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Prove for every $a_n>0$ if $\lim _{n\to \infty }\:\:\left(\sqrt[n]{a_n}\right)=L$ then $\lim _{n\to \infty }\:\:\left(\frac{a_{n+1}}{a_n}\right)=L$ 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?

D.Z.O
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  • That's not correct. A correct counter-example would be a sequence such that $(a_{+1}/a_n)\to\infty$ while $\sqrt[n]{a_n}$ has a finite limit. – Tom-Tom Dec 18 '17 at 13:22
  • @Tom-Tom is there any example like that for $a_n>0$ ? – D.Z.O Dec 18 '17 at 13:36
  • Your example is wrong. With $a_n = n^n$ you have $\sqrt[n]{a_n} = \sqrt[n]{n^n} \to \infty$ and $\frac{a_{n+1}}{a_n} = \frac{\left(n+1\right)^{n+1}}{n^n} \to \infty$. – Martin R Dec 18 '17 at 13:44
  • @MartinR yes i noticed that , i think that Spike Bughdaryan answer is correct because i tried alot to find example to prove it false but didn't find any – D.Z.O Dec 18 '17 at 13:47
  • the question is prove if this statement is true or false – D.Z.O Dec 18 '17 at 13:50
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    @D.Z.O: I misread your question. – The initial statement is wrong, consider the sequence $2, 1, 2, 1, 2, 1, \ldots$ – Martin R Dec 18 '17 at 13:56

2 Answers2

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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.

Martin R
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The theorem says that if both sequences has finite limit, then the two limits are equal.

Spike Bughdaryan
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