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?)
 A: 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.
