Timeline for Arithmetic properties of error terms in divergent series

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Dec 13, 2021 at 21:50	comment	added	François Brunault		The partial sums are essentially a Bernoulli polynomial. We have $\sum_{k=1}^{n-1} k = \frac12 B_2(n) - \frac{1}{12}$ and more generally $\sum_{k=1}^{n-1} k^p = (B_{p+1}(n)-B_{p+1}(0))/(p+1)$, where $B_m(x)$ are the Bernoulli polynomials, see here. This is related to the identity $\zeta(-p) = (-1)^p B_{p+1}/(p+1)$. So yes in some sense you can recover the sum of the divergent series.
Dec 12, 2021 at 19:17	comment	added	Sylvain JULIEN		At first sight it seems rather difficult to say anything relevant about a single polynomial of this form. Perhaps you should consider at once the set of polynomials of the form $6n^2+6n+12k+1$ with $k\in\mathbb{Z}$ and use some ideal theoretic properties to shed light on the proportion of prime values they may take under big conjectures like Schinzel's hypothesis H, Bunyakovsky conjecture, or the like, but it probably requires quite a bunch of efforts.
Dec 12, 2021 at 11:45	comment	added	GH from MO		I think the whole point of regularization is that we don't proceed via the partial sums $\sum_{k=1}^n k$. So the equation in question has little to do with the sequence you talk about. Just my two cents.
Dec 12, 2021 at 11:02	history	edited	YCor	CC BY-SA 4.0	removed capitals from title
Dec 12, 2021 at 10:33	comment	added	Wojowu		Related
Dec 12, 2021 at 9:42	history	asked	David Corwin	CC BY-SA 4.0