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Math Help - Euler's constant.

  1. #1
    MHF Contributor Also sprach Zarathustra's Avatar
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    Euler's constant.

    Hello!

    How to prove that a_n=\Sigma^n_{k=1}\frac{1}{k}-log(n)
    is a bounded sequence. (   0 \leqslant a_n \leqslant 1 )


    Thanks!
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  2. #2
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by Also sprach Zarathustra View Post
    Hello!

    How to prove that a_n=\Sigma^n_{k=1}\frac{1}{k}-log(n)
    is a bounded sequence. (   0 \leqslant a_n \leqslant 1 )


    Thanks!
    Taking into account the basic identity...

    \gamma= \lim_{k \rightarrow \infty} \sum_{n=1}^{k} (\frac{1}{n} - \ln n) = \int_{1}^{\infty} (\frac{1}{\lfloor\ x\ \rfloor} - \frac{1}{x})\ dx (1)

    ... solving the integral in (1) You obtain...

    \gamma= 1 + \sum_{n=2}^{\infty} (\frac{1}{n} - \ln \frac{n}{n-1}) (2)

    Now because for n \ge 2 is...

    \ln \frac{n}{n-1} = \frac{1}{n-1} - \frac{1}{2\ (n-1)^{2}} + ... > \frac{1}{n} (3)

    ... each term of the 'infinite sum' in (2) is <0 and therefore is \gamma<1...

    Kind regards

    \chi \sigma
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    Hint Compare \sum_{n=1}^N1/n with \int_1^Ndx/x=\log N , you'll obtain

    \log (N+1)\leq \sum_{n=1}^N1/n\leq 1+\log N


    Edited: Sorry, I didn't see chisigma's post.
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