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Math Help - How can i prove if a sequence is bounded?

  1. #1
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    How can i prove if a sequence is bounded?

    Given a sequence a(n) = n / [(n^2)+4]
    I know that:
    A sequence is bounded if a K∈R exists so that the the absolute value of the sequence |a(n)| for every n∈N smaller than K is. -- |a(n)| <= K

    How can i prove this? Picking a random K won't do, i guess, so how would you go about doing it?

    thank you
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    Re: How can i prove if a sequence is bounded?

    \frac{n}{n^2 + 4} < \frac{n}{n^2} = \dots ?

    (we can say this because n^2 < n^2 + 4 and n > 0).

    now, can you think of some positive number that will work for K now?
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    Re: How can i prove if a sequence is bounded?

    Hey, thank you for the reply!

    I would have to say K = 2, by calculating the extreme points from the derivative of the sequence..
    if i had to deal just with n / n^2 i would have to say K = 1 or lower, approaching 0..
    or am i approaching this problem the wrong way? :/
    Last edited by nappysnake; November 27th 2011 at 12:18 PM.
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    Re: How can i prove if a sequence is bounded?

    I think that you are missing a most important concept here.
    Theorem: If a sequence has a limit then that sequence is bounded.

    If (x_n)\to L then \left( {\exists N} \right) such that n \geqslant N\; \Rightarrow \;\left| {x_n  - L} \right| < 1.

    From which it follows that n \geqslant N\; \Rightarrow \;\left| {x_n } \right| < 1 + \left| L \right|

    Hence \left( {\forall n} \right)\left[ {\left| {x_n } \right| \leqslant \sum\limits_{k = 1}^N {\left| {x_k } \right|}  + 1 + \left| L \right|} \right]
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  5. #5
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    Re: How can i prove if a sequence is bounded?

    the point is, that n/n^2 = 1/n, and 1/n most certainly is bounded above (by 1), and also bounded below (by 0).

    of course, i'm appealing to the fact that n/(n^2 + 4) is always positive, so we can drop the absolute value signs (we already have a lower bound for it, 0).

    note that K = 1 is not the least upper bound for n/(n^2 + 4), but we're not being asked to find a minimal bound, any bound we find will do (as long as we can

    prove it is indeed a bound). taking the derivative is over-kill, and won't work for alternating series (because (-1)^x isn't even defined for most rational x).

    now, the reason why any convergent sequence is bounded, is because all but a finite number are within epsilon of the limit, and we can always find the maximum

    in absolute value of a finite set, and |L| + \epsilon bounds "the rest of the sequence" , so the bigger of the two is a bound for the entire sequence.

    but with all due respect to Plato, we do not need to prove a sequence convergent, to prove it bounded. if we find a bound, it is bounded, whether or not it converges.

    for example, the sequence: 0,1,0,1,0,1,0,1,0,1,0,1....... is bounded, but not convergent. we can choose K = 1.
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