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Monotone Sequence Theorem

  1. Oct 4, 2008 #1
    The thm says:
    Every bded monotone sequence in R is convergent.

    The proof:
    suppose {x_k} is a bded increasing sequence. Let M be the sup of the set of values {x_1, x_2,....} I claim that x_k -> M.
    Since M is an upper bd, we have x_k <= M for all k. (***)

    on the other hand, since M is the least upper bd, for any epislon >0, there is some K for which x_K > M - epislon. Since the x_k's increase with k, we also have x_k > M - epislon for all k > K.

    therefore M - epislon < x_k <= M for all k > K, and this shows that x_k -> M

    my question is here, I understand most of the proof, but what I want to ask is , why dont we just stop at the line ending with (***) sign above, if we just show M is the least upper bd and x_k is increasing, then we are done, why do we need to show M-epislon < x_k as well?
  2. jcsd
  3. Oct 5, 2008 #2

    Ben Niehoff

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    Because a sequence can be bounded, yet not convergent. Consider the sequence

    [tex]a_n = (-1)^n[/tex]

    Wait, never mind...the sequence must be monotone increasing. Hmm...

    Ah, I see, the reason is this:

    At the (***), we have proved only that x_k <= M for all k. This does not prove convergence (see above). To prove convergence, we use the fact that x_k is monotone increasing, which is your second paragraph. (It so happens that the easiest way to prove convergence is to prove that the sequence converges specifically to its least upper bound.)
  4. Oct 5, 2008 #3


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    Ben, you were right to begin with. Your example didn't work because the sequence you gave was not monotone- and we only use monotone in the proof after the "***".
  5. Oct 27, 2008 #4
    anybody can solve this prob??

    show dat the sequence xn= (n-10)/(n+10) is both bounded & monotone
  6. Oct 27, 2008 #5


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    Yes! But the important thing is can YOU?

    First, do NOT hijack someone else's thread for a new question! It's very easy to start your own thread.

    I will assume that "dat" was supposed to be "that".

    Have you calculated any values? What are x1, x2, x3, ...? If xn is monotone, is it increasing or decreasing? If you have determined that, you can try a "proof by contradiction". That is if you know the function is decreasing, try setting xk> x(k+1) for some k and see if you can't get a contradiction.

    Suppose xn is both bounded and monotone. Then it must converge. What does it converge to? Knowing that will help find a bound on it. You should be able to prove that bound by induction on "n".
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