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Why doesn't $d(x_n,x_{n+1})\rightarrow 0$ as $n\rightarrow\infty$ imply ${x_n}$ is Cauchy?

my question is this:


The following definition is weaker than the definition of Cauchy sequences:

$\forall \; \epsilon > 0, \;\exists N \in \mathbb{N} \;s.t.\; \forall\; n \geq N, \; |a_{n+1}-a_n | < \epsilon.$

Show that this is not equivalent to $(a_n)$ being a Cauchy sequence.


The definition of Cauchy sequence is:

A sequence $(s_n)$ is Cauchy if (and only if) for each $\epsilon > 0$ there exists an integer $N$ with the property that $|s_n-s_m| < \epsilon$ whenever $n\geq N$ and $m \geq N$.

Note that a sequence (of real numbers) is convergent if and only if it is Cauchy.


So I see an (the?) obvious difference between these two in that the Cauchy criteria demands that all values in a sequence above a certain index ($N$) are within a prescribed tolerance of each other, whether adjacent or not. This is where the question is weaker, in that it only requires the immediately adjacent values of the sequence to be within a tolerance of $\epsilon$. This would then allow, by taking successive differences of adjacent values, to accumulate a difference greater than $\epsilon$. This is seen as,

$$\left| \sum_{i=1}^{n+1}\,a_i - \sum_{i=1}^{n}\,a_i\right| < \epsilon,\quad \left|\sum_{i=1}^{n+2}\,a_i - \sum_{i=1}^{n+1}\,a_i\right| < \epsilon,\quad \left| \sum_{i=1}^{n+3}\,a_i - \sum_{i=1}^{n+2}\,a_i\right| < \epsilon,$$ and summing each side of the inequalities gives (after reverting to sequence-notation and employing the triangle inequality), $$ \left(|a_{n+1}-a_n| + |a_{n+2}-a_{n+1}| + |a_{n+3}-a_{n+2}| + \cdots + |a_{K} - a_{K-1}| \right) \leq \left( \epsilon_{1,2} + \epsilon_{2,3} + \epsilon_{3,4} + \cdots + \epsilon_{K-1,K} \right) $$ which implies $$\left( \epsilon_{1,2} + \epsilon_{2,3} + \epsilon_{3,4} + \cdots + \epsilon_{K-1,K} \right)_{\textrm{ by weaker criteria }} \geq |a_n - a_{n+K}|_{\textrm{ by Cauchy criteria }}$$

If I understand these differences correctly, then my main problem is putting these into a formal mathematical proof. Unless this would qualify?

Thanks much for the help and the site!

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marked as duplicate by Kannappan Sampath, David Mitra, Mariano Suárez-Alvarez Feb 13 '12 at 22:00

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

    
Exact Duplicate: math.stackexchange.com/q/107336/21436 –  user21436 Feb 13 '12 at 21:40
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3 Answers 3

up vote 7 down vote accepted

Your analysis is sound: though differences between neighbours may be ultimately small, these differences can build up. However, they do not need to build up: after all, there are Cauchy sequences. So the analysis has to be used to generate specific situations in which the differences do build up, or at least to prove less directly that this can happen.

We exhibit a sequence that satisfies the "weaker" condition and is not Cauchy. The following example is not mine, it was given on MathSE in the fairly recent past.

Look at the sequence $$0, \tfrac{1}{2}, 1, \tfrac{2}{3}, \tfrac{1}{3}, 0, \tfrac{1}{4}, \tfrac{2}{4}, \tfrac{3}{4}, 1, \tfrac{4}{5}, \tfrac{3}{5}, \tfrac{2}{5}, \tfrac{1}{5}, 0, \tfrac{1}{6}, \tfrac{2}{6}, \tfrac{3}{6}, \tfrac{4}{6}, \tfrac{5}{6}, 1, \tfrac{6}{7} \dots$$ (we are travelling back and forth between $0$ and $1$, and each round uses smaller and smaller steps).

The above sequence satisfies your condition: After a while, any two consecutive terms are very close to each other. But the sequence is clearly not Cauchy. It is clear how in this example small differences do build up.

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Hmmm... Nice example. I also knew of $1+1/1 + 1/2 + 1/3 + 1/n $ as a sequence of partial sums. I guess I do not understand why my reasoning does not prove they are not equivalent. Also, is providing a counter-example the only way one would prove they are not equivalent? Thanks and (Thanks also @Jasper) –  nate Feb 13 '12 at 21:30
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@nate: Giving a counterexample may not be the only way. Usually, one shows that $A$ is a proper subset of $B$ by exhibiting an object which is in $B$ but not in $A$. But it is possible to imagine a highly indirect proof. There are some famous such proofs, particularly in Graph Theory. For simplicity, I will give an example that is actually phony, but may be familiar to you. Cantor proved that the reals have uncountable cardinality, and that the set of algebraic numbers is countable. So there exist transcendental numbers. –  André Nicolas Feb 13 '12 at 21:50
    
I am very interested in this indirect proof - it is very interesting, but have no experience at all in graph theory. I do have some experience with Cantor sets though and will think about this some more. Thanks all and sorry for the duplicate - I missed the above previous question. –  nate Feb 14 '12 at 3:33
    
@nate: For genuine and important examples, I was thinking of the [Probabilistic Method] in graph theory. The link will supply some information. –  André Nicolas Feb 14 '12 at 3:40
    
Thank you for the reference to graph theory @Andre, I am taking it next semester! –  nate Feb 14 '12 at 3:47
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Showing that the sum of the epsilons is possibly greater than the single epsilon in the supposedly stronger form does not prove the two forms are not equivalent. What you need is a counterexample.

The sequence $(1,1+\frac{1}{2},1+\frac{1}{2}+\frac{1}{3},\ldots)$ does not converge in $\mathbb R$ and so is not Cauchy. Yet it satisfies the weaker criterion.

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That was the one I was thinking of. (+1) –  robjohn Feb 13 '12 at 22:13
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Your intuition and argument are correct as far as they go, but note that you've only shown that the Cauchy condition is at least as strong, not strictly stronger. You can show "strictly stronger" by finding a counterexample, or by finding a property that all Cauchy sequences have and the weaker formulation doesn't have. A counterexample is probably easier to come by. Think about how this could happen. Can you think of a function $f$ defined for $x>0$ (hence, it is defined for each $n = 1, \; 2, \; 3,\;...$) that approaches $\infty$ as $x \rightarrow \infty$ (hence, $f(n) \rightarrow \infty$ as $n \rightarrow \infty$), but it does so in such a way that its everywhere positive slope approaches $0$ as $x \rightarrow \infty$ (thereby ensuring that $f(n+1) - f(n) \rightarrow 0$)? More specifically, think about how $f(x) = \log x$ or $f(x) = \sqrt x$ could help in light of what I've said.

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