The following is a portion of the proof of the Cauchy-Hadamard theorem given on Wikipedia.
Let $f(z) = \sum_n c_n z^n$ be a power series, where $c_n \in \mathbb{C}$ for each $n \in \mathbb{N}$, and define $R$ by $\frac{1}{R} = \limsup \left( |c_n|^{1/n} \right)$. Assume that $t := \frac{1}{R} \neq, 0, \pm \infty$, and let us observe what happens when $|z| < R$. For any $\epsilon > 0$, there is a finite number of $n$ such that $|c_n|^{1/n} \geq t + \epsilon$, i.e., $|c_n| \leq (t + \epsilon)^n$ holds for all but finitely many $n$. Then this implies that $f$ converges if $|z| < \frac{1}{t + \epsilon}$.
I am not sure how the implication in the last sentence arises.
Let's assume that $|c_n|\le (t+\epsilon)^n$: if $|z|<\frac{1}{t+\epsilon}$ then there exists a $\epsilon_z>0$ such that $$ |z|=\frac{1}{(t+\epsilon)(1+\epsilon_z)}<\frac{1}{t+\epsilon}\iff |z|^n=\frac{1}{(t+\epsilon)^n(1+\epsilon_z)^n}. $$ This in turn implies that $$ \begin{split} |f(z)| &=\Big|\sum_n c_n z^n\Big|\le \sum_n |c_n| |z|^n\\ &\le \sum_n \frac{(t + \epsilon)^n}{(t+\epsilon)^n(1+\epsilon_z)^n}\\ & = \sum_n \frac{1}{(1+\epsilon_z)^n} = \frac{1+\epsilon_z}{\epsilon_z}<+\infty \end{split} $$ thus $f(z)$ converges.