From Chebyshev to Primorials: Establishing the Riemann Hypothesis

1. Introduction

The Riemann Hypothesis, proposed by Bernhard Riemann in 1859, asserts that every non-trivial zero of the Riemann zeta function $\zeta \left(s\right)$ lies on the critical line $\Re \left(s\right)=\frac{1}{2}$. This conjecture is widely regarded as the most important 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 $\zeta \left(s\right)$, initially defined for $\Re \left(s\right)>1$ and analytically continued to the entire complex plane, has trivial zeros at the negative even integers $s=-2,-4,-6,\dots$ and non-trivial zeros confined to the critical strip $0<\Re \left(s\right)<1$.

Beyond its intrinsic theoretical interest, the Riemann Hypothesis has profound consequences for the distribution of prime numbers, a cornerstone of analytic number theory. Numerous equivalent formulations appear in the literature [1]. The one relevant to this work is due to Nicolas [2,3]: the Riemann Hypothesis holds if and only if

$${\displaystyle \frac{N_k}{\phi \left(N_k\right)}}>e^{\gamma }loglog{N}_k\mathrm{for}\mathrm{all}k\ge 1,$$

where $\gamma$ is the Euler–Mascheroni constant, $\phi$ is Euler’s totient function, and $N_k={\prod }_{i=1}^k{p}_i$ is the k-th primorial.

1.1. Overview of the Approach

Our argument is a proof by contradiction structured around the Nicolas ratio

$$R\left(n\right)={\displaystyle \frac{N_n}{\phi \left(N_n\right)loglog{N}_n}},n\ge 1,$$

whose asymptotic behavior is governed by Mertens’ third theorem: $R\left(n\right)\to e^{\gamma }$ as $n\to \infty$ (Proposition 2). By the Nicolas criterion, the Riemann Hypothesis is equivalent to $R\left(n\right)>e^{\gamma }$ failing for no $n\ge 1$.

Assume the Riemann Hypothesis is false. Nicolas’ oscillation result (Proposition 3) then forces $R\left(n\right)<e^{\gamma }$ for infinitely many indices, and an auxiliary function $f\left(p_n\right)=e^{\gamma }/R\left(n\right)$ satisfies $f\left(p_n\right)>1$ for those same indices (Proposition 3).

The core of the proof (Section 3) is the explicit construction of a strictly increasing sequence of indices $n_1<n_2<\dots$ along which:

1.

$R\left(n_j\right)<e^{\gamma }$ for every j (each term lies below the asymptotic value), and

2.

$R\left(n_1\right)>R\left(n_2\right)>\dots$ (the subsequence is strictly decreasing).

A strictly decreasing sequence bounded below converges, by the Monotone Convergence Theorem, to some finite limit $L\ge 0$. However, since $\left(R\left(n_j\right)\right)$ is a subsequence of the convergent sequence $\left(R\right(n\left)\right)$, it must share the same limit: $L=e^{\gamma }$. On the other hand, strict monotonicity forces every term to satisfy $R\left(n_j\right)<R\left(n_1\right)<e^{\gamma }$, so $L\le R\left(n_1\right)<e^{\gamma }$. This is the desired contradiction, proving that the Riemann Hypothesis is true.

2. Materials and Methods

2.1. Definitions

We collect the objects used throughout the paper.

Definition 1

(Chebyshev’s Prime-Counting Function). The Chebyshev 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 primes $p\le x$. Equivalently, $\theta \left(x\right)=log{\prod }_{p\le x}p$. The prime number theorem is equivalent to $\theta \left(x\right)\sim x$ as $x\to \infty$.

Definition 2

(Primorial Numbers). Thek-th primorial $N_k$ is the product of the first k primes:

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

where $p_i$ denotes the i-th prime ($p_1=2,p_2=3,p_3=5,\dots$).

Definition 3

(Nicolas Ratio). For an integer $n\ge 3$, the Nicolas ratio is

$$R\left(n\right)={\displaystyle \frac{N_n}{\phi \left(N_n\right)loglog{N}_n}}.$$

By the Nicolas criterion [2,3], the Riemann Hypothesis holds if and only if $R\left(n\right)>e^{\gamma }$ for all $n\ge 1$.

Definition 4

(Auxiliary Analytic Function). The following real-valued function of $x\ge 2$ appears throughout the proof:

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].

2.2. Key Propositions

We record the known results used in the main proof.

Proposition 1

(Erdos–Szekeres for Infinite Sequences [4]). Let ${\left(a_n\right)}_{n\in \mathbb{N}}$ be an infinite sequence of real numbers. Say the sequence has a strictly decreasing tail if there exists $N\in \mathbb{N}$ such that $a_{n+1}<a_n$ for all $n\ge N$. If $\left(a_n\right)$ has no such tail, then it contains a strictly increasing infinite subsequence ${\left(a_{n_k}\right)}_{k\in \mathbb{N}}$ satisfying $a_{n_1}<a_{n_2}<a_{n_3}<\dots$.

Proposition 2

(Asymptotic limit of $R\left(n\right)$ [5]). As $n\to \infty$, $R\left(n\right)\to e^{\gamma }$.

Proof. 

By Mertens’ third theorem [5],

$$\prod _{p\le x}{\left(1-{\displaystyle \frac{1}{p}}\right)}^{-1}\sim e^{\gamma }logx(x\to \infty ).$$

Setting $x=p_n$ and using ${\displaystyle {\displaystyle \frac{N_n}{\phi \left(N_n\right)}}=\prod _{p\le p_n}{\left(1-1/p\right)}^{-1}}$ gives

$${\displaystyle \frac{N_n}{\phi \left(N_n\right)}}\sim e^{\gamma }log{p}_n,\mathrm{hence}{\displaystyle \frac{N_n}{\phi \left(N_n\right)log{p}_n}}⟶e^{\gamma }.$$

Since the prime number theorem gives $\theta \left(x\right)\sim x$ [6], taking $x=p_n$ yields $p_n\sim \theta \left(p_n\right)=log{N}_n$, and the result follows. □

Proposition 3

(Nicolas’ Theorem [2,3]). Suppose the Riemann Hypothesis is false. Define

$${\mathcal{S}}_{+}=\{n\in \mathbb{N}\mid f\left(p_n\right)>1,R\left(n\right)<e^{\gamma }\},{\mathcal{S}}_{-}=\{n\in \mathbb{N}\mid f\left(p_n\right)<1,R\left(n\right)>e^{\gamma }\}.$$

Both ${\mathcal{S}}_{+}$ and ${\mathcal{S}}_{-}$ are infinite. Moreover, ${\mathcal{S}}_{+}$ splits into two infinite subsets [7]:

•  ${\mathcal{S}}_{+\mathrm{low}}=\{n\in {\mathcal{S}}_{+}\mid \theta \left(p_n\right)<p_n\}$,
•  ${\mathcal{S}}_{+\mathrm{high}}=\{n\in {\mathcal{S}}_{+}\mid \theta \left(p_n\right)>p_n\}$.

3. Main Result

Lemma 1

(Subsequence Extraction). Assume the Riemann Hypothesis is false, and let ${\left({\epsilon }_{n_j}\right)}_{j=1}^{\infty }$ be the subsequence of all indices for which ${\epsilon }_{n_j}=logf\left(p_{n_j}\right)>0$. This sequence has no strictly decreasing tail; hence, by Proposition 1, it contains a strictly increasing subsequence ${\left({\epsilon }_{n_k}\right)}_{k=1}^{\infty }$ satisfying

$${\epsilon }_{n_{k+1}}>{\epsilon }_{n_k}>0\mathit{for}\mathit{all}k\ge 1.$$

Proof. 

Assume the Riemann Hypothesis is false, and let ${\left({\epsilon }_{n_j}\right)}_{j=1}^{\infty }$ be the subsequence of all positive values ${\epsilon }_{n_j}=logf\left(p_{n_j}\right)>0$.

Suppose, for contradiction, that this sequence has a strictly decreasing tail, i.e., there exists $J\ge 1$ such that

$${\epsilon }_{n_j}>{\epsilon }_{n_{j+1}}\mathrm{for}\mathrm{all}j\ge J.$$

We show this is impossible by exhibiting infinitely many indices at which consecutive terms of the positive subsequence increase.

Step 1: Increment formula. For any prime $p_n$, the one-step difference is

$${\Delta }_n=logf\left(p_n\right)-logf\left(p_{n-1}\right)=log\left({\displaystyle \frac{log\theta \left(p_n\right)}{log\theta \left(p_{n-1}\right)}}\left(1-{\displaystyle \frac{1}{p_n}}\right)\right).$$

Since $\theta \left(p_n\right)=\theta \left(p_{n-1}\right)+log{p}_n$, we have

$$log\theta \left(p_n\right)=log\theta \left(p_{n-1}\right)+log\left(1+{\displaystyle \frac{log{p}_n}{\theta \left(p_{n-1}\right)}}\right),$$

so

$${\Delta }_n=log\left(1+{\displaystyle \frac{log\left(1+{\displaystyle \frac{log{p}_n}{\theta \left(p_{n-1}\right)}}\right)}{log\theta \left(p_{n-1}\right)}}\right)-log\left({\displaystyle \frac{p_n}{p_n-1}}\right).$$

Thus ${\Delta }_n\le 0$ if and only if

$${\displaystyle \frac{log\left(1+{\displaystyle \frac{log{p}_n}{\theta \left(p_{n-1}\right)}}\right)}{log\theta \left(p_{n-1}\right)}}\le {\displaystyle \frac{1}{p_n-1}}.$$

(1)

Step 2: Infinitely many local increases. By Proposition 3, the set ${\mathcal{S}}_{+\mathrm{low}}$ is infinite under our assumption. Pick any $(n-1)\in {\mathcal{S}}_{+\mathrm{low}}$, so that $\theta \left(p_{n-1}\right)<p_{n-1}$. Since log is strictly increasing, the left-hand side of (1) satisfies

$${\displaystyle \frac{log\left(1+{\displaystyle \frac{log{p}_n}{\theta \left(p_{n-1}\right)}}\right)}{log\theta \left(p_{n-1}\right)}}>{\displaystyle \frac{log\left(1+{\displaystyle \frac{log{p}_n}{p_{n-1}}}\right)}{log{p}_{n-1}}}.$$

Using $log(1+t)\sim t$ as $t\to 0^{+}$, the right-hand side behaves as ${\displaystyle \frac{log{p}_n}{p_{n-1}log{p}_{n-1}}}$, whereas the right-hand side of (1) behaves as ${\displaystyle \frac{1}{p_n-1}}\sim {\displaystyle \frac{1}{p_n}}$. Since the average prime gap satisfies $p_n-p_{n-1}\sim log{p}_n$, we have $p_n\sim p_{n-1}$, and therefore

$${\displaystyle \frac{log{p}_n}{p_{n-1}log{p}_{n-1}}}\gg {\displaystyle \frac{1}{p_n-1}}\mathrm{as}{p}_n\to \infty .$$

Hence inequality (1) fails for all sufficiently large $(n-1)\in {\mathcal{S}}_{+\mathrm{low}}$, giving ${\Delta }_n>0$, and in fact

$$0<{\epsilon }_{n-1}<{\epsilon }_n.$$

Both values are positive, so they appear as adjacent terms in the filtered subsequence $\left({\epsilon }_{n_j}\right)$, yielding a local increase at infinitely many indices. This contradicts the assumption that the sequence has a strictly decreasing tail.

Since ${\left({\epsilon }_{n_j}\right)}_{j=1}^{\infty }$ has no strictly decreasing tail, Proposition 1 (Erdos–Szekeres) provides a strictly increasing subsequence ${\left({\epsilon }_{n_k}\right)}_{k=1}^{\infty }$ with

$${\epsilon }_{n_{k+1}}>{\epsilon }_{n_k}>0\mathrm{for}\mathrm{all}k\ge 1.$$

This completes the proof. □

Theorem 1

(Main Theorem). The Riemann Hypothesis is true.

Proof. 

Assume, for the sake of contradiction, that the Riemann Hypothesis is false. Set $c:=e^{\gamma }$ for brevity throughout. The proof proceeds in six steps.

Step 1. Infinitely many terms of $R\left(n\right)$ lie below c

By Proposition 3, the set

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

is infinite under our assumption. Let $n_0\in \mathbb{N}$ be a sufficiently large index. Because ${\mathcal{S}}_{+}$ is infinite, we may choose an index $n_1\ge n_0$ in ${\mathcal{S}}_{+}$ such that

$$R\left(n_1\right)<c.$$

This index $n_1$ serves as the base of the inductive construction below.

Step 2. Inductive construction of a strictly decreasing subsequence

We construct a strictly increasing sequence of indices ${\left(n_j\right)}_{j\ge 1}$ such that the corresponding values $R\left(n_j\right)$ are strictly decreasing and all lie below c.

• Base case ($k=1$). Index $n_1$ is chosen so that $n_1\ge n_0$ and $R\left(n_1\right)<e^{\gamma }$.
• Inductive step. Suppose indices $n_1<n_2<\dots <n_k$ with $n_k\ge n_0$ have been constructed so that each satisfies:

  (i)

  $R\left(n_k\right)<e^{\gamma }$, and

  (ii)

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

  We produce an index $n_{k+1}>n_k$ satisfying the same conditions.
  Reduction to a single inequality. Set ${\alpha }_k:=f\left(p_{n_k}\right)$ and ${\alpha }_{k+1}:=f\left(p_{n_{k+1}}\right)$. Since $f\left(p_n\right)=e^{\gamma }/R\left(n\right)$ (Definition 4), the condition $R\left(n_{k+1}\right)<R\left(n_k\right)$ is equivalent to

  $${\alpha }_{k+1}>{\alpha }_k.$$

  (2)
  It therefore suffices to establish (2).
  Expressing ${\alpha }_k$ via the Mertens error. From Definition 4 and $f\left(p_{n_k}\right)>1$,

  $$1<{\alpha }_k=e^{\gamma }log\theta \left(p_{n_k}\right)·\prod _{p\le p_{n_k}}\left(1-{\displaystyle \frac{1}{p}}\right).$$
  Taking logarithms and using $log{(1-1/p)}^{-1}=log(1+1/(p-1))$ gives

  $$log{\alpha }_k=\gamma +loglog\theta \left(p_{n_k}\right)-\sum _{p\le p_{n_k}}log\left(1+{\displaystyle \frac{1}{p-1}}\right).$$
  Since

  $$f\left(p_{n_k}\right)>1\Rightarrow logf\left(p_{n_k}\right)>0,$$

  the quantity

  $${\epsilon }_{n_k}:=\gamma +loglog\theta \left(p_{n_k}\right)-\sum _{p\le p_{n_k}}log\left(1+{\displaystyle \frac{1}{p-1}}\right)$$

  is strictly positive, and ${\alpha }_k=e^{{\epsilon }_{n_k}}$. The identical argument at level $n_{k+1}$ gives ${\alpha }_{k+1}=e^{{\epsilon }_{n_{k+1}}}$ with ${\epsilon }_{n_{k+1}}>0$.
  Establishing ${\epsilon }_{n_{k+1}}>{\epsilon }_{n_k}$. Since the sequence $\left({\epsilon }_{n_j}\right)$ restricted to positive values has no strictly decreasing tail (Lemma 1), Proposition 1 provides a strictly increasing subsequence $\left({\epsilon }_{n_k}\right)$ with ${\epsilon }_{n_{k+1}}>{\epsilon }_{n_k}>0$ for all $k\ge 1$.
  Conclusion. Because the exponential is strictly increasing and ${\alpha }_m=e^{{\epsilon }_{n_m}}$, the inequality ${\epsilon }_{n_{k+1}}>{\epsilon }_{n_k}$ gives

  $${\alpha }_{k+1}=e^{{\epsilon }_{n_{k+1}}}>e^{{\epsilon }_{n_k}}={\alpha }_k,$$

  which is (2), completing the induction.

By 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 conditions a and b at every step, and such that

$$\dots <R\left(n_{j+1}\right)<R\left(n_j\right)<\dots <R\left(n_1\right)<c\mathrm{for}\mathrm{all}j\ge 1.$$

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

Define $a_j:=R\left(n_j\right)$ for $j\ge 1$. By Step 2:

• Strictly decreasing:  $a_1>a_2>a_3>\dots$
• Bounded below by zero: For every primorial index $m\ge 2$, both $\phi \left(N_m\right)>0$ and $loglog{N}_m>0$, so $R\left(n_m\right)=N_m/(\phi \left(N_m\right)loglog{N}_m)>0$; hence $a_j>0$ for all j.

Step 4. Convergence via the Monotone Convergence Theorem

The sequence ${\left(a_j\right)}_{j\ge 1}$ is strictly decreasing and bounded below by zero, so the Monotone Convergence Theorem applies:

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

exists and is finite.

Step 5. Identifying the limit via the subsequence argument

By Proposition 2, the full sequence satisfies $R\left(n\right)\to c$ as $n\to \infty$. Since $\left(a_j\right)=\left(R\left(n_j\right)\right)$ is a subsequence of the convergent sequence $\left(R\right(n\left)\right)$, it must converge to the same limit:

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

Step 6. An $\epsilon$-argument yields a contradiction

We now make the contradiction explicit. Since $a_1=R\left(n_1\right)<c$ by Step 1, the quantity

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

is well-defined and positive.

Because $R\left(n\right)\to c$, there exists $K\in \mathbb{N}$ such that

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

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

• Lower bound (from the limit): $n_{j_0}>K$, so

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

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

Combining the two bounds gives

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

which is impossible for any $\epsilon >0$, since $c-\epsilon <c-\epsilon /2$. More explicitly, the chain of inequalities

$$a_{j_0}>c-\frac{\epsilon }{2}>c-\epsilon =a_1>a_{j_0}$$

yields $a_{j_0}>a_{j_0}$, a clear absurdity.

Since the assumption that the Riemann Hypothesis is false leads to this contradiction, the Riemann Hypothesis must be true. □