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Shocked by a Ramanujan identity

  1. Apr 14, 2009 #1
    i do not remember the webpage i watched this but i remember that they said ' IN chapter 1 of his notebook Ramanujan wrote '

    [tex] \sum_{n=0}^{x}n^{r}= (r+1)^{-1}x^{r+1}+ \zeta (-r) - \sum_{k}B_{2k}\frac{\Gamma (r+1)}{\Gamma (k-2r+2)} [/tex]

    does anyone knows how to get this ??
     
  2. jcsd
  3. Apr 14, 2009 #2
    To begin with, what is B? and in the sum what is the range of k?
     
  4. Apr 14, 2009 #3

    CRGreathouse

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    A Bernoulli number and "0 to infinity", I believe.
     
  5. Apr 15, 2009 #4
    oh, excuse me the lack of notation

    here B_2k means the 2k-th Bernoulli Number, and yes the summation (is finite) goes to k=0 to k= infinite

    i know he used Euler-Maclaurin summation formula but the term [tex] \zeta (-r) [/tex] seems strange to me
     
    Last edited: Apr 15, 2009
  6. Apr 18, 2009 #5
    The "Euler Summation Formula" might give you some insight to why the Bernouli numbers pop up in this formula. The term [tex] \zeta (-r) [/tex] is just [tex] \frac{B_{r+1}}{r+1} [/tex] for a positive integer value of r. Another formula that will compute this sum for a positive integer value of r is :



    [tex] f_{r}(x)= \sum_{n=0} ^{x} n^{r} = r\int f_{r-1}dx + xB_{r} [/tex]

    where

    [tex] f_{1}(x)=\frac{x^2}{2} + \frac{x}{2}[/tex]



    I think one of the Bernoulli brothers could calculate [tex] 1^{10}+2^{10}+3^{10} + .... 1000^{10} [/tex] in about fifteen minutes using one of these formulas. Try it using just pen and paper and see how long it takes you...
     
    Last edited: Apr 18, 2009
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