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Power sum is given by $$ 1^m + \cdots + n^m = \frac1{m+1} \sum_{k=0}^m (-1)^k \binom{m+1}kB_k n^{m+1-k}$$ and negative zeta values are given by $$\zeta(-m) = (-1)^n \frac{B_{m+1}}{m+1}$$ But heuristically, $\zeta(-m) = 1^m + 2^m + \cdots $. So it feels as if taking $\lim_{n \rightarrow \infty}$ in the power sum formula should give us the negative zeta value.

All this is obviously mathematically not rigorous ($\zeta(s)$ with $\text{Re}(s)\le 1$ must be calculated using reflection formula and taking such limit is impossible). But to me, the occurrence of Bernoulli number in both expressions seems too suspicious for a coincidence. Is it really just a coincidence?

finnlim
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  • @SimplyBeautifulArt Well using the Faulhaber's formula to obtain an analytic continuation for $\zeta(s)$, it gives $\zeta(-k) = \sum_{m=0}^k c_{m,k} B_{m+1}$ for some coefficients $c_{m,k}$ which happen to be (some sort of coincidence ?) the ones in the recursion for the Bernoulli numbers so it simplifies to $\zeta(-k) = (-1)^k\frac{B_{k+1}}{k+1}$ – reuns Aug 23 '17 at 02:11

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