Sep 24, 2017 · $\begingroup$ @Saad I think that another way to look at it (which is what this answer is saying in the less formal initial definition) is that if you imagine writing out your entire …

May 22, 2017 · This question is mostly concerned with the infinity of subsequence. Since Wikipedia page regarding subsequence denote the example of subsequence with finite one, …

Dec 14, 2011 · Here is a definition of a subsequence (c.f. with this wikipedia link): A subsequence of a given sequence is a sequence formed from the given sequence by deleting some of the …

Apr 27, 2019 · Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for …

Mar 31, 2016 · However, the opposite way is not easy for me. How can the existence of a convergent sub-subsequence in each subsequence guarantees the convergence of …

Oct 25, 2021 · I am trying to prove that every subsequence of a subsequence of S is itself a subsequence of S. It seems risky to assume convergence or limit the approach to a specific …

Dec 23, 2020 · If a subsequence of $(x_n)$ diverges, it implies that, given any $\epsilon >0$, there are an infinite number of real numbers in the tail that do not lie in an $\epsilon$ …

A subsequence is just some of the terms of the original sequence, kept in order. If your original sequence is $1,2,3,4,5,6\dots $, one subsequence is $1,3,5,7,9 \dots $ Another is $1, 2343, …

Dec 8, 2014 · $\begingroup$ Your reasoning isn't correct, because of the difference between "every subsequence" and "has a subsequence". $\endgroup$ – Jonas Meyer Commented Dec …

$\begingroup$ I don't understand what you're saying. $\alpha := \limsup x_n = \lim_{N \to \infty} sup\{x_n \mid n \geq N \}$ by definition.