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I read in the book Applied Analysis by Hunter and Nachtergale that

the sequence $x(n)=\log(n)$ is not Cauchy since $\log(n)\to\infty$

But that seems to be irrelevant to the definition of a Cauchy sequence which I understand is as follows:

A sequence $x(n)$ is said to be Cauchy if for every $\epsilon > 0$ there exists an $N$ such that $\lvert x(m)-x(n)\rvert < \epsilon$ for all $m,n>N$.

This sequence $(\log n)$ seems to meet the definition. So how come it is not considered Cauchy?

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  • $\begingroup$ yes , thanks much! $\endgroup$ – Alhpa Delta Apr 2 '18 at 13:02
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Hint. Look at the difference $$ \log(2n) - \log(n) . $$

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  • $\begingroup$ isnt $log3>log2$ and thus the requirement of Cauchy sequence satisifed? $\endgroup$ – Alhpa Delta Apr 2 '18 at 12:29
  • $\begingroup$ The definition of a Cauchy sequence requires "less than $\epsilon$ for _all+ $m > n$". What if $m = 2n$? (Use properties of logs.) $\endgroup$ – Ethan Bolker Apr 2 '18 at 12:41
  • $\begingroup$ thanks @ethan! that helps $\endgroup$ – Alhpa Delta Apr 2 '18 at 12:45
  • $\begingroup$ (+1) We have a similar issue with $\sqrt{n}$. Like $\log n$, it grows slowly and the difference between successive terms might vanish, as $\sqrt{n+1}-\sqrt{n}=\frac1{\sqrt{n+1}+\sqrt{n}}$ but clearly we can always choose a sufficiently large $N$ with $\sqrt{N}-\sqrt{n}>\varepsilon$. The relation needs to hold for all $N>n$ but both $\log n$ and $\sqrt{n}$ are unbounded and fail. $\endgroup$ – Jam Feb 22 '20 at 22:17
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The short answer is that $\mathbb{R}$ is complete so every Cauchy sequence has a limit in $\mathbb{R}$. Thus since $log(n)$ does not tend to a real number it cannot possibly be Cauchy.

To see why your idea doesn't work fix $\epsilon=1$

Note $|log(m)-log(n)| = |log(\frac{m}{n})|$

For any N you give me, pick some $n>N$ then for any $m>en$ we have $|log(m/n)|>1=\epsilon$

Edit: in response to comment on original post, we do not need infinity to "count" - every Cauchy sequence has a finite limit. It is easy to see every Cauchy sequence is bounded since $\exists N \in \mathbb{N}: |x_n-x_N|<1 \forall n>N$ Since $x_N$ fixed the sequence is bounded for $n>N$ and the first N terms are a finite sequence so bounded.

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  • $\begingroup$ $m > n$ does not guarantee that inequality on the log of the quotient. $\endgroup$ – Ethan Bolker Apr 2 '18 at 12:28
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    $\begingroup$ You don't need anything about limits or convergence or even boundedness to answer this question. Just the definition of "Cauchy". $\endgroup$ – Ethan Bolker Apr 2 '18 at 12:42
  • $\begingroup$ Yes my answer does not require any of these but OP was confused about why no limit => not Cauchy and a subsequent comment was confused about whether infinity was a valid limit point so I included these in the hope they would be of some help. $\endgroup$ – Mathmop Apr 2 '18 at 12:45
  • $\begingroup$ Patrick , very helpful. $\endgroup$ – Alhpa Delta Apr 2 '18 at 12:45
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It's not clear why you say the sequence $(\log(n))$ "seems" to satisfy the definition of Cauchy sequence, but in fact it does not.

It's trivial from the definition that any Cauchy sequence is bounded; no need to invoke the fact that Cauchy sequences of reals are convergent: Say $(x_n)$ is Cauchy. The definition shows that there exists $N$ such that $|x_n-x_m|<1$ for every $n,m> N$. So in particular $|x_{N+1}-x_n|<1$ for every $n>N$, and now the triangle inequality shows that $$|x_n|<|x_{N+1}|+1\quad(n>N).$$So $x_n$ is bounded.

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  • $\begingroup$ I was proceeding from the definition like this: x(n+1)-x(n)=log(1+1/n). and as n tends to infinity this difference tends to zero. so we could always find an $\epsilon>0$ such that $|x(m)-x(n)|<\epsilon$ So, the definition of Cauchy was satisfied. Where exactly is the fallacy in my logic? $\endgroup$ – Alhpa Delta Apr 2 '18 at 14:39
  • $\begingroup$ The fallacy is that showing $|x_{n+1}-x_n|\to0$ simply does not show that $(x_n)$ is a Cauchy sequence - in fact $x_n=\log(n)$ iis the standard counterexample. You say "so we could always find $\epsilon>0$ such that...". The definition of Cauchy sequence has nothing to do with "finding" $\epsilon$. You need to show that given $\epsilon>0$ yyou can find $N$ so $|x_n-x_m|<\epsilon$ for all $n,m>N$; you haven't done anything remotely like that. $\endgroup$ – David C. Ullrich Apr 2 '18 at 14:46
  • $\begingroup$ ok. So if $\epsilon=1$ then $N=1$; $\epsilon=0.5$, then $N=3$; $\epsilon=0.25$, then $N=5$; $\epsilon=0.125$, then $N=9$; $\epsilon=0.03125$, then $N=33$; and so on... I might be missing the point completely (for which I apologize), but it seems to me that I can always find an $N$ which meets the criteria $|x(n)-x(m)|<\epsilon$. I am sure, I am wrong somewhere, just dont know where.... $\endgroup$ – Alhpa Delta Apr 2 '18 at 15:08
  • $\begingroup$ When you say "if $\epsilon=1$ then $N=1$" all I can guess is that you mean "$|\log(n)-\log(m)|<1$ whenever $n,m>1$". If that's not what you mean, what do you mean? If that is what you mean: That's obviously false - I can't imagine why you think it's true. $\endgroup$ – David C. Ullrich Apr 2 '18 at 15:15
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    $\begingroup$ finally, i got it.log(3)-log(1)=1.1>1 and so on....it does NOT meet Cauchy condition. Given $\epsilon=1$ i am NOT able to get N where it holds for ALL $m,n>N$ $\endgroup$ – Alhpa Delta Apr 2 '18 at 15:57
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Answering late since there does not seem to be a complete proof for your question. For the sake of contradiction, suppose $(\log(n))_n$ is Cauchy. Let $\epsilon>0$. Then there exists $N \in \mathbb{Z}_+$ such that $|\log(n)-\log(m)|<\epsilon$ for $n \geq m \geq N$. Remark $|\log(n)-\log(m)|=|\log\left(\frac{n}{m}\right)|=\log\left(\frac{n}{m}\right)<\epsilon$ for $n \geq m \geq N$. The elimination of the absolute value is because $\log\left(\frac{n}{m}\right)\geq 0$ for any $n,m \in \mathbb{Z}_+$ with $n \geq m$.

Recall the exponential function is increasing, and hence preserves inequality. It follows $\frac{n}{m}<e^\epsilon$. Now suppose $n>\max(N,me^\epsilon,m)$. It follows $\frac{n}{m}>e^\epsilon$. This is a contradiction. Therefore $(\log(n))_n$ cannot be Cauchy. $\square$

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  • $\begingroup$ "The elimination of the absolute value is because $\log\left(\frac{n}{m}\right)\geq 0$ for any $n,m \in \mathbb{Z}_+$." We would also need to specify that $m\le n$. $\endgroup$ – Jam Feb 22 '20 at 21:59
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    $\begingroup$ @Jam oops, fixed it. $\endgroup$ – Spencer Kraisler Feb 22 '20 at 22:14
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Because it can not converge. "$x_n$ is a Cauchy sequence if and only if it is convergennt". ($x_n$ is a real sequence).

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  • $\begingroup$ but (I am sorry for being so obtuse) that is not what the definition of Cauchy sequence says. so why should convergence matter? definition only says "terms should gt close". also, a "stupid question": does covergence to infinity not count? (sorry, I am not a mathematician, but trying to make sure i get the concepts right) $\endgroup$ – Alhpa Delta Apr 2 '18 at 12:26
  • $\begingroup$ Can you write the negation of the definition of Cauchy sequence? $\endgroup$ – Gustave Apr 2 '18 at 12:32
  • $\begingroup$ @AlhpaDelta there is no "convergence to infinity". A sequence that tends to infinity is divergent. $\endgroup$ – Angina Seng Apr 2 '18 at 12:41
  • $\begingroup$ thanks, that definition (of divergent) helps clarify it further. $\endgroup$ – Alhpa Delta Apr 2 '18 at 12:44

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