Convergence of sequence defined with other sequence $a_n$ is a sequence with $\lim\limits_{n \rightarrow \infty}{\frac{a_{n+1}}{a_n}}=L$, $a_n>0$. 
So the task is to show that $c_n:=(a_n)^\frac{1}{n}$ converges and $\lim\limits_{n \rightarrow \infty}{c_n}=L$.
I've been working on this for hours now and don't have any useful result. Please help :)
 A: Let us assume $L > 0$ (otherwise, just use upper bounds).
Let $\epsilon > 0$. There exists $n_0 \in \Bbb N$ such that for every $n \geq n_0$ one has $L - \epsilon \leq \frac{a_{n+1}}{a_n} \leq L + \epsilon$. By an immediate induction this yields
$$a_{n_0} (L-\epsilon)^n\leq a_{n+n_0} \leq a_{n_0}(L+\epsilon)^n$$
so
$$
\frac{\log a_{n_0}}{n+n_0} + \frac{n}{n+n_0}\log (L-\epsilon) \leq \frac{\log a_{n+n_0}}{n+n_0} \leq \frac{\log a_{n_0}}{n+n_0} + \frac{n}{n+n_0}\log (L+\epsilon).
$$
Letting $n\to+\infty$, we deduce that
$$
\log(L-\epsilon) \leq \liminf \frac{\log a_n}{n} \leq \limsup \frac{\log a_n}{n} \leq \log(L+\epsilon)
$$
Finally, let $\epsilon \to 0$ and take exponentials to conclude.
A: http://books.google.com.vn/books?id=WZX4GEpxPRgC&printsec=frontcover&hl=vi#v=onepage&q&f=false
This is a usual result, you can see the proof at Serge lang- Complex Analysis at page  57, in the aove link.
Hope it will help you.
A: We will show that $$\lim\limits_{n \rightarrow \infty}{\frac{a_{n+1}}{a_n}}=\lim_{n\to \infty} \sqrt[n] {a_n}$$.(Because the limits exist).
First we will show that $$\lim\limits_{n \rightarrow \infty}{\frac{a_{n+1}}{a_n}}=L\geq \lim_{n\to \infty} \sqrt[n] {a_n}$$.
Let $ε>0$ then there is a $n_1\in \Bbb N:\frac {a_{n+1}}{a_n}\leq L+\frac {ε}{2}$ for every $n\geq n_1$.
Now write $$a_n=\frac {a_n}{a_{n-1}}\cdot \frac {a_{n-1}}{a_{n-2}}\cdot ...\frac {a_{n_1+1}}{a_{n_1}}\cdot a_{n_1}$$. Thus we have that $$a_n\leq \frac {a_{n_1}}{(L+\frac {ε}{2})^{n_1}}\cdot (L+\frac {ε}{2})^n$$.
Let now $$θ=\frac {a_{n_1}}{(L+\frac {ε}{2})^{n_1}}$$ and by using that $\sqrt[n] {θ}\to 0$ we have that there is a $n_0\geq n_1:\sqrt [n] {a_n}\leq \sqrt[n] {θ}\cdot (L+\frac {ε}{2})<L+ε$ for every $n\geq n_0$ and thus $$\lim_{n\to \infty} \sqrt[n] {a_n}\leq L$$.
Same you do to prove $$\lim_{n\to \infty} \sqrt[n] {a_n}\geq L$$.
A: Hint: For any $\epsilon\gt0$, there is an $N$ so that if $n\ge N$,
$$
|\log(a_{n+1})-\log(a_n)-L|\le\epsilon
$$
Thus, by the triangle inequality
$$
\left|\ \left(\sum_{k=N}^{n-1}\log(a_{k+1})-\log(a_k)\right)-(n-N)L\ \right|\le(n-N)\,\epsilon
$$
Therefore, for $n\ge N$,
$$
\begin{align}
&\left|\ \frac{\log(a_n)}{n}-L\ \right|\\
&\le\left|\ \frac{\log(a_n)-\log(a_0)}{n}-L\ \right|+\left|\ \frac{\log(a_0)}{n}\ \right|\\
&\le\left|\ \frac1n\left(\sum_{k=0}^{N-1}\log(a_{k+1})-\log(a_k)\right)\ \right|\\
&+\;\left|\ \frac1n\left(\sum_{k=N}^{n-1}\log(a_{k+1})-\log(a_k)\right)+\frac1n(n-N)L\ \right|\\
&+\;\frac1nNL\\
&+\;\left|\ \frac{\log(a_0)}{n}\ \right|\\[12pt]
&\to\;0+\epsilon+0+0
\end{align}
$$
