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Application of Bernoulli numbers

  1. Mar 6, 2005 #1
    I need to prove
    1^k+2^k+...+n^k= Sum(j=0, k)[(k choose j)Bj*((n+1)^(k+1-j)/(k+1-j))
    where Bj is the jth Bernoulli number, and k=1,2,...

    We are given
    1+e^z+e^(2z)+...+e^(nz)= ((e^((n+1)z)-1)/z)(z/(e^z-1))
    and told to write each side as a power series to derive the top.

    So far I have proven,
    1+e^z+e^(2z)+...+e^(nz)= Sum(m=0, infinity)[(0^m+1^m+...+n^m)/m!] and
    ((e^((n+1)z)-1)/z)(z/(e^z-1))= Sum(j=0, k)[(k choose j)Bj*((n+1)^(k+1)/(k+1-j))

    But now I am stuck, where do I go from here?
  2. jcsd
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