Is the abscissa of convergence of $s\mapsto\sum_{n>0}(ng_{n}/2)^{-s}$ known? The famous Polignac conjecture posits that the $n$-th prime gap $g_{n}:=p_{n+1}-p_{n}$ attains all even positive integral values infinitely many times, which implies $\displaystyle{\zeta_{Pol}:=s\mapsto\sum_{n>0}(ng_{n}/2)^{-s}}$ has an abscissa of convergence $\sigma_{Pol}$ less or equal to $1$.
Is this abscissa of convergence known unconditionally or under some widely believed conjecture other than Polignac's?
 A: 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 $\sigma_P \ge 1$ it is enough to prove that $\zeta_P(1) = \infty$:
$$\zeta_P(1) \ge \sum_{k=1}^\infty \frac{1}{2^{k+1}}\sum_{n = 2^k}^{2^{k+1}}\frac{1}{g_n} \gg \sum_{k = 1}^\infty \frac{1}{2^k} \frac{2^k}{k} = \infty,$$
where in the second step we used that $\sum_{n = 2^k}^{2^{k+1}} g_n \le p_{2^{k+1}} \ll 2^k \log(2^k)$ and, say, AM-HM inequality.
A: Obviously, abscissa of convergence is at most 1, because the series is dominated by the series for $\zeta(s)$. In fact, it is equal to 1. Indeed, assume that your series converges for some $s=1-\varepsilon$, $\varepsilon>0$. Then the sum
$$
\sum_{X<n<2X} (ng_n)^{-s}
$$
is bounded by constant independent of $X$. Let us prove this is not the case.
Observe that for large integer $X$ we have
$$
\sum_{X<n<2X} g_n=p_{2X}-p_{X}\sim X\log X
$$
By Cauchy-Schwarz we get
$$
\left(\sum_{X<n<2X} g_n\right)\left(\sum_{X<n<2X} (ng_n)^{-1}\right)\geq \left(\sum_{X<n<2X} \frac{1}{\sqrt{n}}\right)^2\geq ((X-1)/\sqrt{2X})^2\gg X,
$$
so that
$$\sum_{X<n<2X} (ng_n)^{-1}\gg \frac{1}{\log X}$$
therefore
$$\sum_{X<n<2X} (ng_n)^{-s}\geq \sum_{X<n<2X} n^\varepsilon (ng_n)^{-1}\gg \frac{X^\varepsilon}{\log X},$$
which contradicts boundedness.
