{\infty} \sum_{k=0}^{n} (n-k+1)z^{n-k} \color{red}{(k+1)}z^k  \\
\end{eqnarray*}
So we need to perform the sum
\begin{eqnarray*}
 \sum_{k=0}^{n} (n-k+1)(k+1) = -\sum_{k=0}^{n} k^2+ n \sum_{k=0}^{n} k +(n+1) \sum_{k=0}^{n} 1.
\end{eqnarray*}
Recall that
\begin{eqnarray*}
 \sum_{k=0}^{n}  k^2= \frac{n(n+1)(2n+1)}{6}
\end{eqnarray*}
and we have
\begin{eqnarray*}
 \sum_{k=0}^{n} (n-k+1)(k+1) = \frac{(n+1)(n+2)(n+3)}{6}.
\end{eqnarray*}
</s>