Give an example of a sequence $(x_n)$ of real numbers, where $\displaystyle\lim_{n\to+\infty}|x_n-x_{n+1}|=0$, but $(x_n)$ is not a Cauchy sequence
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Martin Sleziak
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AdityaGhosh
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9Try $x_n=\sqrt n$. – Davide Giraudo Oct 25 '11 at 19:42
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Alternatively, consider $x_n$ = log($n$). – Grasshopper Oct 25 '11 at 19:47
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6This is not the simplest answer, but you can also $x_n$ to be the $n^{th}$ partial sum of a divergent series whose terms approach $0$. (You can even get examples where the terms are nonnegative and monotonically decreasing.) For e.g., the harmonic series fits the bill. – Srivatsan Oct 25 '11 at 19:49
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2Or take $x_n$ to be the $n$th partial sum of any divergent series $\sum a_k$ such that $a_k\to0$ as $k\to\infty$. – Nick Strehlke Oct 25 '11 at 19:50
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@SrivatsanNarayanan I see you beat me by 35 seconds! :) – Nick Strehlke Oct 25 '11 at 19:51
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@SrivatsanNarayanan That is very good in my opinion. You should post it as an actual answer – mathmath8128 Oct 25 '11 at 19:51
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Please don't just copy your homework problems onto the site; you should perhaps expand on what you've tried or what your thoughts about the problem might be. That way, you also don't seem like you are assigning the reader homework or giving orders, which is what the post reads like right now. – Arturo Magidin Oct 25 '11 at 20:16
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See also: Pseudo-Cauchy sequence, I want an example of a sequence that satisfies $|x(n) - x(n-1)| \to 0$ but not Cauchy, If ${x_n}$ satisfies that $x_{n+1} - x_n$ goes to $0$, is ${x_n}$ a Cauchy sequence?, etc. – Martin Sleziak Dec 03 '16 at 08:05
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I would like to add that one can construct a bounded sequence $\{x_n\}$ satisfying $| x_n - x_{n+1}|\rightarrow 0$ that is not Cauchy. E.g., take: $$ 0, {1\over2}, 1, {2\over3}, {1\over3}, 0 ,{1\over4}, {2\over4}, {3\over4}, 1, {4\over5}, {3\over5}, {2\over5}, {1\over 5}, 0,{1\over6}, \ldots $$
gebruiker
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David Mitra
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Posting my comment as an answer.
You can take $x_n$ to be the $n^{th}$ partial sum of any divergent series $\sum_{k} a_k$ such that $a_k \to 0$ as $k \to \infty$. In fact, we can even produce such an example with the additional restriction that the sequence $(a_k)$ is positive and monontonically decreasing. The canonical example of such a series is the harmonic series $\sum_k \frac{1}{k}$.
Actually, the above description is complete in the sense that if $x_n$ is any sequence satisfying the OP's requirements, then the series $\sum_k a_k$ defined by $a_n = x_n - x_{n-1}$ is such that $a_k \to 0$ as $k \to \infty$ and yet the series is divergent.
Srivatsan
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One problem with your solution - it is very possible that "divergent series" is a concept not familiar to someone asking a homework question about Cacuhy sequences... – Gadi A Oct 25 '11 at 20:45
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@Gadi Thanks for pointing it out. Yes, I too find the solution unsatisfactory, and that's one reason for posting it as CW. But I just felt that the connection is interesting and deserves to be stated somewhere (for other readers, if not for the OP). Also I find the description in terms of series is more intuitive than the condition $x_n - x_{n-1} \to 0$ as $n \to \infty$ (even though these two say the same thing). – Srivatsan Oct 25 '11 at 20:50
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Take $x_n=\sqrt n$. Since $|x_{n+1}-x_n|=\left|\sqrt{n+1}-\sqrt n\right|=\left|\frac {n+1-1}{\sqrt{n+1}+\sqrt n}\right|\leq \frac 1{\sqrt n}$, we have $\displaystyle\lim_{n\to+\infty}x_{n+1}-x_n=0$. But we have $|x_{2n}-x_n|=\sqrt 2\sqrt n-\sqrt n=(\sqrt 2-1)\sqrt n$, hence $(x_n)_n$ cannot be a Cauchy sequence.
Davide Giraudo
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1$x_n$ is unbounded so it cannot be a Cauchy sequence, thus you don't have to do these computations in the second sentence. – Damian Sobota Oct 25 '11 at 21:17
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I agree, but the computations to show this statement are not more complicated than mine. – Davide Giraudo Oct 25 '11 at 21:20