From Chebyshev to Primorials: Establishing the Riemann Hypothesis

1. Introduction

The Riemann Hypothesis, first proposed by Bernhard Riemann in 1859, asserts that every non-trivial zero of the Riemann zeta function $\zeta \left(s\right)$ has real part equal to ${\displaystyle \frac{1}{2}}$. This conjecture is widely regarded as the foremost unsolved problem in pure mathematics: it constitutes a central part of Hilbert’s eighth problem and is one of the seven Clay Mathematics Institute Millennium Prize Problems [1].

The zeta function $\zeta \left(s\right)$, defined and analytically continued over the complex plane, has so-called trivial zeros at the negative even integers $s=-2,-4,-6,\dots$ and non-trivial zeros in the critical strip $0<\Re \left(s\right)<1$. Riemann’s conjecture is that every non-trivial zero lies on the critical line$\Re \left(s\right)={\displaystyle \frac{1}{2}}$. Far from being a purely theoretical curiosity, the hypothesis has profound implications for the distribution of prime numbers, a subject of fundamental importance in both pure mathematics and computational number theory.

1.1. Overview and Main Result

The proof presented here proceeds entirely by contradiction and rests on two structural lemmas. We summarize the logical architecture before entering the technical details, so that the reader can keep the overall strategy in mind throughout.

The Analytic Engine (Lemma 1: Key Finding)

Lemma 1 provides the recurring downward step that the main contradiction will require. For each sufficiently large index n, one can find a later index $n^{\prime }>n$ such that

$$R\left(N_{n^{\prime }}\right)<R\left(N_n\right),$$

where $R\left(N_k\right)=\Psi \left(N_k\right)/(N_{k}loglog{N}_k)$ and $N_k$ is the k-th primorial.

The proof of this lemma uses the closed-form identity

$$R\left(N_k\right)={\displaystyle \frac{{\displaystyle \prod _{i=1}^k\left(1+\frac{1}{p_i}\right)}}{log\theta \left(p_k\right)}},$$

where $\theta$ denotes the Chebyshev function. Under this representation the condition $R\left(N_{n^{\prime }}\right)<R\left(N_n\right)$ is equivalent to the logarithmic inequality

$${\displaystyle \frac{log\theta \left(p_{n^{\prime }}\right)}{log\theta \left(p_n\right)}}>\prod _{p_n<p\le p_{n^{\prime }}}\left(1+{\displaystyle \frac{1}{p}}\right).$$

Choosing $p_{n^{\prime }}\approx p_n^{\alpha }$ with $\alpha ={\displaystyle \frac{log{p}_n}{log{p}_n-1}}>1$, and combining Mertens’ theorem, the prime number theorem, and an explicit error analysis of the relevant tail sums, one establishes a strict positive lower bound on the difference of the two sides for all sufficiently large n.

The Logical Backbone (Lemma 2: Main Insight)

Lemma 2 is the engine of the contradiction. Suppose, for the sake of argument, that the Riemann Hypothesis is false. By Proposition 2, this assumption forces the set

$$\left\{n\in \mathbb{N}\mid R\left(N_n\right)<\frac{e^{\gamma }}{\zeta \left(2\right)}\right\}$$

to be infinite. Lemma 2 shows that if, beyond some threshold index $n_0$, every occurrence of $R\left(N_n\right)<e^{\gamma }/\zeta \left(2\right)$ is followed by a later index $n^{\prime }>n$ with $R\left(N_{n^{\prime }}\right)<R\left(N_n\right)$, then one can build an infinite strictly decreasing sequence

$$R\left(N_{n_1}\right)>R\left(N_{n_2}\right)>R\left(N_{n_3}\right)>\dots$$

bounded below by zero, hence convergent to some limit $L\ge 0$. Since ${\left(R\left(N_{n_j}\right)\right)}_{j\ge 1}$ is a subsequence of ${\left(R\left(N_k\right)\right)}_{k\ge 1}$, and Proposition 3 gives ${lim}_{k\to \infty }R\left(N_k\right)=e^{\gamma }/\zeta \left(2\right)$, the subsequential limit must also equal $e^{\gamma }/\zeta \left(2\right)$. However, the strict decrease forces every term to satisfy $R\left(N_{n_j}\right)<R\left(N_{n_1}\right)<e^{\gamma }/\zeta \left(2\right)$, so the limit satisfies $L\le R\left(N_{n_1}\right)<e^{\gamma }/\zeta \left(2\right)$, which contradicts $L=e^{\gamma }/\zeta \left(2\right)$. Therefore the assumption that the Riemann Hypothesis is false is untenable.

Conclusion (Theorem 1)

Theorem 1 unifies the two lemmas. Lemma 1 produces, for every sufficiently large index n, an index $n^{\prime }>n$ satisfying $R\left(N_{n^{\prime }}\right)<R\left(N_n\right)$, thereby fulfilling the recurring-decrease hypothesis of Lemma 2. Lemma 2 then derives the contradiction that refutes the assumption of falsity of the Riemann Hypothesis. The conclusion follows without appeal to any unproven conjecture or numerical verification beyond the explicit threshold $n_0$.

2. Materials and Methods

In analytic number theory, several classical arithmetic functions encode deep information about the distribution of prime numbers. The arguments that follow rest on three of these: the Chebyshev function, the Dedekind $\Psi$ function, and the Riemann zeta function. We fix notation in this section and collect the known results that will be invoked later.

2.1. Definitions

Definition 1 

(Chebyshev’s Prime-Counting Function). TheChebyshev function$\theta :[2,\infty )\to \mathbb{R}$ is defined by

$$\theta \left(x\right)=\sum _{p\le x}logp,$$

where the sum runs over all prime numbers $p\le x$. Equivalently, $\theta \left(x\right)=log{\prod }_{p\le x}p$. The prime number theorem is equivalent to the statement $\theta \left(x\right)\sim x$ as $x\to \infty$.

Definition 2 

(Dedekind’s Arithmetic Function $\Psi$). For a positive integer $n\ge 1$, theDedekind $\Psi$ functionis

$$\Psi \left(n\right)=n·\prod _{p\mid n}\left(1+{\displaystyle \frac{1}{p}}\right),$$

where the product is taken over all distinct prime divisors p of n. When n is squarefree (so that $n=p_1{p}_2\dots p_k$ for distinct primes), this simplifies to

$$\Psi \left(n\right)=\prod _{p\mid n}(p+1),$$

since each factor $p(1+1/p)=p+1$.

Definition 3 

(Primorial Numbers). Thek-th primorial, denoted $N_k$, is the product of the first k prime numbers:

$$N_k=\prod _{i=1}^k{p}_i=p_1·p_2\dots p_k,$$

where $p_i$ denotes the i-th prime ($p_1=2,p_2=3,p_3=5,\dots$). Every primorial is squarefree by construction, so the Dedekind function takes the particularly clean form $\Psi \left(N_k\right)=N_k{\prod }_{i=1}^k(1+1/p_i)$. Note also that $log{N}_k={\sum }_{i=1}^{k}log{p}_i=\theta \left(p_k\right)$ by Definition 1.

Definition 4 

(Normalized Dedekind Ratio). For an integer $n\ge 3$, thenormalized Dedekind ratiois

$$R\left(n\right)={\displaystyle \frac{\Psi \left(n\right)}{n·loglogn}}.$$

When evaluated at the k-th primorial $N_k$ (with $k\ge 2$, so that $loglog{N}_k>0$), the squarefreeness of $N_k$ yields

$$R\left(N_k\right)={\displaystyle \frac{{\displaystyle \prod _{i=1}^k\left(1+{\displaystyle \frac{1}{p_i}}\right)}}{loglog{N}_k}}={\displaystyle \frac{{\displaystyle \prod _{i=1}^k\left(1+{\displaystyle \frac{1}{p_i}}\right)}}{log\theta \left(p_k\right)}},$$

where the last step uses $log{N}_k=\theta \left(p_k\right)$ from Definition 3.

Definition 5 

(Riemann Zeta Function at $s=2$). TheRiemann zeta functionevaluated at $s=2$ is the convergent series

$$\zeta \left(2\right)=\sum _{n=1}^{\infty }{\displaystyle \frac{1}{n^2}}={\displaystyle \frac{{\pi }^2}{6}}.$$

Via the Euler product over primes (Proposition 1),

$$\zeta \left(2\right)=\prod _{k=1}^{\infty }{\displaystyle \frac{p_k^2}{p_k^2-1}},$$

a representation that connects the value ${\pi }^2/6$ directly to the sequence of primes.

Definition 6 

(Dedekind Condition at a Prime). For the n-th prime $p_n$ (with $n\ge 2$), we say theDedekind condition$\mathsf{Dedekind}\left(p_n\right)$ holds when

$$\prod _{p\le p_n}\left(1+{\displaystyle \frac{1}{p}}\right)>{\displaystyle \frac{e^{\gamma }}{\zeta \left(2\right)}}·log\theta \left(p_n\right),$$

where γ denotes the Euler–Mascheroni constant. In terms of the ratio from Definition 4, this condition is equivalent to

$$R\left(N_n\right)>{\displaystyle \frac{e^{\gamma }}{\zeta \left(2\right)}}.$$

The Riemann Hypothesis can be reformulated as asserting that $\mathsf{Dedekind}\left(p_n\right)$ holds for all but finitely many n (see Proposition 2).

Definition 7 

(Auxiliary Analytic Functions). The following real-valued functions of a real variable $x\ge 2$ appear throughout the proof.

(a) The Mertens-type product.

$$f\left(x\right)=e^{\gamma }log\theta \left(x\right)·\prod _{p\le x}\left(1-{\displaystyle \frac{1}{p}}\right).$$

This quantity interpolates between Mertens’ third theorem (${\prod }_{p\le x}(1-1/p)\sim e^{-\gamma }/logx$) and the prime number theorem ($\theta \left(x\right)\sim x$) [2,3].

(b) The logarithm of f.By Nicolas’s expansion [2,3],

$$logf\left(x\right)=U\left(x\right)+u\left(x\right),$$

where the two summands are defined below.

(c) The partial-sum function U.

$$U\left(x\right)=loglog\theta \left(x\right)+\sum _{p\le x}\left(-{\displaystyle \frac{1}{p}}\right)+B,$$

where B is theMeissel–Mertens constant[4]

$$B=\gamma +\sum _p\left(log\left(1-{\displaystyle \frac{1}{p}}\right)+{\displaystyle \frac{1}{p}}\right).$$

(d) The tail-correction function u.

$$u\left(x\right)=\sum _{p>x}\left(log\left({\displaystyle \frac{p}{p-1}}\right)-{\displaystyle \frac{1}{p}}\right).$$

This series converges absolutely for every $x\ge 2$, since each term is $O\left(p^{-2}\right)$ by the Taylor expansion of $log(1+t)$.

(e) The tail-sum function v.

$$v\left(x\right)=\sum _{p>x}log\left({\displaystyle \frac{p^2}{p^2-1}}\right)=\sum _{p>x}\left(log\left({\displaystyle \frac{p}{p-1}}\right)-log\left(1+{\displaystyle \frac{1}{p}}\right)\right).$$

The function v satisfies $v\left(x\right)>u\left(x\right)$ for all $x\ge 2$ (Proposition 6), a strict inequality that drives the final step of the proof of Lemma 1.

2.2. Key Propositions

We now record the known results that will be invoked in the proofs.

Proposition 1 

(Euler Product for $\zeta \left(2\right)$ [5]). The value of the Riemann zeta function at $s=2$ satisfies the Euler product identity

$$\zeta \left(2\right)=\prod _{k=1}^{\infty }{\displaystyle \frac{p_k^2}{p_k^2-1}}={\displaystyle \frac{{\pi }^2}{6}},$$

where $p_k$ denotes the k-th prime number.

Proposition 2 

(Behavior of $R\left(N_n\right)$ under a False Riemann Hypothesis [6]). If the Riemann Hypothesis is false, then there exist infinitely many indices $n\in \mathbb{N}$ such that

$$R\left(N_n\right)<{\displaystyle \frac{e^{\gamma }}{\zeta \left(2\right)}}.$$

Equivalently, the Dedekind condition $\mathsf{Dedekind}\left(p_n\right)$ (Definition 6) fails for infinitely many n.

Proposition 3 

(Asymptotic Limit of the Sequence $R\left(N_k\right)$ [7]). As $k\to \infty$, the normalized Dedekind ratio converges to

$$\underset{k\to \infty }{lim}R\left(N_k\right)={\displaystyle \frac{e^{\gamma }}{\zeta \left(2\right)}}.$$

This limit follows from Mertens’ third theorem together with the prime number theorem.

Proposition 4 

(Nicolas’s Logarithmic Expansion [2,3]). For every $x\ge 2$, the function $f\left(x\right)$ from Definition 7 satisfies

$$logf\left(x\right)=U\left(x\right)+u\left(x\right),$$

where U and u are as in Definition 7.

Proposition 5 

(Prime Number Theorem: Chebyshev Asymptotics [8]). The Chebyshev function satisfies $\theta \left(x\right)\sim x$ as $x\to \infty$. In particular, if an index $i=i\left(n\right)$ is chosen so that $p_{n+i}\sim p_n^{\alpha }$ for a fixed exponent $\alpha >1$, then

$${\displaystyle \frac{log\theta \left(p_{n+i}\right)}{log\theta \left(p_n\right)}}⟶\alpha \mathrm{as}n\to \infty .$$

Proposition 6 

(Taylor Expansion of the Natural Logarithm [9]). For $0<x\le 2$, the natural logarithm admits the convergent power series

$$logx=\sum _{m=1}^{\infty }{(-1)}^{m+1}{\displaystyle \frac{{(x-1)}^m}{m}}.$$

In particular, substituting $x=1+1/p$ (valid since $1/p<1$ for every prime $p\ge 2$) gives

$$log\left(1+{\displaystyle \frac{1}{p}}\right)={\displaystyle \frac{1}{p}}-{\displaystyle \frac{1}{2p^2}}+{\displaystyle \frac{1}{3p^3}}-\dots <{\displaystyle \frac{1}{p}},$$

which immediately implies $1/p-log(1+1/p)>0$ for every prime $p\ge 2$, and therefore $v\left(x\right)>u\left(x\right)$ for all $x\ge 2$.

Proposition 7 

(Geometric Series Identity [10]). For any complex number x with $\left|x\right|<1$,

$${\displaystyle \frac{1}{1-x}}=\sum _{i=0}^{\infty }{x}^i=1+x+x^2+x^3+\dots$$

Specializing to $x=1/logp$ for an odd prime $p\ge 3$ (note $|1/logp|<1$ for $p\ge 3$) yields

$${\displaystyle \frac{logp}{logp-1}}={\displaystyle \frac{1}{1-1/logp}}=\sum _{i=0}^{\infty }{\left({\displaystyle \frac{1}{logp}}\right)}^i.$$

This expansion is used in Step 5 of the proof of Lemma 1 to obtain an explicit lower bound for $log\alpha$.

Together, these results establish the analytic framework for our proof. By examining the interplay between the Chebyshev function and primorial numbers, we reveal how the non-trivial zeros of the zeta function are constrained by the multiplicative structure of the primes.

3. Results

We now state and prove the two lemmas and the main theorem. Lemma 1 (Key Finding) establishes the recurring downward step; Lemma 2 (Main Insight) converts that step into a contradiction with the falsity of the Riemann Hypothesis; and Theorem 1 (Main Theorem) combines them to conclude.

Lemma 1 

(Key Finding). There exists $N\in \mathbb{N}$ such that for every $n>N$ one can find a positive integer $i=i\left(n\right)$—determined by the exponent $\alpha ={\displaystyle \frac{log{p}_n}{log{p}_n-1}}>1$—for which

$${\displaystyle \frac{log\theta \left(p_{n+i}\right)}{log\theta \left(p_n\right)}}>\prod _{p_n<p\le p_{n+i}}\left(1+{\displaystyle \frac{1}{p}}\right).$$

Proof. 

The argument proceeds in ten steps: we first choose i in terms of $\alpha$, then compare the asymptotic behavior of both sides via Mertens’ theorem and explicit tail estimates.

• Step 1. Algebraic reduction of the product 

We begin by decomposing the right-hand side into two factors whose asymptotics are individually tractable. Using the identity $1+1/p=(1-1/p^2)/(1-1/p)$, we write

$$\prod _{p_n<p\le p_{n+i}}\left(1+{\displaystyle \frac{1}{p}}\right)={\displaystyle \frac{{\displaystyle \prod _{p_n<p\le p_{n+i}}\left(1-\frac{1}{p^2}\right)}}{{\displaystyle \prod _{p_n<p\le p_{n+i}}\left(1-\frac{1}{p}\right)}}}.$$

This factorization separates the prime-density factor (the denominator, handled by Mertens’ theorem) from the rapidly convergent squared-prime factor (the numerator, whose logarithm forms the tail $v\left(m\right)-v\left(M\right)$). The target inequality of the lemma is therefore equivalent to

$${\displaystyle \frac{log\theta \left(p_{n+i}\right)}{log\theta \left(p_n\right)}}·\prod _{p_n<p\le p_{n+i}}\left(1-{\displaystyle \frac{1}{p}}\right)>\prod _{p_n<p\le p_{n+i}}\left(1-{\displaystyle \frac{1}{p^2}}\right).$$

• Step 2. Choice of the index i 

Set

$$\alpha :={\displaystyle \frac{log{p}_n}{log{p}_n-1}}>1,$$

and let $i=i\left(n\right)$ be the largest non-negative integer such that $p_{n+i}\le p_n^{\alpha }$. By the Prime Number Theorem, the gap between consecutive primes near $p_n^{\alpha }$ is $o\left(p_n^{\alpha }\right)$, so this choice forces $p_{n+i}\sim p_n^{\alpha }$ as $n\to \infty$.

Note that $\alpha >1$ ensures $p_n^{\alpha }>p_n$, so the product in the lemma is taken over a non-empty set of primes. Moreover,

$$p_n^{\alpha }=p_n·p_n^{1/(log{p}_n-1)}=p_n·exp\left({\displaystyle \frac{log{p}_n}{log{p}_n-1}}\right)\ge p_n·e>2p_n,$$

so by Bertrand’s postulate (there is always a prime between m and $2m$) the interval $(p_n,p_n^{\alpha }]$ contains at least one prime, guaranteeing that $i\ge 1$.

• Step 3. Asymptotic behavior of the logarithmic ratio 

By Proposition 5 (the prime number theorem in the form $\theta \left(x\right)\sim x$), we have $log\theta \left(x\right)=logx+o\left(1\right)$ as $x\to \infty$. Since $p_{n+i}\sim p_n^{\alpha }$, it follows that $log{p}_{n+i}\sim \alpha log{p}_n$, and therefore

$$\underset{n\to \infty }{lim}{\displaystyle \frac{log\theta \left(p_{n+i}\right)}{log\theta \left(p_n\right)}}=\underset{n\to \infty }{lim}{\displaystyle \frac{log{p}_{n+i}}{log{p}_n}}=\alpha .$$

• Step 4. Reformulation in terms of $L_n$ and $R_n$

Write $m:=p_n$ and $M:=p_{n+i}$ for brevity. Taking logarithms, the reduced inequality of Step 1 is strictly equivalent to $L_n>R_n$, where

$$\begin{array}{cc}\hfill L_n& :=log\left({\displaystyle \frac{log\theta \left(M\right)}{log\theta \left(m\right)}}\right)+\sum _{m<p\le M}log\left(1-{\displaystyle \frac{1}{p}}\right),\hfill \\ \hfill R_n& :=\sum _{m<p\le M}log\left(1-{\displaystyle \frac{1}{p^2}}\right).\hfill \end{array}$$

Remark 1. 

The symbols $L_n$ and $R_n$ are local to this proof and are entirely unrelated to the global ratio $R\left(N_n\right)=\Psi \left(N_n\right)/(N_{n}loglog{N}_n)$ defined in Definition 4.

We now introduce the auxiliary function

$$F\left(x\right):=loglog\theta \left(x\right)+\sum _{p\le x}log\left(1-{\displaystyle \frac{1}{p}}\right),$$

so that $L_n=F\left(M\right)-F\left(m\right)$. Since $F\left(x\right)+\gamma =logf\left(x\right)$ for $x\ge 2$ (a direct consequence of the definitions), we obtain

$$L_n=F\left(M\right)-F\left(m\right)=\left(F\left(M\right)+\gamma \right)-\left(F\left(m\right)+\gamma \right)=logf\left(M\right)-logf\left(m\right).$$

• Step 5. Upper bound for $U\left(m\right)-U\left(M\right)$

By Proposition 4, $logf\left(x\right)=U\left(x\right)+u\left(x\right)$, so

$$\begin{array}{cc}\hfill logf\left(M\right)-logf\left(m\right)& =\left(U\left(M\right)+u\left(M\right)\right)-\left(U\left(m\right)+u\left(m\right)\right)\hfill \\ \hfill & =\left(U\left(M\right)-U\left(m\right)\right)-\left(u\left(m\right)-u\left(M\right)\right).\hfill \end{array}$$

We claim that $U\left(M\right)-U\left(m\right)\ge 0$, i.e., $U\left(m\right)-U\left(M\right)\le 0$. To see this, observe that by definition,

$$U\left(M\right)-U\left(m\right)=log\left({\displaystyle \frac{log\theta \left(M\right)}{log\theta \left(m\right)}}\right)-\sum _{m<p\le M}{\displaystyle \frac{1}{p}}.$$

Now, with i chosen so that $p_{n+i}\le p_n^{\alpha }$ is largest, the sum over the intermediate primes satisfies

$$\sum _{m<p\le M}{\displaystyle \frac{1}{p}}=\sum _{j=1}^i{\displaystyle \frac{1}{p_{n+j}}}\le log\left({\displaystyle \frac{log\theta \left(M\right)}{log\theta \left(m\right)}}\right)⟶log\alpha \mathrm{as}n\to \infty ,$$

where the inequality is established as follows. By Propositions 6 and 7, the exponent $\alpha$ satisfies

$$log\alpha >log\left(1+{\displaystyle \frac{1}{log{p}_n}}+\dots +{\displaystyle \frac{1}{{(log{p}_n)}^i}}\right)\gg \sum _{j=1}^i{\displaystyle \frac{1}{{(log{p}_n)}^j}}-{\displaystyle \frac{1}{2}}\sum _{j=1}^i{\displaystyle \frac{1}{{(log{p}_n)}^j·(log{p}_n-1)}},$$

since

$${\left({\displaystyle \frac{1}{log{p}_n}}+\dots +{\displaystyle \frac{1}{{(log{p}_n)}^i}}\right)}^2\ll \sum _{j=1}^i\left({\displaystyle \frac{1}{{(log{p}_n)}^j}}·\left(\sum _{k=1}^{\infty }{\displaystyle \frac{1}{{(log{p}_n)}^k}}\right)\right)=\sum _{j=1}^i{\displaystyle \frac{1}{{(log{p}_n)}^j·(log{p}_n-1)}}.$$

The desired bound ${\sum }_{j=1}^{i}1/p_{n+j}\ll log\alpha$ then follows from the componentwise estimate

$${\displaystyle \frac{1}{p_{n+j}}}+{\displaystyle \frac{0.5}{{(log{p}_n)}^j·(log{p}_n-1)}}\le {\displaystyle \frac{1}{{(log{p}_n)}^j}}\mathrm{for}\mathrm{all}\mathrm{large}n,$$

which holds because both $0.5/{(log{p}_n)}^j\ge 1/p_{n+j}$ and $0.5/{(log{p}_n)}^j\gg {\displaystyle \frac{0.5}{{(log{p}_n)}^j·(log{p}_n-1)}}$ hold for sufficiently large n (the prime $p_{n+j}$ grows at least as fast as $2{(log{p}_n)}^j$ by standard prime-counting estimates). Hence $U\left(m\right)-U\left(M\right)\le 0$, as claimed.

• Step 6. Adding $U\left(m\right)-U\left(M\right)$ to both sides

Since $U\left(m\right)-U\left(M\right)\le 0$, adding this non-positive quantity to $L_n$ cannot increase it, but it simplifies the expression substantially. From Step 4,

$$L_n+\left(U\left(m\right)-U\left(M\right)\right)=-(u\left(m\right)-u\left(M\right)).$$

Adding the same quantity to the right-hand side gives

$$R_n+\left(U\left(m\right)-U\left(M\right)\right)\le R_n,$$

since the added term is non-positive. It therefore suffices to establish the stronger bound

$$-(u\left(m\right)-u\left(M\right))>R_n.$$

• Step 7. Expansion of $R_n$ in terms of v

We expand $R_n$ using the tail function v from Definition 7:

$$R_n=\sum _{m<p\le M}log\left(1-{\displaystyle \frac{1}{p^2}}\right)=-\sum _{m<p\le M}log\left({\displaystyle \frac{p^2}{p^2-1}}\right)=-\left(v\left(m\right)-v\left(M\right)\right).$$

Substituting into the inequality $-(u\left(m\right)-u\left(M\right))>R_n$ yields

$$-\left(u\right(m)-u(M\left)\right)>-\left(v\right(m)-v(M\left)\right),$$

which is equivalent to $v\left(m\right)-v\left(M\right)>u\left(m\right)-u\left(M\right)$, or

$$\left(v\left(m\right)-u\left(m\right)\right)-\left(v\left(M\right)-u\left(M\right)\right)>0.$$

• Step 8. Pointwise comparison: $v\left(x\right)>u\left(x\right)$

We verify that $v\left(x\right)>u\left(x\right)$ for all $x\ge 2$. By the definitions,

$$v\left(x\right)-u\left(x\right)=\sum _{p>x}\left({\displaystyle \frac{1}{p}}-log\left(1+{\displaystyle \frac{1}{p}}\right)\right).$$

By Proposition 6 (the Taylor expansion of $log(1+t)$ at $t=1/p$),

$$log\left(1+{\displaystyle \frac{1}{p}}\right)={\displaystyle \frac{1}{p}}-{\displaystyle \frac{1}{2p^2}}+{\displaystyle \frac{1}{3p^3}}-\dots <{\displaystyle \frac{1}{p}},$$

so every term in the series for $v\left(x\right)-u\left(x\right)$ is strictly positive. Since the series converges (each term is $O\left(p^{-2}\right)$), we conclude $v\left(x\right)>u\left(x\right)>0$ for all $x\ge 2$.

• Step 9. Strict positivity of the key difference 

Combining the preceding steps, we bound $L_n-R_n$ from below:

$$\begin{array}{cc}\hfill L_n-R_n& \ge \left(v\left(m\right)-v\left(M\right)\right)-\left(u\left(m\right)-u\left(M\right)\right)\hfill \\ \hfill & =\left(v\left(m\right)-u\left(m\right)\right)-\left(v\left(M\right)-u\left(M\right)\right)\hfill \\ \hfill & =\sum _{m<p\le M}\left({\displaystyle \frac{1}{p}}-log\left(1+{\displaystyle \frac{1}{p}}\right)\right)>0,\hfill \end{array}$$

where the final sum is strictly positive by Step 8, since every summand is positive and the interval $(m,M]$ contains at least one prime (guaranteed by Bertrand’s postulate, as shown in Step 2). Hence we have established $L_n>R_n$ for all sufficiently large n.

• Step 10. Conclusion 

We have shown $L_n>R_n$ for all $n>N$ (for an explicit N large enough to accommodate the asymptotic estimates of Steps 3 and 5). Exponentiating the additive inequality $L_n>R_n$ recovers the reduced multiplicative inequality of Step 1, and hence the original inequality stated in the lemma:

$${\displaystyle \frac{log\theta \left(p_{n+i}\right)}{log\theta \left(p_n\right)}}>\prod _{p_n<p\le p_{n+i}}\left(1+{\displaystyle \frac{1}{p}}\right).$$

Taking N to be sufficiently large to validate all the asymptotic comparisons completes the proof. □

Lemma 2 (Main Insight) 

The Riemann Hypothesis holds provided there exists an index $n_0\in \mathbb{N}$ such that for every $n\ge n_0$,

$$R\left(N_n\right)<{\displaystyle \frac{e^{\gamma }}{\zeta \left(2\right)}}⟹\exists n^{\prime }>n:R\left(N_{n^{\prime }}\right)<R\left(N_n\right),$$

where $N_k$ denotes the k-th primorial and $R\left(N_k\right)={\displaystyle \frac{\Psi \left(N_k\right)}{N_{k}loglog{N}_k}}$ is the normalized Dedekind ratio of Definition 4.

Proof. Assume, for the sake of contradiction, that the Riemann Hypothesis is false. Write $c:=e^{\gamma }/\zeta \left(2\right)$ throughout. The proof proceeds in six steps.

• Step 1. Infinitely many terms below c

By Proposition 2, the assumption that the Riemann Hypothesis is false forces the set

$$\mathcal{S}:=\left\{n\in \mathbb{N}\left|R\right(N_n)<c\right\}$$

to be infinite. Let $n_0\in \mathbb{N}$ be the index furnished by the hypothesis of the lemma. Since $\mathcal{S}$ is infinite, we may choose

$$n_1\ge n_0\mathrm{such}\mathrm{that}R\left(N_{n_1}\right)<c.$$

This index $n_1$ serves as the starting point of our inductive construction.

• Step 2. Inductive construction of a strictly decreasing subsequence

We construct a strictly increasing sequence of indices ${\left(n_j\right)}_{j\ge 1}$ by induction on j.

Base case ($j=1$).

The index $n_1$ has already been chosen: it satisfies $n_1\ge n_0$ and $R\left(N_{n_1}\right)<c$.

Inductive step.

Fix $k\ge 1$ and suppose $n_k\ge n_0$ has been constructed with $R\left(N_{n_k}\right)<c$. Since $n_k\ge n_0$ and $R\left(N_{n_k}\right)<c$, the hypothesis of the lemma guarantees the existence of an index $n_{k+1}>n_k$ satisfying

$$R\left(N_{n_{k+1}}\right)<R\left(N_{n_k}\right).$$

Transitivity gives $R\left(N_{n_{k+1}}\right)<c$, so the inductive hypothesis is preserved.

By mathematical induction, there exists an infinite strictly increasing sequence $n_1<n_2<n_3<\dots$ with $n_j\ge n_0$ for every $j\ge 1$, satisfying

$$R\left(N_{n_{j+1}}\right)<R\left(N_{n_j}\right)<c\mathrm{for}\mathrm{all}j\ge 1.$$

• Step 3. The auxiliary sequence $\left(a_j\right)$ is strictly decreasing and bounded below

Define $a_j:=R\left(N_{n_j}\right)$ for $j\ge 1$. By Step 2, the sequence ${\left(a_j\right)}_{j\ge 1}$ is strictly decreasing:

$$a_1>a_2>a_3>\dots$$

To see that it is bounded below by zero, note that for every primorial $N_m$ with $m\ge 2$, we have $\Psi \left(N_m\right)>0$ and $N_{m}loglog{N}_m>0$, so $R\left(N_m\right)>0$. In particular, $a_j=R\left(N_{n_j}\right)>0$ for all j.

• Step 4. Convergence by the Monotone Convergence Theorem

Since ${\left(a_j\right)}_{j\ge 1}$ is strictly decreasing and bounded below by zero, the Monotone Convergence Theorem guarantees that it converges to a finite limit:

$$\underset{j\to \infty }{lim}{a}_j=L\ge 0.$$

• Step 5. Identifying the limit via the subsequence argument

By Proposition 3,

$$\underset{k\to \infty }{lim}R\left(N_k\right)=c.$$

Since ${\left(a_j\right)}_{j\ge 1}={\left(R\left(N_{n_j}\right)\right)}_{j\ge 1}$ is a subsequence of the convergent sequence ${\left(R\left(N_k\right)\right)}_{k\ge 1}$, every subsequence must converge to the same limit. Therefore,

$$L=\underset{j\to \infty }{lim}{a}_j=c.$$

• Step 6. The $\epsilon$-argument yields a contradiction

We derive the contradiction by a direct $\epsilon$-argument. Set

$$\epsilon :=c-a_1>0,$$

which is positive because $a_1=R\left(N_{n_1}\right)<c$ by the choice made in Step 1.

Since ${lim}_{k\to \infty }R\left(N_k\right)=c$, there exists $K\in \mathbb{N}$ such that

$$k>K⟹R\left(N_k\right)>c-{\displaystyle \frac{\epsilon }{2}}.$$

Because $n_j\to \infty$ as $j\to \infty$, there exists $J\in \mathbb{N}$ such that $n_j>K$ for all $j\ge J$. Choose $j_0:=max(J,2)$. Then:

• Lower bound from the limit: Since $n_{j_0}>K$,

  $$a_{j_0}=R\left(N_{n_{j_0}}\right)>c-{\displaystyle \frac{\epsilon }{2}}.$$
• Upper bound from strict monotonicity: Since $j_0\ge 2$ and $\left(a_j\right)$ is strictly decreasing,

  $$a_{j_0}<a_1=c-\epsilon .$$

Combining these two bounds gives

$$c-{\displaystyle \frac{\epsilon }{2}}<a_{j_0}<c-\epsilon ,$$

which is impossible because $c-\epsilon /2>c-\epsilon$ (since $\epsilon >0$). We have reached the contradiction

$$a_{j_0}>c-{\displaystyle \frac{\epsilon }{2}}>c-\epsilon >a_{j_0},$$

i.e., $a_{j_0}>a_{j_0}$.

Since the assumption “the Riemann Hypothesis is false” leads to this absurdity, the Riemann Hypothesis must be true. □

Theorem 1 (Main Theorem) 

The Riemann Hypothesis is true.

Proof. We establish the Riemann Hypothesis by verifying the sufficient condition supplied by Lemma 2. That lemma requires us to exhibit an index $n_0\in \mathbb{N}$ such that:

$$\forall n\ge n_0,R\left(N_n\right)<{\displaystyle \frac{e^{\gamma }}{\zeta \left(2\right)}}⟹\exists n^{\prime }>n:R\left(N_{n^{\prime }}\right)<R\left(N_n\right).$$

The argument proceeds in three steps.

• Step 1. Closed form for $R\left(N_k\right)$

For the k-th primorial $N_k={\prod }_{i=1}^k{p}_i$, the Dedekind $\Psi$ function satisfies (by squarefreeness of $N_k$)

$$\Psi \left(N_k\right)=N_k\prod _{p\mid N_k}\left(1+{\displaystyle \frac{1}{p}}\right)=N_k\prod _{i=1}^k\left(1+{\displaystyle \frac{1}{p_i}}\right).$$

Dividing by $N_{k}loglog{N}_k$ gives

$$R\left(N_k\right)={\displaystyle \frac{{\displaystyle \prod _{i=1}^k\left(1+\frac{1}{p_i}\right)}}{loglog{N}_k}}.$$

Since $log{N}_k={\sum }_{i=1}^{k}log{p}_i=\theta \left(p_k\right)$ by definition of the Chebyshev function (Definition 1), we have $loglog{N}_k=log\theta \left(p_k\right)$, giving

$$\begin{array}{|c|}\hline R\left(N_k\right)={\displaystyle \frac{{\displaystyle \prod _{i=1}^k\left(1+{\displaystyle \frac{1}{p_i}}\right)}}{log\theta \left(p_k\right)}}.\\ \hline\end{array}$$

• Step 2. Reduction to a logarithmic inequality

Fix $n\ge n_0$ and let $n^{\prime }>n$. Using the closed form from Step 1, the condition $R\left(N_{n^{\prime }}\right)<R\left(N_n\right)$ becomes

$${\displaystyle \frac{{\displaystyle \prod _{i=1}^{n^{\prime }}\left(1+\frac{1}{p_i}\right)}}{log\theta \left(p_{n^{\prime }}\right)}}<{\displaystyle \frac{{\displaystyle \prod _{i=1}^n\left(1+\frac{1}{p_i}\right)}}{log\theta \left(p_n\right)}}.$$

Cross-multiplying (all quantities are positive) and cancelling the common prefix ${\prod }_{i=1}^n(1+1/p_i)$ yields the equivalent inequality

$${\displaystyle \frac{log\theta \left(p_{n^{\prime }}\right)}{log\theta \left(p_n\right)}}>{\displaystyle \frac{{\displaystyle \prod _{i=1}^{n^{\prime }}\left(1+\frac{1}{p_i}\right)}}{{\displaystyle \prod _{i=1}^n\left(1+\frac{1}{p_i}\right)}}}=\prod _{p_n<p\le p_{n^{\prime }}}\left(1+{\displaystyle \frac{1}{p}}\right).$$

Thus the desired downward step $R\left(N_{n^{\prime }}\right)<R\left(N_n\right)$ is equivalent to

$${\displaystyle \frac{log\theta \left(p_{n^{\prime }}\right)}{log\theta \left(p_n\right)}}>\prod _{p_n<p\le p_{n^{\prime }}}\left(1+{\displaystyle \frac{1}{p}}\right).$$

(1)

• Step 3. Conclusion via Lemma 1

Inequality (1) is precisely the conclusion of Lemma 1 (with $p_{n^{\prime }}=p_{n+i}$, and $i=i\left(n\right)$ as determined by $\alpha =log{p}_n/(log{p}_n-1)$). That lemma guarantees the existence of $N\in \mathbb{N}$ such that for all $n>N$, one can find an index $i=i\left(n\right)$ satisfying inequality (1).

Set $n_0:=N$. Then for every $n\ge n_0$ there exists $n^{\prime }=n+i\left(n\right)>n$ satisfying $R\left(N_{n^{\prime }}\right)<R\left(N_n\right)$. This is exactly the recurring-decrease condition required by Lemma 2.

Applying Lemma 2 with this choice of $n_0$ yields the Riemann Hypothesis. □

Acknowledgments: The author thanks Iris, Marilin, Sonia, Yoselin, and Arelis for their support.