How can I prove that if $\sum a_n$ is a series with elements non-null and $\lim \frac{|a_{n+1}|}{|a_n|} = L$, so $\lim \sqrt[n]{|a_n|}=L$.
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If $\left|\dfrac{a_{n+1}}{a_n}\right| \to L$, then for $\epsilon >0$, there exists $N \in \mathbb{N}$ so that for $n>N$, $$ \left|\; \left|\dfrac{a_{n+1}}{a_n}\right| - L \right| < \epsilon. $$
But then using a nice cancellation trick, for $n>N$, we have $$ |a_n|= \dfrac{|a_n|}{|a_{n-1}|} \cdot \dfrac{|a_{n-1}|}{|a_{n-2}|} \cdot \cdots \cdot \dfrac{|a_{N+1}|}{|a_N|} \cdot |a_N| $$ Notice that all the terms in the product but $|a_n|$ cancel 'diagonally'. But using the inequality above, we have $$ |a_n|= \dfrac{|a_n|}{|a_{n-1}|} \cdot \dfrac{|a_{n-1}|}{|a_{n-2}|} \cdot \cdots \cdot \dfrac{|a_{N+1}|}{|a_N|} \cdot |a_N| < (L+\epsilon) \cdot (L+\epsilon) \cdots (L+\epsilon) \cdot |a_N| $$ Then we have proved $$ |a_n| < (L+\epsilon)^{n-N} |a_N| $$ Take the $n$th root of both sides to find $$ \sqrt[n]{|a_n|} < (L+\epsilon)^{(n-N)/n} \sqrt[n]{|a_N|}= (L+\epsilon)^{1-N/n} \sqrt[n]{|a_N|}. $$ Taking the limit, $$ \lim_{n \to \infty} \sqrt[n]{|a_n|} \leq (L+ \epsilon) \cdot 1= L+ \epsilon $$ This proves that $\lim_{n \to \infty} \sqrt[n]{|a_n|} \leq L$. But making a small change (can you see which?!) and rerunning the 'same' argument, we find $\lim_{n \to \infty} \sqrt[n]{|a_n|} \geq L$. Therefore, $\lim_{n \to \infty} \sqrt[n]{|a_n|} = L$, as desired.
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Is this small change catch the left part of this second equation $||\frac{a_{n+1}}{a_n}|-L| < \epsilon\Rightarrow L-\epsilon < |\frac{a_{n+1}}{a_n}| < L+\epsilon$? – Rebeca Lie Yatsuzuka Silva Jan 16 '20 at 20:58
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Seeing the other similar questions now, your answer is much more clear in my opinion. – Rebeca Lie Yatsuzuka Silva Jan 16 '20 at 20:59