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Source. Hough, Lemma 7, Appendix A, printed pp. 378–379 of the published paper. The external input is exactly Theorem 6.

Statement. For every integer $n\ge11$, with $q_j=(j+1)^3-j^3$,

$$A_n:=\prod_{e^n<p\le e^{n+1}}\left(1+\frac2{p-1}\right)<\frac65,$$ $$G_n:=\prod_{e^n<p\le e^{n+1}} \left(1+2\sum_{j\ge1}\frac{q_j}{p^j}\right)<\frac{17}{5}, \qquad S_n:=\sum_{e^n<p\le e^{n+1}}\frac1{(p-1)^3} <\frac{22/25}{2ne^{2n}}.$$

The source states the weaker third bound $S_n<1/(2ne^{2n})$. Its proof already establishes the factor $22/25=0.88$ for $n\ge14$; the complete finite certificate establishes it also for $n=11,12,13$.

Complete proof. The cases $n=11,12,13$ are the finite prime calculations proved and executed in the numerical certificate. Now suppose $n\ge14$, put $a=e^n$, $b=e^{n+1}$, and write $E(x)=\theta(x)-x$. Since $e^{14}>678407$, the external theorem gives $|E(x)|<x/(40\log x)$ throughout $[a,b]$.

Set

$$I_n=\int_a^b\frac{d\theta(x)}{x\log x} =\sum_{a<p\le b}\frac1p.$$

Stieltjes integration by parts, using $d\theta=dx+dE$, gives

$$I_n\le\log\frac{n+1}{n} +\frac{|E(b)|}{b(n+1)}+\frac{|E(a)|}{an} +\int_a^b\frac{|E(x)|}{x^2} \left(\frac1{\log x}+\frac1{(\log x)^2}\right)dx.$$

The integral's error is at most $\frac{2}{40n}\log((n+1)/n)$: insert the bound for $E$ and use $\log x\ge n\ge1$. All the resulting positive bounds decrease when $n$ increases. Therefore

$$I_n\le\log\frac{15}{14}+\frac1{40\cdot15^2} +\frac1{40\cdot14^2}+\frac2{40\cdot14}\log\frac{15}{14} <0.0695. \tag{A}$$

Using $\log(1+u)\le u$,

$$\log A_n\le2\sum_{a<p\le b}\frac1{p-1} \le\frac2{1-e^{-14}}I_n<0.14<\log(6/5).$$

For the second product, $q_1=7$ and $3q_j-q_{j+1}=6j^2-4>0$ give $q_j\le7\cdot3^{j-1}$, so $\sum_{j\ge1}q_j/p^j\le7/(p-3)$. Hence

$$\log G_n\le14\sum_{a<p\le b}\frac1{p-3} \le\frac{14}{1-3e^{-14}}I_n <\frac{14\cdot0.07}{1-3e^{-14}}<1<\log(17/5).$$

Finally, since $\log p\ge n$ in this band,

$$S_n\le\frac1{n(1-e^{-n})^3}\int_a^b\frac{d\theta(x)}{x^3}.$$

Writing $d\theta=dx+dE$ again, and integrating the error by parts,

$$\int_a^b\frac{d\theta(x)}{x^3} \le\frac{1-e^{-2}}{2e^{2n}} +\frac1{40ne^{2n}}+\frac1{40(n+1)e^{2(n+1)}} +\frac3{40n}\int_a^b\frac{dx}{x^3}.$$

After division by $n(1-e^{-n})^3$, this is at most

$$\frac1{2ne^{2n}}\frac1{(1-e^{-14})^3} \left(1-e^{-2}+\frac1{20\cdot14} +\frac1{20e^2\cdot15}+\frac3{40\cdot14}\right) <\frac{0.88}{2ne^{2n}}.$$

Here we bounded $1-e^{-2}$ by $1$ only in the last error term and used $n\ge14$ in the other positive factors. Every scalar comparison in (A) and the subsequent displays is enclosed with rational Taylor bounds in the numerical certificate. This proves all three inequalities uniformly for every integer $n\ge11$.

Scope. The ordinary proof is complete relative to the exact Rosser–Schoenfeld estimate. The computation treats three finite bands, not an infinite enumeration. Retaining the factor $0.88$ supplies the uniform numerical margin used in the compiled proof of Theorem 1; this is not presented as an author-issued correction.

Bears on. Problem 2.