Proof of $\lim_{n \to \infty} {a_n}^{1/n} = \lim_{n \to \infty}(a_{n+1}/a_n)$ [duplicate]

Why would $\lim_{n \to \infty} {a_n}^{1/n} = \lim_{n \to \infty}(a_{n+1}/a_n)$ be true where $(a_n)$ is a sequence in $\mathbb{R}$?

Edit: Let all $a_n$ be positive.

Under the hypothesis $\forall n\in\Bbb N: a_n>0$ and $\exists\lim_{n\to\infty}{a_{n+1}\over a_n}>0$, let be $$l=\lim_{n\to\infty}{a_{n+1}\over a_n}.$$ For each $\epsilon>0$, $\exists n(\epsilon)\in\Bbb N$ s.t. $\forall n\ge n(\epsilon)$: $$l-\epsilon\le{a_{n+1}\over a_n}\le l+\epsilon.$$ Multiply the inequalities form $n(\epsilon)$ to $n(\epsilon)+p-1$: $$(l-\epsilon)^p\le{a_{n(\epsilon)+p}\over a_{(n(\epsilon)}}\le(l+\epsilon)^p.$$ I.e., $$(l-\epsilon)^p a_{n(\epsilon)}\le{a_{n(\epsilon)+p}}\le(l+\epsilon)^p a_{n(\epsilon)},$$ and taking $n=p+n(\epsilon)$-root: $$(l-\epsilon)^{(n-n(\epsilon))/n}\root n\of{a_{n(\epsilon)}}\le\root n\of{a_n}\le(l+\epsilon)^{(n-n(\epsilon))/n}\root n\of{a_{n(\epsilon)}}.$$ Now, take $\lim_{n\to\infty}$. Think yourself in the case $\lim_{n\to\infty}{a_{n+1}\over a_n}=0$.

If $a_{n+1}/a_n$ converges in $[0,\infty]$ then $a_n^{1/n}$ does as well and to the same limit, but the converse need not hold. Upon taking logarithms this is a well known statement about Cesaro means of sequences, but I will write out the proof in the present setting.

Suppose $a_{n+1}/a_n\to\ell\in[0,\infty]$. First suppose $0<\ell<\infty$, and let $\alpha>1$. Then there is some $N$ such that $\alpha^{-1}\ell < a_{n+1}/a_n < \alpha\ell$ for all $n\geq N$. Multiplying these together for $n=N,\dots,m-1$ we have $\alpha^{-(m-N)}\ell^{m-N} < a_m/a_N < \alpha^{m-N} \ell^{m-N}$. Taking $m$th roots and the limit $m\to\infty$ shows that $a_m^{1/m} \to \ell$. The cases $\ell=0$ and $\ell=\infty$ are similar, and you might like to try them.

For a counterexample for the opposite implication, consider the sequence $$a_n = \begin{cases} 2&\text{if }n \text{ is odd,}\\ 1/2 &\text{if }n\text{ is even.}\end{cases}$$

This is a particular case of the following chain of inequalities $$\liminf \frac{a_{n+1}}{a_n} \leq \liminf \sqrt[n]{a_n} \leq \limsup \sqrt[n]{a_n} \leq \limsup \frac{a_{n+1}}{a_n}. $$ cf. Theorem 3.37 in Rudin's principles of mathematical analysis. If the limit of the sequence $\frac{a_{n+1}}{a_n}$ exists the outermost expressions coincide, and thus the two in the middle as well.