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Most people know the famous equation $\sum_{k=1}^{\infty} k = -\frac{1}{12}$, justified for example by interpreting the LHS as $\zeta(-1)$.

My question: does the sequence $\{\frac{1}{12}+\sum_{k=1}^n k\}_n$ have any special properties that would hint that it's supposed to tend to zero in some deeper sense?

Multiplying by $12$, that sequence becomes $6n^2+6n+1$. Do the values of this polynomial have any kind of arithmetic properties reminiscent of a sequence tending to $0$? Maybe some kind of pattern in their prime factorizations?

Of course, the original series does not converge $p$-adically for any $p$, so we can't expect higher and higher powers of a particular prime to divide the sequence. But the question is then whether some kind of asymptotic holds for the different primes dividing this sequence.

One could consider something similar like $6n^2+6n+13$ (designed so that it has the same mod $12$ behavior). That series seems to have a larger tendency to be divisible by $5$, which affects the prime factorization a bit. But the idea would be to show that the asymptotic properties of the prime factorization $6n^2+6n+1$ look different from a random polynomial that does not correspond to the error term in a divergent series with a fairly well-defined sum.

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    $\begingroup$ Related $\endgroup$
    – Wojowu
    Commented Dec 12, 2021 at 10:33
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    $\begingroup$ 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. $\endgroup$
    – GH from MO
    Commented Dec 12, 2021 at 11:45
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    $\begingroup$ 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. $\endgroup$
    – Sylvain JULIEN
    Commented Dec 12, 2021 at 19:17
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    $\begingroup$ 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. $\endgroup$
    – François Brunault
    Commented Dec 13, 2021 at 21:50

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