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Is

$a_n= \dfrac{1}{2\sin(n)+0.5\sqrt{n}+1.1}+n$

a Cauchy sequence? If I plot the difference between $a_n$ and $a_{n+1}$ it seems to be shrinking as $n$ goes to infinity, with the difference reaching a limit of 1.

user5402
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Gustavo
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  • It isn't, a Cauchy sequence is bounded. Your observation is also correct: $|a_n-a_{n+1}| = 1 + $terms that have limit $0$. – David Mitra Jul 23 '15 at 11:26
  • cauchy sequence also converge (at least in R) – user190080 Jul 23 '15 at 11:31
  • Thanks David. I have edited the equation to reflect that. It is certainly not bounded, but the difference between terms does get smaller. A property of Cauchy sequences is that they are bounded. The equation above is not bounded. I thought that being Cauchy implied boundedness. – Gustavo Jul 23 '15 at 11:32
  • For a Cauchy sequence, however small a number $\epsilon>0$ be, you can always find a positive integer $N$ such that $|a_m-a_n|<\epsilon\ \forall m,n> N$. Simply speaking for Cauchy sequences the "difference" that you are looking at should converge to $0$ as $n\to \infty$. – Samrat Mukhopadhyay Jul 23 '15 at 11:32
  • Thank you for your comments. I see now that the difference needs to converge to zero and not a constant. – Gustavo Jul 23 '15 at 11:34

2 Answers2

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It's not enough that the difference reaches a limit of $1$. The difference has to become arbitarily small, but it doesn't.

Indeed, you see that $a_{n+1}-a_n = (n+1)-n + \text{junk} = 1+\text{junk}$, where "junk" approaches $0$ as $n\rightarrow\infty$. The sequence is not Cauchy.

Mankind
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The difference between consecutive terms becomes $1$ as $n \to \infty$, this doesn't mean anything, however. For example, consider $a_{n} = n$ which you know obviously diverges, even though the difference between the terms is $1$. So your sequence is most definitely not Cauchy.

A Cauchy sequence requires $|a_n - a_m| < \epsilon$ for all $m,n > N$ for some integer $N$. It has nothing to do with "consecutive" terms and the difference between terms needs to converge to $0$, not $1$.

Zain Patel
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    to show that it is not a cauchy sequence it is enough to show that the difference $|a_{n+1}-a_n|$ doesn't become arbitrary small. but of course, if you want to show it is a c-s then you are completely right! – user190080 Jul 23 '15 at 11:46