Finding the domain of convergence

Question

Find the domain of convergence

$ \sum_{n=0}^{\infty} n (z-1)^n$

My attempt : $-1 < (z-1) <1 \tag 1$

adding $1$ on $(1)$, we have

$0 \le z <2$

so the domain convergence will be $z \in [0,2)$

Is its true ?

Answer 1

If this is really a quastion about Complex Analysis, then your answer doesn't makes sense. Since $\limsup_n\sqrt[n]n=1$, the radius of convergence is $1$. And, if $|z-1|=1$, then you don't have $\lim_nn|z-1|=0$, and therefore the series $\sum_{n=1}^\infty n(z-1)^n$ diverges So, your series converges if and only if $|z-1|<1$.

Answer 2

The series is divergent for $z=0$ as well as $z=2$. So the domain is $(0,2)$.