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Harmonic series diverges and pseudo Cauchy however it's not bounded. So how can I find such a sequence?

A sequence $(s_n)$ is pseudo-Cauchy if, for all $\xi>0$, there exists an $N$ such that if $n ≥ N$, then $|s_{n+1}−s_n| < ξ$.

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    $\begingroup$ What is a pseudo-Cauchy sequence? $\endgroup$ – user75930 Oct 16 '13 at 17:39
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    $\begingroup$ $$\textstyle 0,\,{1\over 2},\, 1,\, {2\over 3},\,{1\over 3},\,{0},\,{1\over 4},\,{2\over 4},\,{3\over 4},\,1,\,{4\over 5},\,{3\over 5},\,{2\over 5},\,{1\over 5}, \,0, \ldots$$ $\endgroup$ – David Mitra Oct 16 '13 at 17:42
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    $\begingroup$ A sequence (sn) is pseudo-Cauchy if, for all ξ> 0, there exists an N such that if n ≥ N, then |sn+1−sn| < ξ $\endgroup$ – user11111 Oct 16 '13 at 17:42
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Try the sequence $$a_n=\sin\sqrt n$$

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  • $\begingroup$ It's bounded and divergent, however it's not pseudo Cauchy. $\endgroup$ – user11111 Oct 16 '13 at 17:52
  • $\begingroup$ @user11111 $\sqrt{n+1}-\sqrt n\rightarrow 0$ as $n\rightarrow\infty$. Nicer than my example. $\endgroup$ – David Mitra Oct 16 '13 at 18:01
  • $\begingroup$ @user11111 My last comment implies Hagen's sequence is indeed pseudo-Cauchy. $\endgroup$ – David Mitra Oct 16 '13 at 18:06
  • $\begingroup$ Be careful here, we want to show |sn+1−sn| < ξ but not n+1 - n .So Sin(n) can be any number when n goes to infinite, thus it's not pseudo Cauchy. $\endgroup$ – user11111 Oct 16 '13 at 18:14
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    $\begingroup$ @user11111 By the Mean Value Theorem, $|\sin\sqrt{n+1}-\sin\sqrt n|\le|\sqrt{n+1}- \sqrt n|{\rightarrow} 0$. $\endgroup$ – David Mitra Oct 16 '13 at 18:18

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