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Math Help - Error in approximate formula

  1. #1
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    Error in approximate formula

    The rth term of a series (for r=1, 2, ...) is \frac{4}{4r^2-1}. Prove that the sum S of the first n terms of the series is given by
    S=\frac{4n}{2n+1}
    Show that 0<\frac{4}{4r^2-1}-\frac{1}{r^2}\leq\frac{1}{4r^2-1}
    Hence prove that the error in the approximate formula
    1^{-2}+2^{-2}+...+n^{-2}\simeq S

    I have found S correctly, my proof of the second part is not quite satisfactory to me, and I don't know how to do the last part.
    Proof:
    \frac{4}{4r^2-1}-\frac{1}{r^2}=\frac{1}{r^2(4r^2-1)}
    For all r, r^2 is positive, hence \frac{1}{r^2(4r^2-1)}is positive.
    When 0<r<\infty, \frac{1}{r^2(4r^2-1)}>0, hence 0<\frac{4}{4r^2-1}-\frac{1}{r^2}
    For all r, \frac{1}{r^2(4r^2-1)}\leq\frac{1}{4r^2-1}
    Hence, 0<\frac{4}{4r^2-1}-\frac{1}{r^2}\leq\frac{1}{4r^2-1}.

    For this last part, I did find the general term, but don't know how to find the sum of the series.
    \sum^n_{r=1} (\frac{1}{r})^2\simeq S

    Thanks!
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  2. #2
    MHF Contributor chisigma's Avatar
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    Is...

    \displaystyle \frac{1}{k^{2}-\frac{1}{4}} = \frac{1}{(k-\frac{1}{2})\ (k+\frac{1}{2})} = \frac{k+\frac{1}{2} - k + \frac{1}{2}}{(k-\frac{1}{2})\ (k+\frac{1}{2})} =

    \displaystyle = \frac{1}{k-\frac{1}{2}} - \frac{1}{k+\frac{1}{2}} (1)

    ... so that is...

    \displaystyle \sum_{k=1}^{n} \frac{4}{4 k^{2} - 1} = \sum_{k=1}^{n} \frac{1}{k^{2} - \frac{1}{4}} = 2 - \frac{2}{3} + \frac{2}{3} - \frac{2}{5}+ ... - \frac{1}{n+\frac{1}{2}} =  2 (1-\frac{1}{2n+1}) =

    \displaystyle \frac{4n}{2n+1} (2)

    End of 'part 1'... to be continued...

    Kind regards

    \chi \sigma
    Last edited by chisigma; August 13th 2010 at 12:22 AM. Reason: not yet answered the question...
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  3. #3
    MHF Contributor chisigma's Avatar
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    Now we have to find the 'error' ...

    \displaystyle e= \sum_{k=1}^{n} (\frac{1}{k^{2}-\frac{1}{4}} - \frac{1}{k^{2}}) (1)

    Is...

    \displaystyle \frac{1}{k^{2}-\frac{1}{4}} - \frac{1}{k^{2}} = \frac{1}{4}\ \frac{1}{k^{4} - \frac{k^{2}}{4}} = \frac{1}{4 k^{2}}\{\frac{1}{k-\frac{1}{2}} - \frac{1}{k+\frac{1}{2}}\}  (2)

    ... so that is...

    \displaystyle e= \frac{1}{2}\ \{(1-\frac{1}{3}) + \frac{1}{4} (\frac{1}{3} - \frac{1}{5}) + \frac{1}{9} (\frac{1}{5} - \frac{1}{7}) + ... + \frac{1}{n^{2}} (\frac{1}{2n-1} - \frac{1}{2n+1}) \} (3)

    At this point the question is: can we further 'simplify' the result (3)?...

    Kind regards

    \chi \sigma
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  4. #4
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    Thank you! it doesn't look very encouraging trying to simplify the result, but I'm supposed to prove that it is less than S/4. I thought it would have something to do with finding the sum of the series, which I don't know how, then finding the difference.
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  5. #5
    MHF Contributor chisigma's Avatar
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    I think that the 'spirit' of the problem is the following. Let suppose we have 'forgotten' that is...

    \displaystyle \sum_{n=1}^{\infty} \frac{1}{n^{2}} = \frac {\pi^{2}}{6} (1)

    ... and we have to find some pratical method to compute the sum. The general term tend to 0 like \frac{1}{n^{2}} and that means that we have to sum a very high number of terms. If however we consider that is...

    \displaystyle \sum_{n=1}^{\infty} \frac{1}{n^{2}} = \sum_{n=1}^{\infty} \frac{1}{n^{2}-\frac{1}{4}} - \sum_{n=1}^{\infty} \frac{1}{4\ n^{4}-n^{2}} (2)

    ... with the first series on the right which we know is equal to 2 and the second which has its general term that tends to 0 like \frac{1}{n^{4}}, our task becomes much less difficult ...

    Kind regards

    \chi \sigma
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