Convergence issues in the identity $\lim a_n^{1/n}=\lim a_{n+1}/a_n$. I discovered that, for the convergence/divergence of series, the ratio test is stronger than the n-th root test. See here for more details. Since both tests should give rise to the same radius of convergence of any power series, I can deduce that
$$
\lim_{n\to \infty} a_n^{1/n}=\lim_{n\to \infty} a_{n+1}/a_n=q
$$
if both limits exist. Of course, all $a_n$ are non-negative.
To prove this directly, for all $\epsilon>0$, there exists $N$ such that $q+\epsilon>a_{n+1}/a_n>q-\epsilon$ for all $n\geq N$. Therefore,
$$
a_{n+r}>(q-\epsilon)^r a_n, a_{n+r}^{1/r}>(q-\epsilon) a_n^{1/r}.
$$
Taking the limit $r\to \infty$ gives
$$
\lim_{r\to \infty} a_{n+r}^{1/r}\geq q-\epsilon.
$$
Similarly,
$$
\lim_{r\to \infty} a_{n+r}^{1/r}\leq q+\epsilon
$$
Now let $k=n+r$, so
$$
\lim_{k\to\infty} a_k^{1/k}\leq\lim_{k\to\infty} a_k^{\frac{1}{k-n}}\leq \lim_{k\to\infty} a_k^{\frac{1}{k}+\frac{n}{k^2}}=\lim_{k\to\infty} a_k^{\frac{1}{k}}a_k^{\frac{n}{k^2}}=\lim_{k\to\infty} a_k^{1/k},
$$
since $\lim_{k\to\infty} a_k^{\frac{n}{k^2}}=1$. So,
$$
\lim_{n\to \infty} a_n^{1/n}=q.
$$
The things that are unclear to me is the convergence of there two limits. I can find examples in which $\lim_{n\to \infty} a_n^{1/n}$ exists, but $a_{n+1}/a_n$ ediverges. However, is it possible that $\lim_{n\to \infty} a_n^{1/n}$ diverge when $a_{n+1}/a_n$ converges? Can I modify the above proof to prove that $\lim_{n\to \infty} a_n^{1/n}$ converges when $a_{n+1}/a_n$ converges?
 A: $$
\varliminf \frac {a_{n+1}}{a_n} \leqslant \varliminf  \sqrt [n ]{a_n} \leqslant \varlimsup \sqrt [n] {a_n} \leqslant \varlimsup \frac {a_{n+1}} {a_n},
$$
so if $\sqrt [n] a_n$ diverges, then $a_{n+1} /a_n$ cannot converge. 
A: Using your approach, it follows that with $n = N+r$ we have
$$(q- \epsilon)^{n-N}a_N \leqslant a_n \leqslant (q+ \epsilon)^{n-N}a_N ,$$
and
$$(q- \epsilon)\underbrace{\left[ \frac{a_N}{(q - \epsilon)^{N}} \right]^{1/n}}_{ \longrightarrow  1 \text{ as } n \to \infty}\leqslant a_n^{1/n} \leqslant (q+ \epsilon)\underbrace{\left[ \frac{a_N}{(q + \epsilon)^{N}} \right]^{1/n}}_{ \longrightarrow  1 \text{ as } n \to \infty}$$
Thus, it follows that for every $\epsilon > 0$,
$$q - \epsilon \leqslant \liminf_{n \to \infty} a_n^{1/n}\leqslant  \limsup_{n \to \infty} a_n^{1/n} \leqslant q + \epsilon,$$
and, hence, that $\liminf_{n \to \infty} a_n^{1/n}= \limsup_{n \to \infty} a_n^{1/n} = \lim_{n \to \infty} a_n^{1/n} = q$.
With convergence of $a_{n+1}/a_n$  we must have convergence of $a_n^{1/n}$ to the same limit. That should address both of your questions.
