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Proof of Euler-Mascheroni Constant

  1. Oct 19, 2011 #1
    Hi PF,

    I have question about harmonic series.

    [tex]\lim_{n\to \infty}(\sum_{k=1}^{\infty}\frac{1}{k}-ln(n))=ln(n)+\gamma +\epsilon _0[/tex]

    I couldn't find any proof for that equality. Do you know its exact proof?

    Thanks for helps.
  2. jcsd
  3. Oct 19, 2011 #2


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    The formula is wrong. It should read sum to n in the brackets and no ln(n) in the RHS.
  4. Oct 19, 2011 #3
    Oh, sorry you're right it should be like that

    [tex]\sum_{k=1}^{n}\frac{1}{k}=ln(n)+\gamma +\epsilon _n[/tex]

    My question is where it comes from? How can we prove it?
    Thanks for your correction.
  5. Oct 19, 2011 #4
    Essentially, it comes from the fact that γ is defined to be [itex]\lim_{n \rightarrow \infty}\sum_{k=1}^n \frac{1}{k} - \ln(n)[/itex], so the only thing that is necessary is to prove that this limit actually exists and is finite. For this, consider the following:

    [tex]\begin{align*} \sum_{k=1}^{n} \frac{1}{k} - \ln(n) &= \sum_{k=1}^{n} \int_{k}^{k+1}\frac{1}{\lfloor x \rfloor} \ dx - \int_{1}^{n} \frac{1}{x} \\ &= \int_{1}^{n}\frac{1}{\lfloor x \rfloor} - \frac{1}{x} \ dx + \int_{n}^{n+1} \frac{1}{\lfloor x \rfloor} \ dx \\ &= \int_{1}^{n} \frac{x-\lfloor x \rfloor}{x \lfloor x \rfloor} \ dx + \frac{1}{n} \end{align*}[/tex]

    Now, as n→∞, 1/n → 0, and the integral on the left converges to:

    [tex]\int_{1}^{\infty} \frac{x-\lfloor x \rfloor}{x \lfloor x \rfloor} \ dx[/tex]

    But this integral is finite, since:

    [tex]\frac{x-\lfloor x \rfloor}{x \lfloor x \rfloor} \leq \frac{1}{\lfloor x \rfloor^{2}}[/tex]


    [tex]\int_{1}^{\infty} \frac{1}{\lfloor x \rfloor^{2}} \ dx = \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} < \infty[/tex]
  6. Oct 19, 2011 #5
    @Citan Uzuki

    Thank you very much for your enlightening response. But what about the term [itex]\epsilon_n[/itex]. Where it comes from? And also how can we calculate γ numerically?
  7. Oct 19, 2011 #6


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    The epsilon is a very small number accounting for the error which disappears when taking the limit.
  8. Oct 19, 2011 #7
    In this http://en.wikipedia.org/wiki/Harmonic_series_(mathematics)#Rate_of_divergence", it says that [itex]\epsilon_k[/itex] is approximately 1/2k so I wonder why Euler inserted this factor into this equality and how he found that it is approximately 1/2k(not 1/5k for example). And is there any special name for that epsilon factor?

    Thanks for all your helps :)
    Last edited by a moderator: Apr 26, 2017
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