Prove for every $a_n>0$ if $\lim _{n\to \infty }\:\:\left(\sqrt[n]{a_n}\right)=L$ then $\lim _{n\to \infty }\:\:\left(\frac{a_{n+1}}{a_n}\right)=L$ Prove for every $a_n&gt;0$ if $\lim _{n\to \infty }\:\:\left(\sqrt[n]{a_n}\right)=L$ then $\lim _{n\to \infty }\:\:\left(\frac{a_{n+1}}{a_n}\right)=L$  is true or false 


  i know that's its true if $\lim _{n\to \infty }\:\:\left(\frac{a_{n+1}}{a_n}\right)=L$ then $\lim _{n\to \infty }\:\:\left(\sqrt[n]{a_n}\right)=L$


but in this case i think that this is false and my example is if $a_n=\sqrt[n]{n^n}$ then $\lim _{n\to \infty }\:\:\sqrt[n]{n^n}=\infty \:$ but $\lim _{n\to \infty \:}\left(\frac{\left(n+1\right)^n}{n^n}\right)=e$

is that correct?
Your first statement does not hold. Consider the sequence $a_n = 1, 2, 1, 2, \ldots$. Then
$\lim_{n \to \infty} \sqrt[n]{a_n} = 1$, but $\lim_{n \to \infty} 
\frac{a_{n+1}}{a_n}$ does not exist.

One always has (compare 
Relationship Between Ratio Test and Power Series Radius of Convergence or Limit of ${a_n}^{1/n}$ is equal to $\lim_{n\to\infty} a_{n+1}/a_n$)
$$
 \liminf _{n \to \infty} 
\frac{a_{n+1}}{a_n} \le \liminf_{n \to \infty} \sqrt[n]{a_n}
 \le \limsup_{n \to \infty} \sqrt[n]{a_n}
 \le  \limsup_{n \to \infty} 
\frac{a_{n+1}}{a_n}
$$
so that the existence of $\lim_{n \to \infty} 
\frac{a_{n+1}}{a_n}$ implies the existence of 
$\lim_{n \to \infty} \sqrt[n]{a_n}$ (with the same limit, which can
be finite, $+\infty$, or $-\infty$),
but not the other way around.
The theorem says that if both sequences has finite limit, then the two limits are equal.