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Math Help - About Euler's constant (3)...

  1. #1
    MHF Contributor chisigma's Avatar
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    About Euler's constant (3)...

    Given the Euler's constant...

    \gamma = \lim_{n \rightarrow \infty} \{\sum_{k=1}^{n} \frac{1}{k} - \ln n \}= .577215664901... (1)

    ... demonstrate that is...

    \gamma = \sum_{k=2}^{\infty} (-1)^{k} \frac{\zeta(k)}{k} (2)

    ... where...

    \zeta(k)= \sum_{n=1}^{\infty} \frac{1}{n^{k}} (3)

    Kind regards

    \chi \sigma
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  2. #2
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    Quote Originally Posted by chisigma View Post
    Given the Euler's constant...

    \gamma = \lim_{n \rightarrow \infty} \{\sum_{k=1}^{n} \frac{1}{k} - \ln n \}= .577215664901... (1)

    ... demonstrate that is...

    \gamma = \sum_{k=2}^{\infty} (-1)^{k} \frac{\zeta(k)}{k} (2)

    ... where...

    \zeta(k)= \sum_{n=1}^{\infty} \frac{1}{n^{k}} (3)

    Kind regards

    \chi \sigma
    for any -1 < x \leq 1 we have \sum_{k=2}^{\infty} \frac{(-x)^k}{k}=x - \ln(1+x), which is proved by integrating \frac{1}{1+x}=1-x + x^2 - \cdots .

    putting x=\frac{1}{n}, \ n \in \mathbb{N}, gives us \sum_{k=2}^{\infty} \frac{(-1)^k}{kn^k} = \frac{1}{n} - \ln \left(1 + \frac{1}{n} \right). therefore:

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

    =\sum_{n=1}^{\infty} \left [ \frac{1}{n} - \ln \left(1 + \frac{1}{n} \right) \right]=\lim_{n\to\infty} \left[\sum_{k=1}^n \frac{1}{k} - \ln(n+1) \right] = \gamma.
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