How to prove this convergence proposition? '$x_n\gt0,$  $\frac{x_{n+1}}{x_n}$converges$\to x_n^\frac1n $ converges'
How to prove this proposition?
What i think is to use
when  $\frac{x_{n+1}}{x_n}=L$   
$ x_{n+1}\le...\le (L+l)^{n-k}x_{k+1} $ something like this
but i failed.
 A: Hint : use the Cesaro Mean applied to $U_n = \ln{(x_{n+1}/x_n)}$.
If needed the proof for Cesaro mean can be found here
A: Suppose that $a_n$ is a sequences of positive terms. Then:
$$\liminf_{n\to \infty} \frac{a_{n+1}}{a_n}\le\liminf_{n\to \infty} a_n^{1/n}\le\limsup_{n\to \infty} a_n^{1/n}\le\limsup_{n\to \infty} \frac{a_{n+1}}{a_n}$$
We have three inequalities  but the inequality in the middle is trivial. So only we concern about the others, note that the proof of the first and last are very similar. 
1) Let $a = \liminf_{n\to \infty} a_{n+1}/a_n$. Plainly $a\ge0$. If $a=0$, there is nothing to prove. So we may assume $a>0$. Let $0<\alpha<a$, then there is a integer $n_0$ such that 
$$\alpha<a_{k+1}/a_k \tag{1}$$ 
for all $k\ge n_0$. Let $n> n_0$, then we get multiplying these inequalities in (1)
$$a_n/a_{n_0}>\alpha^{n-n_0} \; \text{and}\;\left(a_{n_0}\alpha^{-n_0}\right)^{1/n} \alpha< a_n^{1/n}$$
Since this hold for all $n> n_0$, then $ \alpha\le \liminf a_n^{1/n}$. But $\alpha $ is an arbitrary number $<a$. Then $a\le  \liminf a_n^{1/n}$. If $\lim_{n\to \infty} \frac{a_{n+1}}{a_n}=L$ so $\liminf$ and $\limsup$ are equal and the result is trivial by the inequality.
