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Math Help - Limit of a series

  1. #1
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    Limit of a series

    Hi guys, I am currently doing my homework in limits and derivatives, and there is one question that I'm stuck at. I think the correct answer is 0 but still want to ask for you guys' advice and thoughts



    I am quite skeptical about the functions inside the square brackets ... somehow I think they resemble a series, am I right? I wonder if that series has anything to do with this kind of limit. Any suggestions is highly appreciated. Thanks everyone.
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    According to Wolfram, \displaystyle \sum_{k = 1}^{n}{\sin{\left(\frac{k\pi}{n}\right)}} = \cot{\left(\frac{\pi}{2n}\right)}.

    What happens as \displaystyle n \to \infty?
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by sparkyboy View Post
    Hi guys, I am currently doing my homework in limits and derivatives, and there is one question that I'm stuck at. I think the correct answer is 0 but still want to ask for you guys' advice and thoughts



    I am quite skeptical about the functions inside the square brackets ... somehow I think they resemble a series, am I right? I wonder if that series has anything to do with this kind of limit. Any suggestions is highly appreciated. Thanks everyone.
    As n goes to infinity (think Riemann sums here):

    (1/n)[sin(pi/n)+sin(2pi/n+...+sin(npi/n)] -> integral sin(pi x) dx from 0 to 1

    so:

    sin(pi/n)+sin(2pi/n+...+sin(npi/n) = O(n)

    That is the thing diverges.

    (In fact sin(pi/n)+sin(2pi/n+...+sin(npi/n) ~= 2n/pi for large n)

    CB
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    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by sparkyboy View Post
    Hi guys, I am currently doing my homework in limits and derivatives, and there is one question that I'm stuck at. I think the correct answer is 0 but still want to ask for you guys' advice and thoughts



    I am quite skeptical about the functions inside the square brackets ... somehow I think they resemble a series, am I right? I wonder if that series has anything to do with this kind of limit. Any suggestions is highly appreciated. Thanks everyone.
    The first step is the computation of...

    \sigma_{n}= \sum_{k=1}^{n} \sin (k \frac{\pi}{n}) = \frac{1}{2i}\ \sum_{k=1}^{n} (e^{i k \frac{\pi}{n}} - e^{-i k \frac{\pi}{n}})=

    = \frac{1}{i}\ (\frac{e^{i \frac{\pi}{n}}}{1-e^{i \frac{\pi}{n}}}- \frac{e^{-i \frac{\pi}{n}}}{1-e^{-i \frac{\pi}{n}}}) = \frac{1}{i}\ (\frac{1}{1-e^{i \frac{\pi}{n}}}- \frac{1}{1-e^{-i \frac{\pi}{n}}}) =

    = \frac{1}{2} \frac{e^{i \frac{\pi}{2n}} + e^{-i \frac{\pi}{2n}}} {\sin \frac{\pi}{2n}}} = \frac {\cos \frac{\pi}{2n}}{\sin \frac{\pi}{2n}}= \cot \frac{\pi}{2n}} (1)

    The second step [just a bit more confortable...] is to evaluate the \lim_{n \rightarrow \infty} \sigma_{n}...

    Kind regards

    \chi \sigma
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