I would like to verify that my proof of the following exercise is correct:
Determine the area of convergence and the sum-function for the power series $$ \sum_{n \geq 0} n(z-2)^n. $$ Proof: Firstly, we notice that this is a power series with coefficient $c_n = n$. Hence, we compute the radius of convergence as $$ \lim_{n \to \infty} \left| \frac{c_n}{c_{n+1}} \right| = \frac{n}{n+1} = 1. $$ So this series converges for all $z \in \mathbb{C}$ with $|z-2| < 1$, in other words, in the open ball $B(2,1)$.
Now we compute the sum function. Firstly, notice that the Cauchy-product of the series $\sum_{n=0}^\infty (z-2)^n$ with itself is given by $$ \sum_{n=0}^\infty \sum_{k=0}^n a_k b_{n-k} = \sum_{n=0}^\infty \sum_{k=0}^n (z-2)^k (z-2)^{n-k} = \sum_{n=0}^\infty (n+1)(z-2)^n = \sum_{n=0}^\infty n(z-2)^n + \sum_{n=0}^\infty (z-2)^n. $$ Now by the geometric series we have in the area of convergence $|z-2| < 1$, $$ \sum_{n=0}^\infty (z-2)^n = \frac{1}{1-(z-2)} = \frac{1}{3-z}. $$ Moreover, we know that the cauchyproduct of two series converges to the product of the limit of both series. So from this we know that $$ \sum_{n=0}^\infty n(z-2)^n + \sum_{n=0}^\infty (z-2)^n = \frac{1}{(3-z)^2} $$ Hence we compute $$ \sum_{n=0}^\infty n(z-2)^n = \frac{1}{(3-z)^2} - \sum_{n=0}^\infty (z-2)^n = \frac{1}{(3-z)^2} - \frac{1}{3-z} = \frac{z-3}{(3-z)^2}. $$