A few years ago, I have asked about the nature of the following series: $$\sum \limits _{n\geq 1}\frac{2^n\mod n}{n^2}$$ It was proven by user https://math.stackexchange.com/users/9340/sangchul-lee Does the sum $\sum_{n \geq 1} \frac{2^n\operatorname{mod} n}{n^2}$ converge? that the series diverges. A similar question was answered here: Sum of the series $\sum_{n\geq1} \frac{P(n) \bmod Q(n)} {n^2}$ and a more general statement was analyzed here: Convergence of the series $\sum_ {n\geq1} \frac {(f(n) +P(n)) \pmod {Q(n)}} {D(n)}$ .
I am now asking if:
$$\sum \limits _{n\geq 1}\frac{(2^n\mod P(n))\mod n}{n^2}$$ diverges, for:
- P(n)=n^2+2*n+5
- P(n)=n^2+3*n-2
The difference between the 2 is that the first has a positive remainder and the second has a negative remainder. In both cases we work under the intuitive heuristic that $2^n$ is Equi distributed in 0..P(n)-1. In the first case the remainder of P(n) divided by n is 5, so we shall have a slightly higher chance to get 0,1,2,3,4 which are constant. The sum of these constants over $n^2$ should be finite but as n increases the likelihood of getting them also decreases to 0. In the second case where the remainder of P(n) divided by n is -2 and the 0..n-3 values are a bit more likely than the last 2 values, but this only move the n/2 average a little lower. So, by these arguments assuming the heuristics hold the series should still diverge. Am I correct? Is there a reason why the heuristics should not hold?
