In Schaum's complex variables book, I cannot think of an explanation of one step made in the solutions for problem 6.19, although I feel that it should be basic.
Let $f(z) = \sum _{n=1}^\infty a_n z^n$ be a power series with radius of convergence $R$.
Why does $\lim_{n \to \infty} a_n z^n \ne 0$ for $\lvert z \rvert > R$ hold?
I wanted to start by assuming $\lim _{n \to \infty} a_n z^n = 0$ and find a contradiction because $\sum _{n = 1}^\infty a_n z^n$ does not converge, but I cannot.
This can be explained in two ways with Hadamard's radius of convergence formula $$ \frac{1}{R} = \varlimsup_{n \to \infty} \sqrt[n]{|a_n|}. $$
Method 1: show that when $|z| > R$, so $1/|z| < 1/R$, we have $1 < |a_nz^n|$ for infinitely many $n$. (You may need to treat the case $R = 0$ separately.)
Method 2: show $|a_nz^n|$ is unbounded when $|z| > R$ by showing that if $|a_nz^n| \leq B$ for some $B > 0$, then $|z| \leq R$. (You may need to treat the case $R = 0$ separately.) Note $\lim_{n \to \infty} |a_nz^n|$ doesn't exist (it has an unbounded subsequence), which is stronger than saying the limit is not $0$.
