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I have never seen this terminology before, so I will provide the given definition.

A Pseudo-Cauchy sequence is : A sequence $(a_n)$ if for any $\epsilon > 0$ there exists $N \in \mathbb{N}$ such that $|a_{n+1} - a_n | \leq \epsilon \space \forall \space n \geq N$

So then my question is that is a pseudo-cauchy sequence always converging?

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Take $$a_n = \sum_{k=1}^n {1\over k}.$$

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    $\begingroup$ excuse my ignorance, but since $\frac{1}{k} \to 0$ as $k \to \infty$, does this sequence not converge? I know the harmonic series is divergent, but the question is in particular asking for a pseudo-cauchy sequence which does not converge. $\endgroup$ – elbarto Apr 16 '15 at 19:48
  • $\begingroup$ @elbarto The sequence $a_n$ is Pseudo-Cauchy because $|a_{n+1} - a_n| = \frac{1 }{n+1}$. The key is that you're only comparing every $n\geq N$ to $n+1$, rather than an arbitrary $m$ as is in the Cauchy Sequence definition. $\endgroup$ – user38858 Apr 16 '15 at 20:45
  • $\begingroup$ I think you have misunderstood my question. I agree it is Cauchy, but it doesn't diverge. The sequence $(a_n) = \frac{1}{n} \to 0 \space \text{as} \space n \to \infty$ However, I quickly realized mathsadist was referring to the sequence of the partial sums can be used as an example, which is fine :) $\endgroup$ – elbarto Apr 17 '15 at 3:18
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Consider the sequence $$0,1,\frac{1}{2},0,\frac{1}{3},\frac{2}{3},1,\frac{3}{4},\frac{2}{4},\frac{1}{4},0,\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5},1,\frac{5}{6},\frac{4}{6},\frac{3}{6},\frac{2}{6},\frac{1}{6},0, \frac{1}{7},\frac{2}{7},\cdots.$$ This is a pseudo-Cauchy sequence, and every real number between $0$ and $1$ is the limit of a subsequence of our sequence.

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  • $\begingroup$ Is there a precise way to quantify the rule of this sequence $\endgroup$ – elbarto Apr 16 '15 at 19:50
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    $\begingroup$ You mean find an explicit formula? Yes, it can be done, but is unenlightening. $\endgroup$ – André Nicolas Apr 16 '15 at 20:12
  • $\begingroup$ Yes, that is what I meant. Do you just have an arsenal of "useful" sequences which you call upon, or did you think of this one at the top of your head. I wonder because I am unsure if I should have been able to intuitively derive such a sequence, or if I was unprepared in not building my arsenal. I ask this as a student studying undergraduate real analysis 1, sitting an exam in 7 weeks. $\endgroup$ – elbarto Apr 17 '15 at 17:27
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    $\begingroup$ This one I made up quite a while ago. Of course many thousands have made up something similar, but I do not remember reading it. May be just poor memory. Yes, one does acquire a stock of examples that perhaps with modification are useful in a new setting. The standard example for your question is the partial sums of the harmonic series. In my example above, convergence fails much more spectacularly. For explicit formula, it is a bit more convenient to work with numbers of the form $\frac{k}{2^n}$, where $n$ ranges over the positive integers, and $k$ ranges from $0$ to $2^{n}-1$. $\endgroup$ – André Nicolas Apr 17 '15 at 18:09
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For the record it doesn't even work if you ask $a_n$ to be bounded. Take $a_n = \sin(\sum_1^n 1/k)$. Since $\sum_1^n 1/k → ∞ $ and also the difference $\left|\sum_1^{n+1} 1/k - \sum_1^n 1/k\right| = \frac{1}{n+1} \to 0$, $a_n$ oscillates between 1 and -1 forever. But as $\sin$ is Lipschitz, $$|a_{n+1} - a_{n}| \leq \left|\sum_1^{n+1} \frac{1}{k} - \sum_1^n \frac{1}{k} \right| = \frac{1}{n+1} → 0$$

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    $\begingroup$ Your conclusion is correct but incomplete. Define $b_n = \sin(\sum_1^n \pi)$. Then $\sum_1^n \pi \to \infty$, but $b_n$ does not oscillate between $1$ and $-1$ forever. $\endgroup$ – Umberto P. Apr 16 '15 at 16:17
  • $\begingroup$ Thank you, I have added to the argument. $\endgroup$ – Calvin Khor Apr 16 '15 at 17:34
  • $\begingroup$ Thanks @CalvinKhor this is a very interesting example. I feel it will be useful in future :) $\endgroup$ – elbarto Apr 17 '15 at 17:27
  • $\begingroup$ You're welcome :) $\endgroup$ – Calvin Khor Apr 17 '15 at 18:37
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Sequence $$a_n = \sqrt n$$ is pseudo-Cauchy but divergent.

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