# Solving summation; ( double sum).

I found the expression to the sum of powers long ago and ofcours I think it is true but i don't know for sure, the problem is, it's little though for me to test and try it out. Also i'd like to know how should i prove it.

for k>=0 $$\sum_{m=1}^{n} m^{k+1} =\sum^{k}_{s=0}\sum^{s}_{i=0} {s \choose i}{n+1 \choose s+2}(n-i)^{k} (-1)^{s+i}$$

I revised this because i was looking for additional ways to get the zeta function, this was the formula were i needed it: $\zeta(-s)= 1/(2^s-1)*(2^s\sum_{n=1}^{m/2} n^s +2^s\sum_{n=1}^{(m-1)/2} n^s-\sum_{n=1}^{m} n^s$)

After using both formulas is there a way to simplify it?

• For me, I always remember the formula for sum of powers as $\sum_{k=0}^x k^p = \frac{B_{p+1}(x+1) - B_{p+1}(0)}{p+1}$ where $B_p(x)$ is the Bernoulli polynomials. If your formula is correct, one should be able to derive it from the generating function of the Bernoulli polynomials pretty easily. Commented Jun 20, 2015 at 23:14
• Well thats more or less the question. Since i'm not too familiar myself with the bernoulli numbers/ polynomials i would love to see how you would create them out of this formula, again if it's correct ofcours. But i start looking into it and they seem really usefull. Thanks inadvance! Commented Jun 21, 2015 at 2:46
• If this makes sense, this is how i got the summation $$^{s}\sum^{n}_{m=1}m^k=n .^{s}\sum^{n}_{m=1}m^{k-1}-s*.^{s+1}\sum^{n}_{m=1}(m-1)^{k-1}$$ I hope i wrote this correct down, since i got this before i was used to summation notation at all. The little s before the sum means it's the s'th sum. so the s'th sum till n, over a lineair function would be (n+s+1)!/(n-1)!/(s+1)!. I bet you get what i mean with the s'th sum. And this formula above i use to derive the rest, since I write my formula in terms of this basic know summation. Commented Jun 21, 2015 at 19:37