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Math Help - Sequence

  1. #1
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    Sequence

    Does the following sequence converge or diverge? If it converges what does it converge to?

    a$_{n}$ = $\cos (\frac {6\pi n^{2}}{3n+1})$

    This is my first use of LaTex. It's really cool!
    Last edited by browni3141; March 14th 2012 at 07:59 PM. Reason: Fixed LaTex Syntax
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  2. #2
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    Re: Sequence

    Quote Originally Posted by browni3141 View Post
    Does the following sequence converge or diverge? If it converges what does it converge to?

    a$_{n}$ = $\cos (\frac {6\pi n^{2}}{3n+1})$

    This is my first use of LaTex. It's really cool!
    \lim_{n \to \infty} a_n=\lim_{n \to \infty} \cos \left(\frac{6 \pi n^2}{3n+1}\right)=\cos \left({\lim_{n \to \infty} \frac{6 \pi n^2}{3n+1}}\right)

    \lim_{n \to \infty} a_n=\cos \left(6\pi \cdot \lim_{n \to \infty} \frac{ n^2}{3n+1}\right)=\cos \left(\4 \pi \cdot \lim_{n \to \infty} n\right)

    -1 \leq \cos \left(\4 \pi \cdot \lim_{n \to \infty} n\right) \leq 1 , therefore sequence diverge .
    Thanks from mash
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  3. #3
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    Re: Sequence

    Thank you for the reply, but showing that the limit does not exist does not prove that the sequence diverges. I actually know that it converges and I know what it converges to, but I don't understand how to prove it mathematically.
    I constructed this problem intentionally so that it would converge while the limit does not exist, but I made a bad assumption about limits and basically got lucky. It converges to -1/2, but I don't know how to show this.
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  4. #4
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    Re: Sequence

    Quote Originally Posted by browni3141 View Post
    showing that the limit does not exist does not prove that the sequence diverges.
    That is a truly an absurd statement. It has an internal contradiction: a violation of the definition of divergent.

    Quote Originally Posted by browni3141 View Post
    I actually know that it converges and I know what it converges to, but I don't understand how to prove it mathematically.
    How can you know that something which is demonstrably false is actually true?
    Surely you know that {\lim _{n \to \infty }}\cos (n) does not exist,
    If you have doubts about that then you have no business trying this question.

    If you do understand that fact, then note that \left( {\frac{{6\pi {n^2}}}{{3n + 1}}} \right) \to \infty
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  5. #5
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    Re: Sequence

    Quote Originally Posted by Plato View Post
    That is a truly an absurd statement. It has an internal contradiction: a violation of the definition of divergent.


    How can you know that something which is demonstrably false is actually true?
    Surely you know that {\lim _{n \to \infty }}\cos (n) does not exist,
    If you have doubts about that then you have no business trying this question.

    If you do understand that fact, then note that \left( {\frac{{6\pi {n^2}}}{{3n + 1}}} \right) \to \infty
    The problem with what you and princeps have said is that these limits are taken over the real numbers, but n is an integer. The limit over integers may still exist, and it does in this case. Surely you can see that a_{n} converges to -1/2 by observing values of a_{n} as n gets large.
    a_{5} = -.3827
    a_{6} = -.4017
    a_{7} = -.4154
    a_{20} = -.4700
    a_{100} = -.4940
    a_{1000} = -.4994
    a_{10000} = -.4999
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  6. #6
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    Re: Sequence

    Whatever, if I'm wrong, then explain why it appears that a_{n} converges to -1/2.
    I don't see why it's so hard for you to understand what I'm asking. It is clear that a_{n} is approaching -1/2 based on the data provided above, so perhaps you should come back and give the problem more than a second of thought.
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  7. #7
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    Re: Sequence

    By "long division",
    \frac{3n^2}{3n+1} = n - \frac{1}{3} + \frac{1}{3(3n+1)}
    so
    \cos \left( 2 \pi \cdot \frac {3n^2}{3n+1} \right) = \cos \left( 2 n \pi - \frac{2 \pi}{3} + \frac{2 \pi}{3(3n+1)} \right) =  \cos \left(- \frac{2 \pi}{3} + \frac{2 \pi}{3(3n+1)} \right)
     \to \cos \left( - \frac{2 \pi}{3} \right) =  - \frac{1}{2}
    as n \to \infty
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  8. #8
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    Re: Sequence

    Quote Originally Posted by awkward View Post
    By "long division",
    \frac{3n^2}{3n+1} = n - \frac{1}{3} + \frac{1}{3(3n+1)}
    so
    \cos \left( 2 \pi \cdot \frac {3n^2}{3n+1} \right) = \cos \left( 2 n \pi - \frac{2 \pi}{3} + \frac{2 \pi}{3(3n+1)} \right) =  \cos \left(- \frac{2 \pi}{3} + \frac{2 \pi}{3(3n+1)} \right)
     \to \cos \left( - \frac{2 \pi}{3} \right) =  - \frac{1}{2}
    as n \to \infty
    This seems to be exactly what I'm looking for. Thank you for understanding what I was asking.
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