Prove that if $\lim\limits _{ n\rightarrow \infty }{ \frac { { x }_{ n+1 } }{ { x }_{ n } } =a } $ exist then... I should prove that for series $\sum _{ n=1 }^{ \infty  }{ { x }_{ n } } $, $(x_n > 0)$, if $\lim\limits _{ n\rightarrow \infty  }{ \frac { { x }_{ n+1 } }{ { x }_{ n } } =a } $ exists then $\lim\limits_{ n\rightarrow \infty  }{ \sqrt [ n ]{ { x }_{ n } } =a } $ also exists. 
Converse is not true always.
How can show it? I don't how to start. Thanks beforehand.
 A: $$
\lim_{n\to\infty}\frac{x_{n+1}}{x_n}=a\tag{1}
$$
is equivalent to
$$
\lim_{n\to\infty}\left(\vphantom{\lim_{n\to\infty}}\log(x_{n+1})-\log(x_n)\right)=\log(a)\tag{2}
$$
Applying the Stolz-Cesàro Theorem to $(2)$ says
$$
\lim_{n\to\infty}\frac{\log(x_n)}{n}=\log(a)\tag{3}
$$
which is equivalent to
$$
\lim_{n\to\infty}\sqrt[\large n]{x_n}=a\tag{4}
$$
To show that the other direction is not true, consider the sequence
$$
x_n=4^{-\lfloor n/2\rfloor}\tag{5}
$$
A: The result follows from Cesàro mean lemma applied to $(\log(x_n))_n$. If you are in need of some more details, let me know.
A: To show the converse is not necessarily true, choose the sequence $\{x_n\}$ as follows.
For each $n\in \mathbb{N}$, $$x_n =
\begin{cases}
na^n   & \text{if $n$ is even} \\[2ex]
a^n & \text{if $n$ is odd .}
\end{cases}$$
Then it is easy to see that $\lim\limits_{ n\rightarrow \infty  }{ \sqrt [ n ]{ { x }_{ n } } =a }$. But $\lim\limits _{ n\rightarrow \infty  }{ \frac { { x }_{ n+1 } }{ { x }_{ n } } }$ dose not exist.
