Consider the polynomial $S_k(x) \in \mathbb{Q}[x]$ such that $S_k(n)=\sum_{i=1}^{n}i^k, \forall n \in \mathbb{N}$. Now if i recall correctly the definition it should be that $S_n(x)=\frac{B_{n+1}(x)-B_{n+1}}{n+1}$ where $B_{n+1}(x)$ is the (n+1)-th Bernoulli polynomial and $B_{n+1}$ is the $(n+1)$-th Bernoulli number.
I saw that a certain question i'm working on, reduces to the following: for wich $n$ odd, $S_n(\frac{-1}{2})=0$? That is: for wich n even $B_n(\frac{-1}{2})=B_n$?
Of course for n even(resp odd) this is always the case.
Is this something known from the existing literature about Bernoulli polynomials/numbers?
Thanks for any help!
According to pag 3 of the following article http://ac.els-cdn.com/0021904587900712/1-s2.0-0021904587900712-main.pdf?_tid=aec020e0-7282-11e4-a9ba-00000aab0f6c&acdnat=1416686813_61c8897ff66e12d7ed1920c64b6f87eb
We get, putting $z=\frac{-1}{2}$, $B_{2k}(\frac{-1}{2})\frac{({2\pi})^{2k}}{2(2k)!}(-1)^{k+1} \to -1$.
Putting $z=0$ we get $B_{2k}(0)\frac{({2\pi})^{2k}}{2(2k)!}(-1)^{k+1} \to 1$.
So making the quotient we get that $B_{2k}(0)$ and $B_{2k}(-\frac{1}{2})$ have opposite sign for k large enough(more precisely the quotient converge to -1).
So $S_{2k+1}(\frac{-1}{2}) \neq 0 $ for k large enough.
Since it seems a very delicate argument because it works just because we ahave different signs in the same expression i would be glad if someone confirms me that it's everything correct and this answer my original question.