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How to prove that the limit of a sequence is a Wallis product?

  1. Apr 15, 2012 #1
    1. The problem statement, all variables and given/known data

    Let t_1=1, t_(n+1)= [1- 1/4(n^2)]*t_n for n>=1.

    The book says the limit is a Wallis product 2/pi, but I don't know where to start. I've been searching, but I'm lost. Could you point me in the right direction?
     
    Last edited: Apr 15, 2012
  2. jcsd
  3. Apr 15, 2012 #2
    correction the limit is two over pi

    my keyboard is messing up so i can not correct

    I think I got it. I don't know if it's a proof, but I've found how it's the limit.
     
    Last edited: Apr 15, 2012
  4. Apr 15, 2012 #3
    Ok. I've found the form of a Wallis product...

    lim n--> infinity (2n/ 2n-1) * (2n / 2n+1) = pi/2

    so my sequence is pretty much the reciprocal of the equation that's why it's 2/pi... but I still don't know why it's 2/pi.

    I saw some website saying this comes from some difficult integral, something about using method of reduction. I don't think I've ever done that in a calculus course before.

    All help is appreciated! Thanks, folks.



    I calculated up to t_7 to get

    t_7= (143/144)*(99/100)*(63/64)*(35/36)*(15/…

    so looking at the form of a Wallis product I changed it into

    t_7= (11*13/ 12*12) * (9*11/10*10) * (7*9/8*8)* (5*7/6*6) * (3*5/4*4) *(1*3/2*2)

    so that means t_n= (2n-1)*(2n+1)/ (2n)(2n), so the lim n-- infinity (t_n)= 2/pi.

    Is this enough to consider this problem solved?
     
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