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Limit of sequence

  1. Jan 20, 2010 #1
    1. Let a0 and a1 be positive real numbers, and set an+2 = sqrt(an+1) + sqrt(an) for n [tex]\geq[/tex] 0.
    (a) Show that there is N such that for all n [tex]\geq[/tex] N, an [tex]\geq[/tex] 1.
    (b) Let en = |an −4|. Show that en+2 [tex]\leq[/tex](en+1 +en)/3 for n[tex]\geq[/tex] N.
    (c) Prove that this sequence converges.




    Can someone please give me some hints to start with a)? Thank you in advanced.
     
  2. jcsd
  3. Jan 20, 2010 #2

    vela

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    If [itex]0<x<1[/itex], is [itex]\sqrt{x}[/itex] bigger or smaller than [itex]x[/itex]?
     
  4. Jan 20, 2010 #3
    if 0 < an < 1 then an x an < an ==>
    an < sqrt(an)

    So you mean i should prove part a by contradiction....
     
  5. Jan 20, 2010 #4

    vela

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    Perhaps. I don't know. But you can definitely show now that [itex]a_{n+2} > a_{n+1}+a_n[/itex] if [itex]a_n, a_{n+1} < 1[/itex], which may be useful.
     
  6. Jan 20, 2010 #5
    So
    a) Assume that for all n [tex]\geq[/tex] 0, 0 < an < 1
    then sqrt(an) > an

    ie, an+2 > an+1 + an
    an is increasing sequence

    I dont know how to show the contradiction here, but there is no assumption of increasing sequence if you choose a0 to start with

    ---> there is N st aN [tex]\geq[/tex] 1
    Assume for all n [tex]\geq[/tex] N, an+1 = an + an - 1 > 1

    therefore, by induction it is true for all n [tex]\geq[/tex] N, an [tex]\geq[/tex] 1
     
  7. Jan 20, 2010 #6

    jgens

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    Edit: My previous post was so incomprehensible that I don't think that it would have been much help. I'll post again later if I can get my thoughts together, but anyway, good luck!
     
    Last edited: Jan 20, 2010
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