How to prove $\lim_{n \to \infty}a_n=1 \rightarrow \lim_{n \to \infty}\sqrt[n] a_n=1$

Question

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?

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)? — dalastboss, Dec 29 '14 at 00:36
@dalastboss I do not know the name but it is as describe here: http://math.stackexchange.com/questions/76743/limit-of-fraca-n1a-n — gbox, Dec 29 '14 at 00:39
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 $a_n$. — dalastboss, Dec 29 '14 at 00:45
yoyo, that limit goes to 1, as does the ratio of successive terms. — dalastboss, Dec 29 '14 at 00:59

score 6 · Accepted Answer · edited Apr 13 '17 at 12:20

6

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.

edited Apr 13 '17 at 12:20

Community

1

answered Dec 29 '14 at 00:44

user2345215

16,422

In other words, it follows from the Squeeze Theorem. – Greg Martin Dec 29 '14 at 01:24
@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. – user2345215 Dec 29 '14 at 01:28
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. – Greg Martin Dec 29 '14 at 03:40
@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 $\sqrt[n]{a_n}$. – user2345215 Dec 29 '14 at 03:48
which is the proof of the Squeeze Theorem! :) I feel like we're arguing the same side here....

Greg Martin

Dec 29 '14 at 04:57

score 3 · Answer 2 · answered Dec 29 '14 at 00:39

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.

quid · Answer 3 · 2014-12-29T01:12:37.790

1

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.

edited Dec 29 '14 at 01:12

answered Dec 29 '14 at 00:53

quid

42,135

This is also in a comment. But I think it should be an answer too. – quid Dec 29 '14 at 01:00
He could use the approach from other answers, or he could prove the lemma... – Thomas Andrews Dec 29 '14 at 01:09
@ThomasAndrews that's quite true. :-) – quid Dec 29 '14 at 01:11

score 0 · Answer 4 · answered Dec 29 '14 at 00:46

0

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.

answered Dec 29 '14 at 00:46

DeepSea

77,651

$|\sqrt[n]{a_n} - 1| = \dfrac{|a_n-1|}{|(\sqrt[n]{a_n})^{n-1} + (\sqrt[n]{a_n})^{n-2}+\cdots +\sqrt[n]{a_n}+1|}$ what Theorem is this? – gbox Dec 29 '14 at 00:51
$x^n-1 = (x-1)(x^{n-1}+x^{n-2}+\cdots+1)$ – DeepSea Dec 29 '14 at 00:53
1

This does not answer the question as asked. – quid Dec 29 '14 at 01:02

How to prove $\lim_{n \to \infty}a_n=1 \rightarrow \lim_{n \to \infty}\sqrt[n] a_n=1$

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