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If a monotone increasing sequence is not convergent, then is the sequence unbounded?

I know that a monotone increasing sequence that is convergent must be bounded.

But $P \to Q$ being true doesn't necessarily mean $-P \to -Q$ must be true.

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  • $\begingroup$ This is the "completeness axiom" for the real numbers $\endgroup$ Feb 19, 2017 at 22:37
  • $\begingroup$ A monotone increasing sequence is convergent if and only if it is bounded above $\endgroup$
    – joeb
    Feb 19, 2017 at 22:40
  • $\begingroup$ @joeb And below too $\ddot \smile$ $\endgroup$
    – Git Gud
    Feb 19, 2017 at 22:40
  • $\begingroup$ @user3000482 What is $P$ and what is $Q$ to you? $\endgroup$
    – Git Gud
    Feb 19, 2017 at 22:41
  • $\begingroup$ Consider rational numbers $1, \frac75, \frac{41}{29}, \ldots$ where if a term is $q$ the next is $\frac{4+3q}{3+2q}$. This is monotone increasing and is clearly bounded above by $2$, but does not converge to a rational number $\endgroup$
    – Henry
    Feb 19, 2017 at 22:52

2 Answers 2

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Let $(a_n)$ an increasing and divergent sequence of real numbers.

If we suppose that this sequence has an upper bound, being increasing and having an upper bound, it would converge to some real limit : a contradiction. So this sequence doesn't have any upper bound (and in particular, it's unbounded).

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The monotone convergence theorem states that if $(a_n)$ is monotone then it converges if and only if it is bounded.

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