limits - Elementary way to show $lim_{n rightarrow infty} sqrt[n]{a_n} = lim_{n rightarrow infty} frac{a_{n+1}}{a_n}$?

Let $a_n \gt 0$ for $n \in \mathbb{N}$. The convergence radius of the series $\sum_{n=0}^\infty a_n z^n$ is $\frac{1}{q}$ with $q = \lim_{n \rightarrow \infty} \sqrt[n]{a_n}$ or $q = \lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n}$, if these limits exist. Therefore the $\lim$s must be identical.

However I was wondering whether there exists a more elementary way to show this identity? (Or generally any other way?)

Answer

The general relation is

$$\liminf_{n \to \infty} \frac{a_{n+1}}{a_n} \le \liminf_{n \to \infty} \sqrt[n]{\mathstrut a_n} \le \limsup_{n \to \infty} \sqrt[n]{\mathstrut a_n} \le \limsup_{n \to \infty} \frac{a_{n+1}}{a_n} \quad \text{(*)}$$
provided that all $a_n$ are positive. It follows that if
$\lim_{n \to \infty} \frac{a_{n+1}}{a_n}$ exists then
$\lim_{n \to \infty} \sqrt[n]{\mathstrut a_n}$ exists as well and they are equal.

To prove the rightmost inequality, define $S := \limsup_{n \to \infty} \frac{a_{n+1}}{a_n}$. For every $\epsilon > 0$ there is a $N \in \mathbb N$
such that

$$\frac{a_{n+1}}{a_n} < S + \epsilon \quad \text{ for } n \ge N \, .$$
$$\implies a_n < a_N \, (S + \epsilon)^{n - N} \quad \text{ for } n \ge N$$
$$\implies \sqrt[n]{\mathstrut a_n} < (S + \epsilon) \sqrt[n]{ a_N (S + \epsilon)^{-N}}$$
$$\implies \limsup_{n \to \infty} \sqrt[n]{\mathstrut a_n} \le S + \epsilon \, .$$

The leftmost inequality can be proved in the same way, or by taking
the reciprocals.

That the convergence radius $R$ of a power series $\sum_{n=0}^\infty a_n z^n$ can be
determined with the root test or
with the ratio test
is a consequence of the above relation (*), not the other way around.

But note that the tests are slighty different:

$$\frac 1R = \limsup_{n \to \infty} \sqrt[n]{\mathstrut |a_n|}$$
holds even if the limit does not exists.
$$\frac 1R = \lim_{n \to \infty} \bigl| \frac{a_{n+1}}{a_n} \bigr|$$
is only valid if the limit on the right-hand side exists. You cannot
generally replace the $\lim$ by $\limsup$ in the ratio test.