# Proof verification. Show that if $\{x_n\}$ diverges then there must be a sequence $\{p_n\}\subset\Bbb N$ such that $\lim(x_{n+p_n} - x_n)\ne 0$

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!

• 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. – Michał Miśkiewicz Dec 26 '18 at 18:03
• A visualization of the idea from @RafayAshary answer. Just for the case – 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 $$I^-=\{n\in\mathbb N:x_n are disjoint and infinite. So for each $$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$$
• in your counter example, you mean $$x_n=\sum_{i=1}^n \frac{1}{i}$$ – John Doe Dec 26 '18 at 18:01