Let $E$ be a fixed infinite sequence of primes such that $\sum_{p \in E} \frac{1}{p} = \infty$. Assume that $\sigma > 1$ depends on some parameter $x \rightarrow \infty$ in such a way that $\sigma \rightarrow 1^+$. Consider the ratio $R_2(\sigma) := (\sigma-1)^2\sum_{p \in E} \frac{\log^2 p}{p^{\sigma}}$ (morally, this quantity corresponds, up to constants, to the ratio of the 2nd derivatives of $\zeta_{E}(s) := \prod_{p \in E} \left(1-\frac{1}{p^s}\right)^{-1}$ with $\zeta(s)$, at $s = \sigma$). Suppose we define $\sigma' := 1+\frac{1}{2}(\sigma-1)$. Since $(\sigma'-1) \asymp (\sigma-1)$ and the series is decreasing in $\sigma$, it is easy to see that $R_2(\sigma) \ll R_2(\sigma')$. Is it possible to show that $R_2(\sigma') \ll R_2(\sigma)$ as well?
My motivation for this is in comparing $R_2(\sigma)$ with $R_3(\sigma) := (\sigma-1)^3\sum_{p \in E} \frac{\log^3 p}{p^{\sigma}}$. By some relatively simple manipulations, I can show that $R_3(\sigma) \ll R_2(\sigma) \log\left(\frac{1}{R_2(\sigma)}\right)$, but this is not sufficiently strong for my application. (It is, in fact best possible if we allow $E$ to change in relation to $\sigma$, e.g., $E:= \{p > e^{\frac{A}{\sigma-1}}\}$, where $A > 1$ is a sufficiently large constant, but since $E$ is fixed this counterexample is unimportant to us. It can also be shown that the sum over primes also does not diverge in this case.)
Noting that $te^{-\frac{1}{2}t}$ is decreasing when $t > 2$, setting $t = (\sigma-1)\log p$ gives rise to \begin{equation*} (\sigma-1)^3\sum_{p \in E \atop p > e^{\frac{2}{\sigma-1}}} \frac{\log^3 p}{p^{\sigma}} = (\sigma-1)^2\sum_{p \in E \atop p > e^{\frac{2}{\sigma-1}}} \frac{\log^2 p}{p^{\sigma'}}\left((\sigma-1)\log pe^{-\frac{1}{2}(\sigma-1)\log p)}\right) \leq 8e^{-1}R_2(\sigma'), \end{equation*} since $(\sigma-1)^2 = 4(\sigma'-1)^2$, while \begin{equation*} (\sigma-1)^3\sum_{p \in E \atop p \leq e^{\frac{2}{\sigma-1}}} \frac{\log^3 p}{p^{\sigma}} \leq 2(\sigma-1)^2\sum_{p \in E \atop p \leq e^{\frac{1}{\sigma-1}}} \frac{\log^2 p}{p^{\sigma}} \leq 2R_2(\sigma). \end{equation*} In other words, I have $R_3(\sigma) \ll R_2(\sigma) + R_2(\sigma')$. Thus, an affirmative answer to the question above solves my problem.
Any insights would be helpful.