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Math Help - series and convergence

  1. #1
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    series and convergence

    This might seem super easy, but could anyone give me some hint how to show the following An is convergent?
    1.n!/n^n
    2.(1-1/n)^(n^2)
    3. 1-cos (1/n)
    I've tried the limit test/ratio test, but not working. Maybe comparison test?
    Also, could anyone help calculating the series using taylor expansion of cos?
    the sum of all (cos n)/(2^n) where n goes from 0 to infinity?
    Any help is greatly appreciated!
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  2. #2
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    Use the root test for these
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  3. #3
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    thanks, I've figured out how to do the convergence tests. Could anyone please give me a hint how to calculate the sum of cos n/2^n?
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by nngktr View Post
    thanks, I've figured out how to do the convergence tests. Could anyone please give me a hint how to calculate the sum of cos n/2^n?
    \sum_{n=0}^{\infty}\frac{\cos(n)}{2^n}=\text{Re}\s  um_{n=0}^{\infty}\frac{e^{in}}{2^n} and since \left|\frac{e^i}{2}\right|<1 we see that this is equal to \text{Re}\frac{1}{1-\frac{e^i}{2}}=\text{Re}\frac{2}{2-e^i}=\cdots=\frac{2(\cos(1)-2)}{4\cos(1)-5}

    I've rigorously justified nothing. I leave that to you.


    EDIT: P.S. that is not the Taylor expansion for \cos(x). Try the Maclaurin \sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{(2n)!}
    Last edited by Drexel28; May 18th 2010 at 09:00 PM.
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  5. #5
    MHF Contributor chisigma's Avatar
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    Because is...

    \cos x = \sum_{k=0}^{\infty} (-1)^{k} \frac{x^{2k}}{(2k)!} (1)

    ... is also...

    1 - \cos \frac{1}{n} = \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{n^{2k} (2k)!} (2)

    ... so that is...

    \sum_{n=1}^{\infty} (1-\cos \frac{1}{n}) = \sum_{n=1}^{\infty} \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{n^{2k} (2k)!} = \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{(2k)!} \sum_{n=1}^{\infty} \frac{1}{n^{2k}} = \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{(2k)!} \zeta (2k) (3)

    For k\ge 2 is 1 < \zeta (k) < 2 so that the series in second term of (3) converges... and converges also very quickly! ...

    Kind regards

    \chi \sigma
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