Proof of $\lim_{n \to \infty} {a_n}^{1/n} = \lim_{n \to \infty}(a_{n+1}/a_n)$ 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.
 A: 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 $\lim_{n\to\infty}{a_{n+1}\over a_n}=0$.
A: 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
$$a_n = \begin{cases} 2&\text{if }n \text{ is odd,}\\ 1/2 &\text{if }n\text{ is even.}\end{cases}$$
A: 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.
A: Let $a_n \in \mathbb{R}, a_n > 0$, and let $b_n = \log(a_n)$. Since $\log(x)$ is a continuous function, $b_n$ converges if, and only if, $a_n$ converges, and $\lim_{n \to \infty} {b_n} = \log(\lim_{n \to \infty} {a_n})$. 
Now, $\log({a_n}^{1/n}) = \frac{1}{n}\log({a_n}) = \frac{1}{n}b_{n}$, and $\log(a_{n+1}/a_n) = \log(a_{n+1}) - \log(a_n) = b_{n+1}-b_{n}$. Hence, if $b_n$ converges to some positive number, then $b_{n+1}-b_{n}$ tends to zero, and so does $\frac{1}{n}b_{n}$. Then you can remove the logarithms and you get your proof. Note that the converse is not true, in general, since you can have $b_n$ bounded, but not convergent, so that $\frac{1}{n}b_{n}$ converges to zero, but $b_{n+1}-b_{n}$ does not have a limit.
