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Let $a_n\geq0$

Prove/disprove: $$\lim_{n \to \infty}a_n=1 \rightarrow \lim_{n \to \infty}\sqrt[n] a_n=1$$

Proof: By definition a sequence $\displaystyle\lim_{n \to \infty}\sqrt[n] b_n=L$ iff $\displaystyle\lim_{n \to \infty}\frac{b_{n+1}}{b_n}=L$ since $\displaystyle\lim_{n \to \infty}a_n=1$ $\displaystyle\lim_{n \to \infty}\frac{a_{n+1}}{a_n}=1$ and therefore $\displaystyle\lim_{n \to \infty}\sqrt[n] a_n=1$

Am I right?

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  • $\begingroup$ Is $a_n>0$ for all $n\in\mathbb{N}$? $\endgroup$
    – user60887
    Dec 29, 2014 at 0:32
  • $\begingroup$ @user60887 Yes, Sorry I will add it $\endgroup$
    – gbox
    Dec 29, 2014 at 0:32
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    $\begingroup$ Why should $\lim\limits_{n \rightarrow \infty} \sqrt[n]{b_n} = L \iff \lim\limits_{n \rightarrow \infty} \frac{b_{n+1}}{b_n} = L$ be true "by definition" (if at all)? $\endgroup$
    – dalastboss
    Dec 29, 2014 at 0:36
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    $\begingroup$ If that is a result you may use in this context, I agree the proof follows from your argument modulo a few details. For example, it is also probably worth showing that $\lim\limits_{n \rightarrow \infty} a_n = 1 \implies \lim\limits_{n \rightarrow \infty} \frac{a_{n+1}}{a_n} = 1$. This should only require a couple lines of justification using limit rules, and also be careful to explain why it is safe to divide $a_n$. $\endgroup$
    – dalastboss
    Dec 29, 2014 at 0:45
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    $\begingroup$ yoyo, that limit goes to 1, as does the ratio of successive terms. $\endgroup$
    – dalastboss
    Dec 29, 2014 at 0:59

4 Answers 4

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I don't see where does the fact you use come from (certainly not from the definition). From the link you provided it seems at least the "if" part is true, so I guess you can prove it like that.

But there's also a quick and easy way to see it: \begin{align*}1\le\sqrt[n]x\le x&\text{ if }x\ge1\\1\ge\sqrt[n]x\ge x&\text{ if }x\le1\end{align*} Therefore $\sqrt[n]{a_n}\to1$ because all its members are closer to $1$ than in the original sequence.

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  • $\begingroup$ In other words, it follows from the Squeeze Theorem. $\endgroup$ Dec 29, 2014 at 1:24
  • $\begingroup$ @GregMartin It follows directly from the definition. One could somehow apply the squeeze theorem ($1\pm|1-a_n|$?), but it seems a little bit forced to me. $\endgroup$ Dec 29, 2014 at 1:28
  • $\begingroup$ The Squeeze Theorem can be applied with the three sequences $\min\{1,a_n\}$, $\{\sqrt[n]{a_n}\}$, $\max\{1,a_n\}$ - doesn't seem that forced to me. And "all its members are closer to 1 than in the original sequence" is an intuitive idea (a good one), the rigorous realization of which is in fact this application of the Squeeze Theorem. $\endgroup$ Dec 29, 2014 at 3:40
  • $\begingroup$ @GregMartin The "intuitive idea" is exactly in the definition of a limit! The sentence you quote says $|\sqrt[n]{a_n}-1|\le |a_n-1|$, so from the convergence of $a_n$ we get $\forall\varepsilon>0\exists n_0\forall n\ge n_0, |\sqrt[n]{a_n}-1|\le |a_n-1|<\varepsilon$, which is the convergence of $\sqrt[n]{a_n}$. $\endgroup$ Dec 29, 2014 at 3:48
  • $\begingroup$ - which is the proof of the Squeeze Theorem! :) I feel like we're arguing the same side here.... $\endgroup$ Dec 29, 2014 at 4:57
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I'm not convinced by your proof; even if it's true, it's certainly not 'by definition'.

This problem a bit easier to think about if you take logarithms. The statement is equivalent to $$\lim_{n \to \infty} a'_n = 0 \quad \Rightarrow \quad \lim_{n \to \infty} \frac{1}{n} a'_n = 0$$ where $a'_n=\log a_n$.

Hopefully you can convince yourself that this is true, prove the above statement, and then take exponentials and deduce (from continuity of $\exp$) the result you seek.

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You need to be careful about the implications and how you phrase things. You use a result, this is not "by definition" and the result you quote also proves just one implication, not an equivalence. You could state things like this:

We know that if $\displaystyle\lim_{n \to \infty}\frac{b_{n+1}}{b_n}=L$ and $b_n > 0$ for all sufficiently large $n$, then $\displaystyle\lim_{n \to \infty}\sqrt[n] b_n=L$.

Since $\displaystyle\lim_{n \to \infty}a_n=1$, we have $\displaystyle\lim_{n \to \infty}\frac{a_{n+1}}{a_n}=1$ and $a_n >0$ for all sufficiently large $n$, and therefore $\displaystyle\lim_{n \to \infty}\sqrt[n] a_n=1$.

Commentary: This assumes you are free to use the result you quoted so that the "We know" is justified. Otherwise use an approach from other answers, or prove the result you use too.

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  • $\begingroup$ This is also in a comment. But I think it should be an answer too. $\endgroup$
    – quid
    Dec 29, 2014 at 1:00
  • $\begingroup$ He could use the approach from other answers, or he could prove the lemma... $\endgroup$ Dec 29, 2014 at 1:09
  • $\begingroup$ @ThomasAndrews that's quite true. :-) $\endgroup$
    – quid
    Dec 29, 2014 at 1:11
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Observe that: $|\sqrt[n]{a_n} - 1| = \dfrac{|a_n-1|}{|(\sqrt[n]{a_n})^{n-1} + (\sqrt[n]{a_n})^{n-2}+\cdots +\sqrt[n]{a_n}+1|} < |a_n-1|$. This implies the answer.

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  • $\begingroup$ $|\sqrt[n]{a_n} - 1| = \dfrac{|a_n-1|}{|(\sqrt[n]{a_n})^{n-1} + (\sqrt[n]{a_n})^{n-2}+\cdots +\sqrt[n]{a_n}+1|}$ what Theorem is this? $\endgroup$
    – gbox
    Dec 29, 2014 at 0:51
  • $\begingroup$ $x^n-1 = (x-1)(x^{n-1}+x^{n-2}+\cdots+1)$ $\endgroup$
    – DeepSea
    Dec 29, 2014 at 0:53
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    $\begingroup$ This does not answer the question as asked. $\endgroup$
    – quid
    Dec 29, 2014 at 1:02

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