Let $\{x_n\}$ denote a non-convergent sequence. Show that there exists a sequence of natural numbers $\{p_n\} \subset \Bbb N$ such that: $$ \lim_{n\to\infty}(x_{n+p_n} - x_n) \ne 0 $$

Suppose that: $$ \lim_{n\to\infty}(x_{n+p_n} - x_n) = 0 $$

Clearly $n+p_n > n$. Denote $n+p_n = m_n$: $$ \lim_{n\to\infty}(x_{m_n} - x_n) = 0 \iff \forall \epsilon > 0\ \exists N \in\Bbb N: \forall m_n > n > N \implies |x_{m_n} - x_n| < \epsilon $$ which denotes a Cauchy Criterion for the sequence $x_n$. If $x_n$ is fundamental then it must converge to some limit, but from the problem statement $x_n$ is divergent and hence we've arrived at a contradiction.

Therefore for $x_n$ to be divergent there must exist some sequence $\{p_n\}$ for which: $$ \lim_{n\to\infty}(x_{n+p_n} - x_n) \ne 0 $$

Please let me know whether there is anything wrong with the proof or whether it's fine. Thank you!

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    $\begingroup$ As Rafay Ashary noted, there's a problem with the reformulation. Something like "for every sequence $p_n$" should appear on both sides of the equivalence. Here, it seems to appear in the wrong place in the RHS, and it doesn't appear in the LHS at all. $\endgroup$ – Michał Miśkiewicz Dec 26 '18 at 18:03
  • $\begingroup$ A visualization of the idea from @RafayAshary answer. Just for the case $\endgroup$ – roman Dec 26 '18 at 18:36

I don't see why $$\lim_{n\to\infty}(x_{m_n} - x_n) = 0 \iff \forall \epsilon > 0\ \exists N \in\Bbb N: \forall m_n > n > N \implies |x_{m_n} - x_n| < \epsilon$$ Implies that $\{x_n\}_{n\in\mathbb N}$ converges. (What if $m_n=n+1$ and $x_n=\sum_{i=1}^n \tfrac{1}{i}$?)

An alternative approach is to take $x^+=\limsup_{n\to\infty}(x_n)$ and $x^-=\liminf_{n\to\infty}(x_n)$. Then by assumption (specifically the non-convergence of $x_n$) we have that $x^+\neq x^-$. Let $x^+-x^-=\delta>0$. By definition the two sets $I^+=\{n\in\mathbb N:x_n>x^+-\tfrac{1}{3}\delta\}$ and $I^-=\{n\in\mathbb N:x_n<x^-+\tfrac{1}{3}\delta\}$ are disjoint and infinite. So for each $n\in I^-$ we may choose $n+p_n\in I^+$, in which case $$x_{n+p_n}-x_n>(x^+-\tfrac{1}{3}\delta)-(x^-+\tfrac{1}{3}\delta)=\tfrac{1}{3}\delta$$ And this situation occurs infinitely often, so $\lim_{n\to\infty}(x_{n+p_n}-x)$ cannot possibly exist $\Box$

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    $\begingroup$ in your counter example, you mean $$x_n=\sum_{i=1}^n \frac{1}{i}$$ $\endgroup$ – John Doe Dec 26 '18 at 18:01
  • $\begingroup$ Yes, my apologies :) $\endgroup$ – Rafay Ashary Dec 26 '18 at 18:19
  • $\begingroup$ That's a nice way to approach the proof, thank you! $\endgroup$ – roman Dec 26 '18 at 18:37

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