The following is the definition of a pseudo-Cauchy sequence:

A sequence $a_n$ is pseudo Cauchy if $\forall$ $\epsilon > 0$, $\exists \ N \in \mathbb N$ such that whenever $n \ge N$, $|a_{n+1} - a_n| \lt \epsilon$.

I don't know if this is a valid proof or not, but this is how I tackled it:


I noticed that the definition of pseudo-Cauchy is similar to that of Cauchy, which is defined as:

A sequence $a_n$ is Cauchy if $\forall$ $\epsilon > 0$, $\exists \ N \in \mathbb N$ such that whenever $m,n \ge N$, $|a_m - a_n| \lt \epsilon$.

Cauchy seems more liberal in the sense that any elements in the set have to be less than $\epsilon$, whereas the pseudo-Cauchy has a more strict requirement that neighbors and only neighbors must satisfy the requirement. However, I said that if we let $m = n+1$, that is, just choose the other element in the set to be the neighbor, that Pseudo-Cauchy sets actually are Cauchy. It is well known that Cauchy sequences are bounded, therefore pseudo-Cauchy sequences are bounded because they can be moulded into the definition of Cauchy.

  • $\begingroup$ Every Cauchy sequence is a pseudo-Cauchy sequence, not the other way around. You can't just let $m=n+1$, in order to show that a sequence is Cauchy you will have to show it holds for all $m$ not just for $m=n+1$. $\endgroup$ – Jorik Nov 18 '15 at 15:41
  • $\begingroup$ I remember someone writing that, historically, there were some mathematicians who suggested that a necessary and sufficient condition for convergence of a sequence would be this condition that you call "pseudo-Cauchy." As you can see this is false. I think this is the only point of the exercise. I doubt you will ever see the term "pseudo" Cauchy again as the idea proves to be useless. $\endgroup$ – B. S. Thomson Nov 18 '15 at 16:44
  • $\begingroup$ This question was continued a day later by the OP who hasn't yet sorted out the connection between "bounded" and "convergent" and "Cauchy." See math.stackexchange.com/questions/1536274/…. $\endgroup$ – B. S. Thomson Nov 19 '15 at 20:31

Define $a_n = \sum_{i=0}^n \frac{1}{i+1}$ (in the reals).

Then $|a_{n+1} - a_n| = \frac{1}{n+2}$, so the sequence is pseudo-Cauchy.

But it is a divergent sequence, as is well known (harmonic series).

So no, not all pseudo-Cauchy sequences are Cauchy. And this sequence is unbounded.

  • 2
    $\begingroup$ As an interesting extension of this example, it can be used (with a few negative signs) to produce a bounded pseudo-Cauchy sequence that diverges. So pseudo-Cauchy are not necessarily Cauchy, nor necessarily bounded. Even if they are bounded they need not converge. $\endgroup$ – B. S. Thomson Nov 18 '15 at 19:41
  • $\begingroup$ How is the sequence unbounded? $\endgroup$ – Gooshi Roy Nov 19 '15 at 2:40
  • $\begingroup$ The sequence $\{a_n\}$ that Dr Brandsma constructed for you is strictly increasing. If it were bounded it would converge and the harmonic series would then be convergent. It is hard to construct an example of unbounded pseudo-Cauchy sequence if you don't recall what you learned about series in the previous year. The harmonic series should come immediately to mind, since each term adds just a wee bit to the partial sums and yet the series diverges. $\endgroup$ – B. S. Thomson Nov 19 '15 at 2:49
  • $\begingroup$ @B.S.Thomson, ah thank you. I naiively slipped up on some definitions. $\endgroup$ – Gooshi Roy Nov 19 '15 at 3:20

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