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Need hint regarding Euler's constant question

  1. Jun 8, 2010 #1
    1. The problem statement, all variables and given/known data

    Show that Euler's constant is 0 < [tex]\gamma[/tex] < 1

    2. Relevant equations

    According to my book, [tex]\gamma[/tex] = lim((1+1/2+....+1/n) - log n) as n approaches infinity

    3. The attempt at a solution

    At first glance I was thinking about proving by contradiction.

    First I assume [tex]\gamma[/tex] is equal to 0.

    Then I will get lim(1+1/2+....+1/n) = lim (log n) as n approaches infinity.

    However 1+1/2+....+1/n is a harmonic series which diverges, so the equality does not hold.

    So [tex]\gamma[/tex] does not equal 0.

    Next I assume [tex]\gamma[/tex] is less than 0.

    Since log function never yields negative result, this implies that lim(1+1/2+....+1/n) < lim(log n) as n approaches infinity.

    Again due to the divergent nature of harmonic series, the inequality does not hold.

    These are all I have for now.


    I feel I have logic error in the attempt to solve the problem, and I hope that someone will give me a hint on the correct solution.
  2. jcsd
  3. Jun 8, 2010 #2


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    Keep in mind that

    [tex]\lim_{x\to a} [f(x)-g(x)]=\lim_{x\to a}f(x)-\lim_{x\to a}g(x)[/tex]

    Is only true if both the limits on the RHS exist.

    I haven't tried the problem yet, but you might consider using the squeeze theorem.
  4. Jun 8, 2010 #3


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    For proving the positivity of [tex]\gamma[/tex] you should think about Riemann sums
  5. Jun 9, 2010 #4


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    just to add to you first post, log(n) also diverges, so the harmonic argument doesn't quite hold
  6. Jun 10, 2010 #5

    Gib Z

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    Same for showing it is less than 1. Very simple and nice when you see the idea.
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