4

Bernouli's Formula for sum of kth powers of first n natural numbers is given by: $$f_k(n)=\frac{1}{k+1}\sum_{j=0}^k{k+1\choose j}B_j(n+1)^{k+1-j}$$ where $Bj$ is the $j^{th}$ Bernoulli Number and is in a sense recursively given by:$$B_j=-\frac{1}{j+1}\sum_{i=0}^{j-1}{j+1 \choose i}B_i$$.

I did find a generalized proof of this for Generalized case where powers can be complex numbers. I am looking for simpler proofs. Do you have any idea if this can be proved by induction.

Thank you.

PS. I am not sure of tags and appreciate if they are corrected.

Added By simpler I mean that do involve only integer powers.

007resu
  • 1,991
  • Hint: use Gamma function evaluad on Gauss interger! – Elias Costa Jan 05 '13 at 19:03
  • 3
    Section 7.6 of Graham, Knuth, & Patashnik, Concrete Mathematics, has a proof via exponential generating functions that isn’t too ugly. – Brian M. Scott Jan 05 '13 at 19:05
  • @Elias, I am not sure what you mean. I am not looking for proofs involving complex powers. and Brian, thanks, I will try to have a look at it. – 007resu Jan 05 '13 at 19:20

1 Answers1

4

The proof using the generating function: $$\frac{t}{e^t-1}$$ Can be found here. In essence, the proof follows from noting that: $$e^{kt}=\sum k^m \frac{t^m}{m!}$$ So that the sum $\ \sum k^m$ is related to the sum of the geometric series which is in turn related to the generating function.

Nathaniel Bubis
  • 33,018