Results 1 to 4 of 4

Math Help - Euler-Mascheroni Constant

  1. #1
    Newbie
    Joined
    Oct 2011
    Posts
    2

    Euler-Mascheroni Constant

    OK can somebody show me how sequence 1+1/2+1/3+....+1/n-ln(n) converges.Also how did euler compute this constant,how can I compute it?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5

    Re: Euler-Mascheroni Constant

    Quote Originally Posted by armansuper View Post
    OK can somebody show me how sequence 1+1/2+1/3+....+1/n-ln(n) converges.Also how did euler compute this constant,how can I compute it?
    Have you attempted to find answers to your questions using Google.

    There is also an excellent texbook: Amazon.com: Gamma: Exploring Euler's Constant (Princeton Science Library) (9780691099835): Julian Havil, Freeman Dyson: Books

    I suggest you try to find and borrow it from a library (perhaps at the institute you study at).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5

    Re: Euler-Mascheroni Constant

    Quote Originally Posted by armansuper View Post
    OK can somebody show me how sequence 1+1/2+1/3+....+1/n-ln(n) converges.Also how did euler compute this constant,how can I compute it?
    The Euler-MCLaurin summation has been illustrated in...

    http://www.mathhelpforum.com/math-he...es-188663.html

    ... and it can be used for an approximate computation of the so called harmonic numbers...

    H_{n} \sim \gamma + \ln n + \frac{1}{2n}\ \sum_{k \ge 1} \frac{B_{2k}}{2 k\ n^{2 k}} (1)

    ... where \gamma is the Euler's constant and B_{2k} the Bernoulli numbers. A formula like (1) allows great precision if an 'optimal' value of k is choosen. In 1736 Euler used the formula (1) with k=7 ...

    \gamma \sim H_{n} - \ln n - \frac{1}{2 n} + \frac{1}{12\ n^{2}} -  \frac{1}{120\ n^{4}} +  \frac{1}{252\ n^{6}}-  \frac{1}{240\ n^{8}} +  \frac{1}{132\ n^{10}}-

     -  \frac{691}{32760\ n^{12}} +  \frac{1}{12\ n^{14}} (2)

    ... and n=10 to compute the first sixteen digit of \gamma obtaining...

    H_{10}= 2.9289682539682539...

     \ln 10 = 2.302585092994045684...

     \gamma = .5772156649015329...

    Kind regards

    \chi \sigma
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Sep 2011
    Posts
    70
    Thanks
    2

    Re: Euler-Mascheroni Constant

    This is illustrated in Apostol's Calculus Vol. I, and I am on this section right now so it would be helpful for me to try and explain it to you. The basics that you need to understand are Leibniz's Rule (also known as The Alternating Series Test).

    Let
    a_1=1, \qqaud a_2=\int_1^2 \frac {dx} x, \qquad a_3=\frac 1 2, \qquad a_4=\int_2^3 \frac {dx} x, \hdots
    where, in general,
    a_{2n-1}=\frac 1 n, \qquad a_{2n}=\int_n^{n+1} \frac {dx} x

    We want to apply Leibniz's Rule to the series \textstyle \sum (-1)^{n-1} a_n , so we must prove (A) that the series is monotonic decreasing, and (B) that it's limit as n increases without bound is equal to 0.

    To verify (A), note that, by the mean value theorem for integration, \frac 1 c = \int_n^{n+1} \frac {dx} x for some c\in[n,n+1]. Therefore, \frac 1 n \ge \frac 1 c\ge \frac 1 {n+1}, and so a_{2n-1}\ge a_{2n}\ge a_{2n+1}. This covers all possibilities, so \{a_n\} is monotonically decreasing.

    To verify (B), consider even and odd terms separately.
    a_{2n-1} = \frac 1 n, so clearly \lim_{n\to\infty}a_{2n-1}=0
    a_{2n}=\int_n^{n+1} \frac {dx} x=\log (\frac{n+1} n), so \lim_{n\to\infty} a_{2n} = 0.

    Therefore, by Leibniz's Rule, the series \sum(-1)^{(n-1)}a_n converges. Denote it's sum by \gamma and it's nth partial sum by s_n. Now we inspect the (2n-1)st partial sum:

    s_{2n-1}=1-\int_1^2 \frac {dx} x + \frac 1 2 - \int_2^3 \frac {dx} x + \cdots + \frac 1 {n-1} - \int_{n-1}^n \frac {dx} x + \frac 1 n
    Rearranging...
    s_{2n-1}=1 + \frac 1 2 + \cdots + \frac 1 n - \log n

    Since s_{2n-1} \rightarrow C as n \rightarrow \infty, we have
    \lim_{n\to\infty}\left(1+\frac 1 2 + \cdots + \frac 1 n -\log n \right)=\gamma

    This \gamma is the Euler-Mascheroni constant. The relation above can also be expressed as

    \sum_{k=1}^n \frac 1 k = \log n + \gamma + o(1)


    This entire post is very similar to the text in Apostol's Calculus, so if there is any concern of copyright infringement (although the book was written in 1969, and up to 10% is reproducible for educational purposes) please feel free to delete this post.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. About Euler's constant (4)...
    Posted in the Math Challenge Problems Forum
    Replies: 4
    Last Post: November 16th 2009, 08:37 AM
  2. About Euler's constant (3)...
    Posted in the Math Challenge Problems Forum
    Replies: 1
    Last Post: November 11th 2009, 08:02 AM
  3. About Euler's constant (2)...
    Posted in the Math Challenge Problems Forum
    Replies: 3
    Last Post: November 9th 2009, 04:30 AM
  4. About Euler's constant (1)...
    Posted in the Math Challenge Problems Forum
    Replies: 13
    Last Post: November 5th 2009, 12:10 PM
  5. Euler–Mascheroni constant
    Posted in the Calculus Forum
    Replies: 3
    Last Post: June 14th 2009, 04:39 AM

Search Tags


/mathhelpforum @mathhelpforum