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Calc analysis - monotone sequences

  1. Oct 11, 2005 #1
    I think I'm having some trouble on this. First I'll state what the question is then I'll show what i have and my reasoning. Determine if the sequence{b sub(n)} is convergent by deciding on monotonicity and boundness. given:
    b sub(n)=n^2/2^n

    First I plugged in numbers for n starting with 1. after n=3 the sequence started taking shape and it appears to be approaching zero as n gets large. my feeling at this point is that its monotone decreasing with zero as the lower bound. I guess to prove this i would have to show that b sub(n+1)< b sub(n) thus impling that the lim b sub(n) = 0 or at least exists as n goes to infinity. my first step was to show that b sub(n+1) exists by way of induction. using n=1 in the given i find it true. here is where i start to question myself. b sub(n+1) = (n+2)^2/2^(n+1). I'm not sure how to simplify this but as i plug in numbers I can easily see that 0< b sub(n+1)< b sub(n) therfore leading me to conclude that there is a lim b sub(n) = 0 as n goes to infinity. Am I flawed in this thought process, or are there holes. In my induction part are there ways of simpliying the equation? help and comments are greatly appreciated as i am rusty and many years removed from my last calc class.
     
  2. jcsd
  3. Oct 12, 2005 #2

    EnumaElish

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    I think you need to show 0 < bn+1 < bn for general n, using induction.

    (n+1)^2/2^(n+1) < n^2/2^n
    (n+1)^2/n^2 < 2^(n+1)/2^n
    n^2 + 2n + 1 < 2n^2
    etc...
     
  4. Oct 12, 2005 #3

    HallsofIvy

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    I'm puzzled by this: " my first step was to show that b sub(n+1) exists by way of induction."

    Is there any question that n2/2n exists for all n?

    Since it is also obvious that n2/2n is positive for all n (so that 0 is a lower bound), to use "monotone convergence" to show that it converges, you need only show that this is a decreasing sequence.

    Generally speaking the best way to show that a sequence is decreasing (or increasing) is to compare successive terms. If bn+1< bn of all n then we must have bn+1- bn< 0 or, alternatively, bn+1/bn< 1.

    In this case, because bn is defined as a fraction, I would try the latter of those: bn+1/bn= ((n+1)2/2n+1)(2n/n2)= (1/2)((n+1)2/n2). Can you show that (n+1)2/n2< 2? Well, that certainly isn't true for n= 1! What n is it true for?
     
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