If $\limsup|a_n|^{\frac{1}{n}} = \frac{1}{R}$ and $\frac{1}{r} > \frac{1}{R}$, why is there an $N$ such that $|a_n|^{1/n}<1/r$ for all $n\ge N$?

Question

I'm reading Conway's Functions of One Complex Variable and I didn't understand this proof on page $31$:

I didn't understand why $\frac{1}{r}>\frac{1}{R}$ implies there is an integer $N$ such that $|a_n|^{\frac{1}{n}}<\frac{1}{r}$ for all $n\ge N$.

Any help is welcome.

Answer 1

Suppose there is no such $N$. Then we can find a subsequence $a_{n_k}$ such that $|a_{n_k}|^{\frac{1}{n_k}} \geq \frac{1}{r}$. It follows that

$$\frac{1}{r} \leq \limsup\left(|a_{n_k}|^{\frac{1}{n_k}}\right) \leq \limsup \left(|a_n|^{\frac{1}{n}}\right) = \frac{1}{R}$$

which is a contradiction.

Here we've used the fact that the limit supremum of a subsequence is less than or equal to the limit supremum of the original sequence. This follows immediately from the characterisation of the limit supremum as the supremum of the subsequential limits.