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In the past I asked here of the series of type $$\sum_ {n = 1}^\infty \frac {(f(n) +P(n)) \pmod{ Q(n)}} {D(n)}$$ where $f(n)= \displaystyle \sum_{r=2}^{k} a_{r} r^n$ with $ \{a_{r} \}_{r=2}^k$ real numbers and k is a natural number equal or larger to two. Also $P(n), Q(n)$ and $D(n)$ are polynomials with $\text{deg}(Q)=\text{deg}(D)-1$. I have also asked here if such a series diverges when taken only over the prime numbers. In the spirit of that now I'm asking if, with the notations and conditions above, the series: $$\sum_ {n = 1}^\infty \frac {(f(n) +P(n)) \pmod{ Q(n)}} {D(n)*\ln(n+1)}$$ also diverges. My first instinct is that this problem is very hard. But if solved I think it can shed some light on the prime numbers series above. Any ideas on how to find the nature of the last series?

Dominique
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Aurelian Florea
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  • I'm a bit surprised, I must admit, that nobody has any comments. I think the heuristics mentioned in the first link should hold here, too as after all, if there is a constant distribution for Q(n) then we are dealing only with a constant that doesn't really matter. Am I wrong to believe that? An argument pro or against this heuristic, should suffice for an answer, I don't expect a full solving of the problem, because as I have said I consider this problem very hard! – Aurelian Florea Feb 12 '25 at 09:10
  • @Joshua P. Swanson Dear sir maybe you have something to say here! That, because you are the one that has discussed the heuristic in question. – Aurelian Florea Feb 17 '25 at 14:37

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Not a full answer, but too long for a comment:

$P(n)\pmod{Q(n)}$ is a polynomial in $n$ of degree less than the degree of $Q$ and therefore at least $2$ less than the degree of $D$. Thus $\sum_ {n = 1}^\infty \frac {P(n) \pmod{ Q(n)}} {D(n)\ln(n+1)}$ is majorized by $\sum_ {n = 1}^\infty \frac 1{n^2}$, which converges.Therefore your series converges or diverges with $\sum_{n = 1}^\infty \frac {f(n) \pmod{ Q(n)}} {D(n)\ln(n+1)}$.

Now $f(n) \sim k^{n+1}$, so the convergence of that series is the same as $\sum_{n = 1}^\infty \frac {k^{n+1} \pmod{ Q(n)}} {D(n)\ln(n+1)}$. Since the numerator is always $< Q(n)$ whose degree is $1$ less than that of $D$, that series is majorized by $\sum_{n = 2}^\infty \frac1{n\ln n}$. But by the integral test, that series diverges. So this by itself is insufficient to settle the matter, but does suggest that $\sum_{n = 1}^\infty \frac {f(n) \pmod{ Q(n)}} {D(n)\ln(n+1)}$ is likely to diverge as well.

If one can show that for any $m > 0$ there exist values $0< r \le 1, 0\%< s\% \le 100\%$ such that $k^{n+1}\pmod{n^m}\ge rn^m$ for at least $s\%$ of all $n$, that would be sufficient to leverage the divergence of $\sum_{n = 2}^\infty \frac1{n\ln n}$ to show $\sum_{n = 1}^\infty \frac {f(n) \pmod{ Q(n)}} {D(n)\ln(n+1)}$ diverges as well.

Paul Sinclair
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  • But why Q(n) is one degree higher than P(n) mod Q(n)? P(n) mod Q(n) can be Q(n)-1 which has the same degree as Q(n). Maybe you confused the polinomial division with numbers division. P(n) is the value in n. Or am I wrong in writting the problem. Have I confused you? – Aurelian Florea Feb 13 '25 at 15:23
  • @SangchullLee responded here about this case in particular, but with a less restrictive denominator: https://math.stackexchange.com/questions/1965123/sum-of-the-series-sum-n-geq1-fracpn-bmod-qn-n2#1974003 – Aurelian Florea Feb 13 '25 at 15:26
  • I did not confuse the divisions. However I did overlook that if you evaluate the polynomials in the Euclidean algorithm for polynomials at $n$, you get a similar reduction as the Euclidean algorithm for integers, except you may get negative remainders and keep proceeding, in violation of the integer algorithm. The case of $P$ being of the same degree as $Q$ is easily handled by other means (the $P$-alone series diverges in this case). I'm not entirely sure yet what happens for higher degree $P$. – Paul Sinclair Feb 13 '25 at 15:58
  • I think you are on to something. In the first link in the question an heuristic argument is brought in support of the divergence of the series presented there. But, as a matter of fact maybe the average of the remainder is not Q(n)/2. That is because as you say sometimes the polinomial does not matter, and then the problem is reduced at the sum of the exponentials ie f(n). If this is in turn "enough" times like a convergent series then we are in big trouble relative to finding a general solution. And then the problem can only be solved on case by case scenario. Honestly I'd love that! – Aurelian Florea Feb 13 '25 at 18:03
  • In the second link I have mentioned (the one with the prime number series) someone in the comments section suggests that that series could be either convergent or divergent on case by case numbers. It is possible that that happens here also. – Aurelian Florea Feb 13 '25 at 18:07
  • You seem to be the only one I have sparked an interest in this problem. Your answer has wrong parts as we have discussed above. If you can improve it I'd be grateful and you would win a small but we'll deserved reputation boost. I was wrong too saying that P(n) mod Q(n) can be anything. It is f(n)+P(n) that seemingly yield remainders front 0 to Q(n)-1. As I have said before though the problem is very hard, so any small progress is good. – Aurelian Florea Feb 15 '25 at 09:35
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    Even without the modulus error, this post falls short of being worthy of such a reward. I am happy that it was still of use to you, though. I have no need of more rep points, and would feel bad receiving them for such a weak effort. Alas, I hadn't done anything about the error as I thought I could provide a quick fix, but it wasn't so quick after all. I still suggest that it might be best to consider $f$ and $P$ separately. If you can bound one and show the other is unbounded, or show that both are positive and one is unbounded, the issue is solved. – Paul Sinclair Feb 17 '25 at 19:59