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Proving that a sequence is within certain bounds

  1. Sep 9, 2013 #1
    1. The problem statement, all variables and given/known data
    Define a1=1, and for every n>1, an+1 = an + [itex]\frac{1}{an}[/itex]. Prove that 20 < a200 < 24


    3. The attempt at a solution

    I tried a few things to no avail. First, I showed that this is an increasing function by showing an+1 > an. I tried finding a limit, by saying if lim[itex]_{n\rightarrow∞}[/itex]an = L then lim[itex]_{n\rightarrow∞}[/itex]an+1 = L. Plugging in L, and solving but that didn't help. I feel like I need to find an upper and lower bound and work from there but I'm not sure how
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 10, 2013 #2
    Is this the complete question?
     
  4. Sep 10, 2013 #3
    Yeah it is.
     
  5. Sep 10, 2013 #4

    haruspex

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    The first step is to get some idea of how fast an grows. If you think of the recurrence relation as a PDE, Δa = an+1 - an = 1/a then you can integrate to get a as a function of n. You'll then see how the n=200 relates to the lower bound of 20. Next, need to make up some inductive hypothesis for how far the sequence can deviate from the analytic solution.
     
  6. Sep 12, 2013 #5

    haruspex

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    Hi Elysian... Have you solved this now, got too busy elsewhere, or given up? Did you understand my hint? I can guide you to the solution.
     
  7. Sep 12, 2013 #6
    I did actually solve it, thank you for checking back. My bad on not replying. I found an upper and lower bound of the sequence, and with this I plugged in 200 into the lower bound of sqrt(2n), and the upper bound of sqrt(13n/6) and then got 20 and 20.866 respectively.
     
  8. Sep 12, 2013 #7

    CompuChip

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    Of course 20.866 is certainly smaller than 24, but it seems a bit tight given the number in the question. This makes me a bit suspicious. Are you sure about your calculation? :-)
     
  9. Sep 12, 2013 #8
    Pretty sure. To give you an overview of what I did,

    I took an and looked at an+12. Which gave a formula an2 + 2 + [itex]\frac{1}{ an+12}[/itex]. I compared an2 to 2n, and said that an2 > 2n for n[itex]\geq[/itex]3. Then by induction proved that this is the case for an+12 [itex]\geq[/itex] 2(n+1). Then I had to find a value greater than 2n, so I approximated that it would be (2+z)n, where z is a small number. Finding z to be 1/6, I plugged that in, got 13/6 as 2+z, and then continued the same process as before.
     
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