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Math Help - A limit that tends infinity...

  1. #1
    Newbie Flamuri's Avatar
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    A limit that tends infinity...

    \lim_{n->\infty } (1-\frac{1}{2^{2}})(1-\frac{1}{3^{2}})...(1-\frac{1}{n^{2}})

    Can someone please help me out here? Have never find the limit of a function like this. I've learned limits but not this kind, if someone could tell me the first steps in those kind of tasks, would be great.

    Thanks,
    Flamuri
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  2. #2
    Junior Member
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    Re: A limit that tends infinity...

    well, the comment before is wrong....use trig function to find the limit
    Last edited by piscoau; January 16th 2012 at 06:34 AM.
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  3. #3
    MHF Contributor chisigma's Avatar
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    Re: A limit that tends infinity...

    Quote Originally Posted by Flamuri View Post
    \lim_{n->\infty } (1-\frac{1}{2^{2}})(1-\frac{1}{3^{2}})...(1-\frac{1}{n^{2}})

    Can someone please help me out here? Have never find the limit of a function like this. I've learned limits but not this kind, if someone could tell me the first steps in those kind of tasks, would be great.

    Thanks,
    Flamuri
    Remembering the 'infinite product'...

    \sin \pi x = \pi x\ \prod_{n=1}^{\infty} (1-\frac{x^{2}}{n^{2}}) (1)

    ... You obtain first...

    \frac{\sin \pi x}{\pi x}=\prod_{n=1}^{\infty} (1-\frac{x^{2}}{n^{2}}) (2)

    ... and from (2)...

    \prod_{n=2}^{\infty} (1-\frac{1}{n^{2}})= \lim_{x \rightarrow 1} \frac{\sin \pi x}{\pi x (1-x^2)}= \frac{1}{2} (3)

    Kind regards

    \chi \sigma
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