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I have no idea if it is known, but here is the proof: First of all, we can remove the coprimality condition by factoring out the gcd and just prove that $S=\sum_{a, b = 1}^\infty \frac{1}{ab(a+b)} = 2 …

The answer to your question is yes, and it is a pretty well-understood topic. First of all $X$ is more or less irrelevant for the bounds in 4) so let us take $X = Y$, say, for convenience. Second, w …

I will prove below that your bound $\frac{\exp\left(C\frac{\log N}{\log \log N}\right)}{A^3}$ (which follows from $\sum_{d\mid N, d > A} \frac{1}{d^3} \le \frac{d(N)}{A^3}$) is optimal at least in the …

To get a bound which is worse than the one of GH from MO asymptotically, but which doesn't require any case checking, we can do the following: if $\gcd(t, N) = k$, then $\frac{N}{k} = d$ which is an i …

Well, the abscissa of convergence is $1$, though I see no way to deduce anything about it from the Polignac conjecture. $\sigma_P \le 1$ is obvious from $|\zeta_P(s)| \le \zeta(\Re s)$, while for $\si …

Since I was asked to in the comments, let me record here some very simple observations. They boil down to the following inequality: for all $k \ge N$ we have $$I_{k - N} \min_{x\in \mathbb{R}} h_N(x) …