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Math Help - Riemann Integrals, Limits

  1. #1
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    Riemann Integrals, Limits

    Prove that a_n has a limit:

    a_n=\sum_{i=1}^{2^n} sin\frac{i}{2^{2n}}cos\frac{i}{2^{2n}}cos\frac{i}{  2^{2n-1}}...cos\frac{i}{2^{n+1}}

    The subject is integrals, Riemann's definition and all that... I have no idea how to look for a solution

    Thanks!
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  2. #2
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    Quote Originally Posted by adam63 View Post
    Prove that a_n has a limit:

    a_n=\sum_{i=1}^{2^n} sin\frac{i}{2^{2n}}cos\frac{i}{2^{2n}}cos\frac{i}{  2^{2n-1}}...cos\frac{i}{2^{n+1}}

    The subject is integrals, Riemann's definition and all that... I have no idea how to look for a solution

    Thanks!
    Step 1::

    Take d = i/2^(n+1)

    Use 2 sin a . cos a = sin 2a

    For example,

    2 sin i/2^k . cos i/2^k = sin i/2^(k-1)

    Step 2 ::

    Then use Comparision Test with series 1/2^(n+1)
    Last edited by ques; May 5th 2010 at 03:14 PM.
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  3. #3
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    Quote Originally Posted by ques View Post
    Step 1::

    Take d = i/2^(n+1)

    Use 2 sin a . cos a = sin 2a

    For example,

    2 sin i/2^k . cos i/2^k = sin i/2^(k-1)

    Step 2 ::

    Then use Comparision Test with series 1/2^(n+1)
    ---

    Well.. it looks like an optimal solution, but I am looking for a solution that involves use of integrals, and the Riemann's definition of integrals. Something like \sum_{i=1}^n f(x_i)\Delta x_i
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  4. #4
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    Can anyone please help me with that? I need it urgently.
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by adam63 View Post
    Can anyone please help me with that? I need it urgently.
    Let's see some work. What does the summand simplify to?
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