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Source: published paper, printed pp. 413–414 (PDF pp. 3–4), Theorem 3.

Statement

For every integer $k\ge2$, with $p_1=2$,

$$p_k\ge k(\log k+\log\log k-1). \tag{1}$$

This is the inequality actually displayed in Theorem 3. The title and abstract say “greater than”; the compilation does not silently replace the displayed $\ge$ by a strict all-$k$ conclusion. The proof below has strict margins when $p_k\ge10^{11}$. The smaller range is imported at its stated weak scope. Equation numbers on this page are its own; the print's (1) and (2) on p. 413 are Robin's estimate for $\theta(p_k)$ and Schoenfeld's final-range estimate.

Full proof relative to the exact external estimates

The cited Robin finite-range result proves (1) for $3\le p_k\le10^{11}$. This includes every $k\ge2$ whose prime lies in that range. No fresh enumeration of that range is asserted.

For the remaining ranges put

$$t=\log k,\quad q=\log p_k,\quad a=2.1454,\quad c=0.0077629.$$

They have $k\ge6$, so all required inequalities from Lemma 1 apply. Since $p_k\ge k+1$, we have $t<q$. The bound $p_k\le k\log p_k$ also gives

$$t\ge q-\log q. \tag{2}$$

First range: $10^{11}\le p_k\le e^{500}$. We have $t<q\le500$. If $t\le20$, Lemma 1 would give

$$p_k\le e^t(t+\log t) \le e^{20}(20+\log20)<10^{11},$$

where monotonicity is valid for $t\ge\log6>1$, and the last inequality is the certified endpoint bound. Thus $20<t<500$.

Schoenfeld's first-range estimate and $p_k\le kq$ give

$$p_k\ge\theta(p_k)-c\frac{p_k}{q}\ge\theta(p_k)-ck.$$

Insert Robin's lower estimate for $\theta(p_k)$ to obtain

$$p_k\ge k\left(t+\log t-1+\frac{\log t-a}{t}-c\right).$$

The complete calculus bound $g(t)>c$ on $[20,500]$ proves (1), with strict inequality in this range.

Intermediate range: $e^{500}<p_k<e^{1800}$. Since $q-\log q$ is increasing for $q>1$, equation (2) and $\log500<7$ give $t>493$; also $t<q<1800$. By Theorem 2 and $\theta(x)\le\psi(x)$,

$$\theta(p_k)-p_k<\varepsilon p_k,\qquad \varepsilon=0.905\cdot10^{-7}.$$

Combining this with Robin's estimate and $p_k\le k(t+\log t)$ gives

$$p_k> k\left(t+\log t-1+ \frac{\log t-a}{t}-\varepsilon(t+\log t)\right).$$

The correction term is positive because

$$\frac{\log t-a}{t(t+\log t)} >1.6\cdot10^{-6}>\varepsilon.$$

Thus (1) again holds strictly.

Final range: $p_k\ge e^{1800}$. The external Schoenfeld estimate, with $\eta_4=16570000$, yields

$$|\theta(p_k)-p_k| \le\eta_4\frac{p_k}{q^4} \le\eta_4\frac{k}{q^3} \le\frac{\eta_4}{1800^2}\frac{k}{t},$$

using $q\ge1800$ and $q>t>0$. Equation (2) gives $t\ge1800-\log1800>1792$, hence $\log t>7.49$. Robin's estimate now implies

$$p_k\ge k\left( t+\log t-1+ \frac{\log t-a-\eta_4/1800^2}{t}\right).$$

The numerator of the last fraction exceeds $7.49-7.26>0$ by the certified rational constant inequality. This proves strict inequality in the final range and completes the weak all-$k$ statement (1).

Every essential same-paper step is supplied at the linked pages. The analytic explicit formula, finite zero verification, Robin estimates, Schoenfeld estimates and Lemma 1 are the exact external boundaries.

Use in the covering-system termination proof

The BBMST termination induction needs only (1). In its notation $\lambda_i=\log i+\log\log i-3$, the strict increment $i(\lambda_i-\lambda_{i-1})>1$ combines with the weak prime bound to give

$$p_i-1\ge i(\lambda_i+2)-1>i(\lambda_{i-1}+2).$$

Thus no stronger external inequality is needed for that strict inductive step. This identifies the interface; the canonical BBMST proof is not duplicated or modified here.

Bears on

• Problem 2: the weak bound (1) is the external prime input of BBMST's Theorem 6.1, on which their bound $616000$ for the minimum modulus of a distinct cover rests.
• Problem 7: through the same Theorem 6.1, (1) is an input to the 2021 square-free obstruction, which covers only the square-free special case.