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Infinite Product

  1. Jan 9, 2006 #1
    Hi, can someone tell me how I would go about evaluating the following infinite product? I've included the answer and it's taken from mathworld.

    \prod\limits_{n = 2}^\infty {\frac{{n^2 - 1}}{{n^2 + 1}}} = \pi \cos ech\left( \pi \right)

    I think I'm supposed to be write an expression for the first n products and take the limit. I've tried doing that but I don't know what to do with the expression.

    [tex]\left( {\frac{{2^2 - 1}}{{2^2 + 1}}} \right)\left( {\frac{{3^2 - 1}}{{3^2 + 1}}} \right)...\left( {\frac{{n^2 - 1}}{{n^2 + 1}}} \right)[/tex]

    Any help would be good thanks.
  2. jcsd
  3. Jan 10, 2006 #2


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    Do you know the product form for sine? It's usually the first target for writing as an infinite product, and the product you have follows fairly easily from it.
  4. Jan 10, 2006 #3
    Ok, thanks for the hint. I'll go look up the product form of sine and see what I can come up with.
  5. Jan 11, 2006 #4


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    I love infinite products!

    [tex]\prod\limits_{n = 2}^\infty {\frac{{n^2 - 1}}{{n^2 + 1}}} = \pi \mbox{cosh} \left( \pi \right)[/tex]

    The above product is from the infinite product page at mathworld, their products were presumably derived using the infinite product for the Euler Gamma Function. I will use the sine product instead.

    The infinite product for the sine function is (for complex z):

    [tex]\sin \left( z \right) = z\prod\limits_{n = 1}^{\infty} {\left( 1-\frac{z^2}{\pi ^2 n^2} \right)} [/tex]

    Recall that [tex]\sin (iz) = i \mbox{sinh}(z)[/tex] so that [tex]\mbox{csch}(z) = \frac{1}{\mbox{sinh}(z)}= \frac{i}{\sin (iz)}[/tex], and hence

    [tex]\mbox{csch} \left( z \right) = \frac{i}{(iz)\prod\limits_{n = 1}^{\infty} {\left( 1-\frac{(iz)^2}{\pi ^2 n^2}\right) } }=\frac{1}{z}\prod\limits_{n = 1}^{\infty} {\left( 1+\frac{z^2}{\pi ^2 n^2}\right) ^{-1}}=\frac{1}{z}\prod\limits_{n = 1}^{\infty} {\frac{\pi ^2 n^2}{\pi ^2 n^2+z^2}} [/tex]

    it follows that

    [tex]\pi \mbox{csch} \left( \pi z \right) = \frac{1}{z}\prod\limits_{n = 1}^{\infty} {\frac{n^2}{n^2+z^2}} [/tex]

    plugging z=1 into the above gives

    [tex]\pi \mbox{csch} \left( \pi \right) = \prod\limits_{n = 1}^{\infty} {\frac{n^2}{n^2+1}} [/tex]

    I can't quite get to what they had, but it is close.

    Last edited: Jan 11, 2006
  6. Jan 11, 2006 #5
    That looks quite interesting. Is there a 'model' infinite product which is used to evaluate other infinite products (like what shmoe appears to have alluded to) . If so, is there a way to evaluate such a product, using fairly elementary formulas and algebra?
  7. Jan 11, 2006 #6
    You are almost done.Note that n^2-1 is (n-1)(n+1)
    So product(n^2-1) for n=2 to n is 1.2.3^2.4^2.....=product(n^2)/2
    Product(n^2+1) for n=1 to n =2.Product(1+n^2) for n=2 to n

    So from the above expression
    for n=2 to infty
  8. Jan 11, 2006 #7


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    Yes! The Gamma Function

    Yes, use the Euler Gamma function, it has several infinite product forms:

    [tex]\Gamma (z) = \lim_{n\rightarrow\infty} \frac{n!n^{z-1}}{z(z+1)\cdot\cdot\cdot (z+n-1)} = \frac{1}{z}\prod_{k=1}^{\infty}\left[ \left( 1 + \frac{1}{k}\right) ^{z} \left( 1 + \frac{z}{k}\right) ^{-1}\right] = \left[ze^{\gamma z} \prod_{j=1}^{\infty} \left( 1+ \frac{z}{j}\right)e^{-\frac{z}{j}} \right] ^{-1}[/tex]

    where [itex]\gamma = -0.577...[/itex] is Euler's constant. See my paper for a more complete discussion of the above infinite products and the Euler Gamma function.
    Last edited: Jan 11, 2006
  9. Jan 11, 2006 #8


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    A notable identitiy is the Euler reflection formula:

    [tex]\Gamma (z) \Gamma (1-z) = \frac{\pi}{\sin (\pi z)}[/tex]

    try deriving it via the above products...
  10. Jan 11, 2006 #9


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    It's odd that mathworld page only mentions Hadamard/Weiestrass product as a footnote. If you want to learn where the product form of these functions comes from, look for "hadamard factorization theorem", "hadamard products", "functions of finite order" or some variations. They'll be in more complex analysis books, and also in analytic number theory books (comes up with the zeta function).

    If you look at the product form of sine, it's essentially factoring sine as though it were an "infinite polynomial", it's a product over its zeros (paired up to simplify convergence). This is possible for some analytic functions by hadamards work. In some ways, sine is the prototype simple case, up to normalization it's zeros are nicely placed on the integers.

    Gamma has no zeros, but it has poles at 0, -1, -2, .... so 1/Gamma is then the other "prototype case, of this type of factorization, having zeros at the natural numbers (up to a simple change of variables).
  11. Jan 11, 2006 #10
    Thanks for the information benorin and shmoe. If I have time before the end of my summer break then I'll have look at some of those topics.
  12. May 2, 2009 #11
    HI Guys....

    I am trying to figure out the infinite product :


    Can anyone help? Benorin, you say you love infinite products... have a go!

    I would be very grateful!!
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