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Homework Help: Convergence of an infinite product

  1. Jun 19, 2010 #1
    1. The problem statement, all variables and given/known data
    Find what the infinite product (2N-1)(2N) converges too.

    Example. 1/2*3/4*5/6


    2. Relevant equations

    That's it.

    3. The attempt at a solution

    I have several strategies for this, all of which are similar, and none of which I know how to show.I think it converges to zero.

    The idea, is that I show that at any N, I can find a finite amount of N after that, when multiplied together, are below 1/2, or really, any other set fraction. Example.

    3/4*5/6*7/9 is less than 1/2. If I can find this to be true for an infinite amount of times, it will converge to zero.


    Another idea I have is to take the log of it, and than show that it diverges to negative infinity. Because this product is bounded by zero, if would mean it converges to zero.
     
  2. jcsd
  3. Jun 19, 2010 #2
    I think I got it.

    I used the integral test. The ln sum diverged to infinity.

    Because of that, this converges to zero.
     
  4. Jun 19, 2010 #3

    Gib Z

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    Homework Helper

    You mean diverged to negative infinity?

    You could also have found an expression for the n-th partial product and applied Stirling's Approximation to get [tex]\prod_{k=1}^{n} \frac{2n-1}{2n} = \frac{ (2n)! }{2^{2n} (n!)^2} \sim \frac{1}{\sqrt{ \pi n}} \rightarrow 0[/tex]
     
  5. Jun 19, 2010 #4
    Hmm, I am not aware of those methods.

    Can you point me to an online textbook/ a few links that have the material and some challenging but doable examples?
     
  6. Jun 19, 2010 #5

    Gib Z

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    Homework Helper

    You mean of the methods I used?

    The idea is to stop thinking of it as an infinite product so much and re-view it as just the limit of a sequence. Then if we can just what the terms of the sequence are, we can take the limit and find it's value. To find the term, I wrote out in full the general n-th term;
    [tex]\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\cdots \frac{2n-3}{2n-2}\cdot\frac{2n-1}{2n}\cdot[/tex],

    and then I noticed I could produce some factorial terms, a) by "completing" the missing terms in the numerator by multiplying by 2*4*6...*2n, and then dividing by that again. Then b) in the denominator, we get factorial terms by pulling out a factor of 2 from each factor.

    So that gives us the term I posted before, and thats just algebra so I can't really refer you to anywhere for that. The final part of this method is taking the limit. It's a good idea to have a bag of tricks to do limits with and being able to do them well and find harder ones pays off. In this case, since factorials aren't so easy to take limits of otherwise, there's a nice tool called Stirling's Approximation. You can find it here: http://en.wikipedia.org/wiki/Stirling's_approximation.

    The part we are interested in is : [tex]\lim_{n\to\infty} \frac{n!}{\sqrt{2\pi n} (\frac{n}{e})^n} = 1[/tex].
    This tells us that if we have the limit of n! in some expression and it's going to infinity, we can put the denominator of that limit in place of n! and the limit won't change. So then if you do that with our general term, it simplifies very nicely and the limit comes out.
     
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