I have the power $\sum_{n=0}^\infty a_n z^n$ for which $a_n \rightarrow a \neq 0$, and am asked to conclude that the radius of convergence is equal to or smaller than $1$.
I tried using the following fact, but couldn't see its utility:
$\lim \sup x_n = \ell \in \mathbb{R}$ if and only if $\forall \epsilon > 0$, there exists an $N \in \mathbb{N}$ such that $\forall n \ge N$ we have $x_n > \ell - \epsilon$, and that there exist finitely many terms for which $x_n > \ell + \epsilon$.
I have tried proving the contrapositive as well as doing a proof by contradiction, but I cannot see the way. How should I proceed?
If the radius of convergence is greater than $1$ then taking $z=1$ shows that $\sum_{n=1}^{\infty}a_n$ converges, but this is impossible because $\lim_{n\to\infty}a_n=a\neq 0$.