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Proving that a sequence is within certain bounds

  • Thread starter Elysian
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Homework Statement


Define a1=1, and for every n>1, an+1 = an + [itex]\frac{1}{an}[/itex]. Prove that 20 < a200 < 24


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

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
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Is this the complete question?
 
  • #3
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Yeah it is.
 
  • #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.
 
  • #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.
 
  • #6
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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.
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.
 
  • #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? :-)
 
  • #8
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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? :-)
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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