If $\frac{a_{n+1}}{a_n}$ converges to a real value L, then $\sqrt[n]{a_n}$ converges to $L$. To prove:
If {$a_n$} is a sequence of positive real numbers, and the sequence $\frac{a_{n+1}}{a_n}$ converges to a real value L, then $\sqrt[n]{a_n}$ converges to $L$.
Now I'm having trouble proving this. I tried the following:
Suppose that $\sqrt[n]{a_n}$ does NOT converge to value L. Then there is an $\epsilon$ such that there exists an $n_1>n_0$ for any $n_0$ such that:
$|\sqrt[n]{a_n}-L|\ge\epsilon$.
Then for that same value $n_1$ we still know:
$|\frac{a_{n+1}}{a_n}-L|<\epsilon$
It follows $|\sqrt[n]{a_n}-L|>|\frac{a_{n+1}}{a_n}-L|\ge\frac{a_{n+1}}{a_n}-L$
Now take a look at 2 different situations: 
a) $\sqrt[n]{a_n}$ converges to a value greater than $L$: then 
$\sqrt[n]{a_n}-L>\frac{a_{n+1}}{a_n}-L$ for certain $n_1$ that still satisfies the above conditions. 
$\sqrt[n]{a_n}>\frac{a_{n+1}}{a_n}$
$a_n^{n+1}>a_{n+1}^n$
Now as $n$ approaches $\infty$, $a_{n+1}\rightarrow La_n$.
So 
$a_n\lim\limits_{x \to \infty} a_n^{n} \ge \lim\limits_{x \to \infty} L^na_n^n$
$a_n \ge L^n$
I have no idea what I'm doing anymore.. Can someone please help :(
Also this was an exercise in a chapter concerning Cauchy's root and ratio test, but I don't see a way to employ that.
 A: $\newcommand{\eps}{\varepsilon}$
Here is a more direct approach. Take $\eps>0$ and suppose $L-\eps<\frac{a_{n+1}}{a_n}<L+\eps$ for $n\geq N$. Then
$$a_n=a_1\frac{a_2}{a_1}\frac{a_3}{a_2}\dots\frac{a_{N}}{a_{N-1}}\frac{a_{N+1}}{a_N}\dots\frac{a_n}{a_{n-1}}.$$
We can bound all but a fixed number of terms. Show that after taking $n$th roots  the other terms become insignificant.
A: Hint
$$\lim_{n\to+\infty}\frac{a_{n+1}}{a_n}=L$$
$$\implies \lim_{n\to+\infty}\ln(\frac{a_{n+1}}{a_n})=\ln(L)$$
$$\implies \lim_{n\to+\infty}\frac{\sum_{k=1}^n\ln(\frac{a_{k+1}}{a_k})}{n}=\ln(L)$$
using Cesaro sequence.
$$\implies \lim_{n\to+\infty}\frac{\ln(\frac{a_{n+1}}{a_1})}{n}=\ln(L)$$
Yes, you can conclude.
A: Let us first prove the following:

Lemma. If $(a_n)$ is convergent to $\ell$, then $\displaystyle\left(\frac{1}{n}\sum_{k=0}^{n-1}a_k\right)$ is also convergent to $\ell$.

Proof. Let $\varepsilon>0$, there exists $N_1\in\mathbb{N}$ such that: $$\forall n\geqslant N_1,|a_n-\ell|\leqslant\varepsilon.$$
Let $n\geqslant N_1$, using triangle inequality, one has: $$\left|\frac{1}{n}\sum_{k=0}^{n-1}a_k-\ell\right|=\left|\frac{1}{n}\sum_{k=0}^{n-1}(a_k-\ell)\right|\leqslant\frac{1}{n}\sum_{k=0}^{N_1-1}|a_k-\ell|+\frac{(n-N_1+1)}{n}\varepsilon.$$
Since $\displaystyle\frac{1}{n}\underbrace{\sum_{k=0}^{N_1-1}|a_k-\ell|}_{\textrm{constant}}\rightarrow 0$, there exists $N_2\in\mathbb{N}$ such that: $$\forall n\geqslant N_2,\frac{1}{n}\sum_{k=0}^{N_1-1}|a_k-\ell|\leqslant\varepsilon.$$
Furthermore, since $\displaystyle\frac{n-N_1+1}{n}\rightarrow 1$, there exists $N_3\in\mathbb{N}$ such that: $$\forall n\geqslant N_3,\frac{n-N_1+1}{n}\leqslant 2.$$
Finally, for all $n\geqslant\max(N_1,N_2,N_3)$, one has: $$\left|\frac{1}{n}\sum_{k=0}^{n-1}a_k-\ell\right|\leqslant 3\varepsilon.$$
Whence the result. $\Box$
From this lemma, one derives the following:

Proposition. If $(a_n)$ is such that $(a_{n+1}-a_n)$ converges to $\ell$, then $\displaystyle\left(\frac{a_n}{n}\right)$ converges to $\ell$.

Proof. Applying the lemma, one has that: $$\left(\frac{1}{n}\sum_{k=0}^{n-1}(a_{n+1}-a_n)\right)=\left(\frac{a_n}{n}-\frac{a_0}{n}\right)$$
converges to $\ell$. Whence the result, since $\displaystyle\left(\frac{a_0}{n}\right)$ converges to $0$ and $\displaystyle\left(\frac{a_n}{n}\right)=\left(\frac{a_n}{n}-\frac{a_0}{n}\right)+\left(\frac{a_0}{n}\right)$. $\Box$
Can you take it from here?
