Can a decreasing sequence be bounded below but not above? Does it make sense to have such a sequence? I was looking at a Theorem: if a sequence is bounded and decreasing, then it's limit is going to be the greatest lower bound. So, let's say the sequence starts off with some number, then it starts decreasing, it makes sense for the limit to be the greatest lower bound. I also understand why we need the bounded condition (since bounded =bounded above and bounded below): because if it wasn't bounded below we couldn't find the limit right? Could we say in this example that the sequence is bounded below rather than saying its bounded above too? (and consequently saying it is bounded?)

I'm sorry if this is a confusing-looking question. Thank you in advance.


1 Answer 1


A nonincreasing sequence (a fortiori a decreasing one) $(x_n)_{n\geq 1}$ is always bounded above by $x_1$. So it is bounded if and only if it is bounded below.

  • $\begingroup$ ok! I just wanted to be sure! thank you :) $\endgroup$
    – phoenix
    Commented Mar 30, 2013 at 19:55
  • 1
    $\begingroup$ @user65165 You're welcome. That's why in the monotone convergence theorem, one can simply say: a monotone sequence converges if and only if is bounded. One side of the boundedness is always true. $\endgroup$
    – Julien
    Commented Mar 30, 2013 at 19:57
  • $\begingroup$ what about the sequence $a_n = \frac{1}{n}$? starting at $n = 0$ it technically would not be bounded above, right? $\endgroup$
    – DanZimm
    Commented Mar 30, 2013 at 20:48
  • $\begingroup$ @DanZimm It is not defined at $n=0$... $\endgroup$
    – Julien
    Commented Mar 30, 2013 at 20:51
  • $\begingroup$ right, so wouldn't you need the condition that $x_1$ exists? $\endgroup$
    – DanZimm
    Commented Mar 30, 2013 at 20:52

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