Prove that $\lim_{n\to\infty}a_n^{1/n}=\lim_{n\to\infty}\frac{a_{n+1}}{a_n}$ I am stuck with this question which asks me to prove that for all sequences of real numbers $(a_n)$ $$\lim_{n\to\infty}a_n ^{1/n}=\lim_{n\to\infty}\frac{a_{n+1}}{a_n}$$ provided that the rightmost limit exists. How would I do this? I cannot use L'Hopital's Rule as $(a_n)$ is a sequence, not a function. Is there a direct way to prove this?
 A: I will assume the sequences we take are nonnegative.
Let $L=\lim\frac{a_{n+1}}{a_n}$.
Given $\epsilon>0$, we have, for $n$ big
$$L-\epsilon\leq\frac{a_{n+1}}{a_n}\leq L+\epsilon$$
or equivalently
$$(L-\epsilon)a_n\leq a_{n+1}\leq(L+\epsilon)a_n$$
Now iterate! For all $n$ large and all $k$,
$$(L-\epsilon)^ka_n\leq a_{n+k}\leq(L+\epsilon)^k a_n.$$
Fix $n$ for which these inequalities hold. Equivalently, taking $(n+k)$-th roots,
$$(L-\epsilon)^{k/(n+k)}a_n^{1/(n+k)}\leq a_{n+k}^{1/(n+k)}\leq (L+\epsilon)^{k/(n+k)}a_n^{1/n+k}$$
Let $k\to\infty$ and make a sandwich.
A: Assume $a_n>0$ and suppose the limit on the right hand side exists and is equal to $l.$ The only thing you know is that for $\epsilon>0$, there is an integer $N>0$ such that if $n>N$, then $\left|\frac{a_{n+1}}{a_n}-l\right |<\epsilon$. Now use a trick: write $a_n=a_N\prod^{n}_{i={N+1}}\frac{a_i}{a_{i-1}}$ and note that each factor in the product satisfies $l-\epsilon<\frac{a_i}{a_{i-1}}<l+\epsilon.$ Therefore, $a_N(l-\epsilon)^{n-N}<a_n<a_N(l-\epsilon)^{n-N}.$ To finish, take the $n^{\text{th}}$ root across the inequality and let $n\to \infty.$
Remark: a slick way to do this is to take logs and use the Stoltz-Cesaro theorem. 
A: Note that 
$$a_n^\frac{1}{n}= e^\frac{\ln(a_n)}{n}$$
Now, set $b_n= \ln(a_n)$ and prove instead that, if $\lim_n b_{n+1} -b_n$ exists, then
$\lim_n \frac{b_n}{n}$ exists and 
$$\lim_n \frac{b_n}{n}= \lim_n b_{n+1}-b_n$$
This is sometimes referred to as Cezaro average (for $c_n:=b_{n+1}-b_n$) and can be proven with Stolz--Cezaro, but it is an easy exercise.
