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So if a certain sequence $a_n$ is convergent then its bounded.So from Bolzano-Weierstrass $a_n$ has a convergent sub-sequence, but where do I continue from here?

CSDude101
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    Assuming this is a sequence of real numbers, then there must be either an infinite number of terms of the convergent subsequence which are at least the limit, or an infinite number of terms of the subsequence which are at most the limit. Can you take it from here? – MPW May 05 '15 at 02:51
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    See this question and links there. –  May 05 '15 at 02:56

2 Answers2

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In fact, every sequence $\{a_n\}_{n=1}^{\infty}$ of real numbers has a monotone subsequence. Here's how to prove it:

Call $n\in \mathbb{N}$ a peak point if $a_n > a_m$ for all $m > n$. If the sequence $\{a_n\}$ has infinitely many peak points, then the peak points form a monotonically decreasing subsequence.

Otherwise, there are only finitely many peak points. This means precisely that, for some sufficiently large $K \in \mathbb{N}$, whenever $n > K$, there is $m>n$ such that $a_m > a_n$. Choosing $n_1 > K$, then $n_2> n_1$ such that $a_{n_2} > a_{n_1}$, then $n_3 > n_2$ such that $a_{n_3} > a_{n_2}$, etc., we have found a monotonically increasing subsequence.

In either case, a monotone subsequence has been found.

Sameer Kailasa
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    I think you need a weak inequalities in your definition of peak point. Otherwise, a sequence like sin x would cause you to "run out of runway." – Student96 Jun 10 '19 at 21:02
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Not only every convergent sequence, but also every sequence in $\mathbb{R}$ has a monotone subsequence. Look at the first lemma here: http://en.wikipedia.org/wiki/Bolzano%E2%80%93Weierstrass_theorem

Mehdi Jafarnia Jahromi
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