# Application of Bernoulli numbers

1. Mar 6, 2005

### Rex85

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?