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Math Help - When taking the limit of an infinite series...

  1. #1
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    When taking the limit of an infinite series...

    When you \int_0^1 cos[n(x+1)] as n-> \infty, at one point you say lim n-> \infty 1/n [sin2n -sin(n)] \leq lim n-> \infty 2/n = 0 .

    Why is it okay to say \leq here? How do you know that the integral is 0, and not something less than 0.
    Last edited by gummy_ratz; September 28th 2011 at 07:41 PM.
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  2. #2
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    Re: When taking the limit of an infinite series...

    Quote Originally Posted by gummy_ratz View Post
    When you integrate cos[n(x+1)] as n->00 from 0 to 1, at one point you say lim n->00 1/n[sin2n -sin(n)] <= lim n->oo 2/n = 0.

    Why is it okay to say <= here? How do you know that the integral is 0, and not something less than 0.
    Hi gummy_ratx,

    It is not clear what your problem is. Can you use a bit of latex to write your expression. http://www.mathhelpforum.com/math-he...orial-266.html
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