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A question regarding a new solution

  1. Jan 25, 2008 #1
    i recently posted a queston about prooving a convergence of a series
    and to find the limit

    i have found a new way to proove
    i showed it in the link

    is this method ok??

    the problem is in the second step
    when i am trying to prove that the series doesnt go bellow 0.61

    i used the splitting method
    of the fraction but its inconclusive

    how do i proove the second part
    ???

    do i need to proove anything else inorder to complete the objective???

    http://img134.imageshack.us/my.php?image=img8220ez9.jpg

    (there is a typing miste in the link its a n+1>0.61)
     
    Last edited: Jan 25, 2008
  2. jcsd
  3. Jan 25, 2008 #2
    This looks like a sequence to me!
     
    Last edited: Jan 25, 2008
  4. Jan 25, 2008 #3
    Why don't u post the original question first?
     
  5. Jan 25, 2008 #4
    the original question and how i tried to solve it in the new link

    on the top of the first page
     
  6. Jan 25, 2008 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    You say, at one point, "[itex]a_k> 0.61[/itex] given". Well, you aren't "given" that. And I don't see where you have proved it. But then you say "we need to prove that ak+1< 0.61" which isn't true. Are you saying that ak> 0.61 for some k? Well, ak+1 < 0.061 is never true, any way.

    I think it is simpler to prove the contrapositive. Suppose that, for some k, ak+1> ak. Then
    [tex]a_{k+1}= \frac{a_k+ 1}{a_k+ 2}> a_k[/itex]
    Multiply on both sides by the positive value [itex]a_k+ 2[/itex]. Then you are saying that [itex]a_k+ 1> a_k^2+ 2a_k[/itex] so that [itex]a_k^2+ a_k-1< 0[/itex]. That's a parabola opening upward. For what values of x is x2+ x- 1< 0? Are those possible values of ak?
     
  7. Jan 25, 2008 #6
    the parabula shows how the series goes

    if the line after some point goes is on the negative part of the y axes
    (bellow the x axes) then each next member will decrease in value

    thats how i overrided the prooving
    of

    an+1<an inequality

    you showed yourself the parula formula in the numanator
    so i think it is the same.
    is it ok??

    now i want to proove that
    its not going bellow 0.61

    by the induction method

    we presume that
    an>0.61

    and using that we proove that

    an+1>0.61

    as i showed in the link
    i am having trouble to solve it
    ??
     
    Last edited: Jan 25, 2008
  8. Jan 25, 2008 #7
    Since [itex]a_0=1[/itex] then

    [tex]a_{k+1}= \frac{a_k+ 1}{a_k+ 2}> 0[/tex]

    by induction, i.e. [itex]a_k>0 \,\forall\, k[/itex]. Thus the limit is

    [tex]l=\frac{-1+\sqrt{5}}{2}[/tex]
     
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