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.
$$ \lim_{n\to\infty}\frac{x_{n+1}}{x_n}=a\tag{1} $$ 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 $$ x_n=4^{-\lfloor n/2\rfloor}\tag{5} $$
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.
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.
