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Math Help - Cauchy Criterion.

  1. #1
    MHF Contributor Also sprach Zarathustra's Avatar
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    Cauchy Criterion.

    Prove that the following infinite sum is converges using Cauchy Criterion.

    \Sigma^{\infty}_{n=1} \frac{cos{nx}-cos{(n+1)x}}{n}

    x-Constant

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  2. #2
    A Plied Mathematician
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    Just to clarify, you're trying to show that

    \displaystyle{\sum_{n=1}^{\infty} \frac{\cos(nx)-\cos((n+1)x)}{n}}}

    converges, correct? And you're required to use the Cauchy criterion?

    What have you done so far?
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  3. #3
    MHF Contributor Also sprach Zarathustra's Avatar
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    Yes.

    I say the next thing:

    \Sigma^{\infty}_{n=1} \frac{cos{nx}-cos{(n+1)x}}{n}=2sin(\frac{1}{2}x)\Sigma^{\infty}_  {n=1}\frac{sin(\frac{2n+1}{2}x}{n}

    So, I need to prove that for every \epsilon >0, exist N(\epsilon), for all n, n>N(\epsilon), and for all p natural:

    |S_{n+p}-S_n|<\epsilon

    I choose p=n.

    |S_{2n}-S_n|=|\frac{sin(\frac{2(n+1)+1}{2}x)}{n+1}+...+\fr  ac{sin(\frac{4n+1}{2}x)}{2n}|<|\frac{sin(\frac{2(n  +1)+1}{2}x)}{n+1}|+...+|\frac{sin(\frac{4n+1}{2}x)  }{2n}|<\frac{1}{n+1}+...+\frac{1}{2n}<n\frac{1}{n+  1}<\frac{n}{n}<1


    My \epsilon is not containing n...
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  4. #4
    Senior Member Dinkydoe's Avatar
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    How about this then...
    Choose \epsilon >0 and write:

    x_n= \cos(nx)
    a_n = \frac{1}{n}(x_n-x_{n+1})

    We need to show that for all \epsilon>0 we can find N such that for all n>N and p\geq 1 we have A_{n,p}:= |a_n+\cdots +a_{n+p}|< \epsilon

    We can write: A_{n,p}= |\frac{x_n}{n}-\frac{x_{n+1}}{n(n+1)}-\frac{x_{n+2}}{(n+1)(n+2)}-\cdots - \frac{x_{n+p}}{(n+p-1)(n+p)}-\frac{x_{n+p+1}}{n+p}|

    since -1\leq \cos(nx)\leq 1 for all n we may assume (with triangle inequality)

    A_{n,p}\leq \frac{1}{n}+\frac{1}{n(n+1)}+\frac{1}{(n+1)(n+2)}+  \cdots + \frac{1}{(n+p-1)(n+p)}+\frac{1}{n+p}

    Now it can't be too hard to show that we can find N such that for all n>N and p\geq 1 we have

    \frac{1}{n}+ \frac{1}{n+p}< \frac{\epsilon}{2}

    and

     \frac{1}{n(n+1)}+\frac{1}{(n+1)(n+2)}+\cdots + \frac{1}{(n+p-1)(n+p)} < \frac{\epsilon}{2}
    Last edited by Dinkydoe; July 29th 2010 at 04:26 AM. Reason: some index-mixups
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