Can I prove convergence of and show the sum $\sum_{n=1}^{\infty} (z-1){z}^{n}$ where $|z|<1$ by assuming $z$ to be a complex number?

Question

While $z$ usually represents a complex number, in this case it isn't given if $z \in \mathbb{C}$, so would it make any difference if $z \in \mathbb{R}$?

Answer 1

Apply the root test: if $a_n = (z-1)$, then $\limsup {|a_n|}^{\frac{1}{n}} = \limsup {|z-1|^{\frac{1}{n}}} = \lim {|z-1|^{\frac{1}{n}}} = 1$. So the radius of convergence of the power series $\sum a_nz^n$ is 1. Hence $\sum(z-1)z^n$ converges for $|z|<1$. Further, this power series converges absolutely in the open disk $B_0(1)$ and uniformly in any compact subset of $B_0(1)$.