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An Elementary Problem Equivalent to the Riemann Hypothesis

This paper shows the equivalence of the Riemann hypothesis to an sequence of elementary inequalities involving the harmonic numbers H_n, the sum of the reciprocals of the integers from 1 to n. It is a modification of a criterion due to Guy Robin.


Introduction
We consider the following problem. with equality only for n = 1.
The function σ(n) = d|n d is the sum of divisors function, so for example σ(6) = 12. The number H n is called the n-th harmonic number by Knuth, Graham and Patashnik [9,Sect. 6.3], who detail various properties of harmonic numbers.
The 'E' in Problem E might stand for either 'easy' or 'elementary'. Perhaps 'H' for 'hard' would be a better letter to use, since our object is to show the following equivalence. Theorem 1.1 Problem E is equivalent to the Riemann hypothesis.
The Riemann hypothesis, stated by Riemann [17] in 1859, concerns the complex zeros of the Riemann zeta function. The Riemann zeta function ζ(s) is defined by the Dirichlet series which converges for ℜ(s) > 1, and it has an analytic continuation to the complex plane with one singularity, a simple pole with residue 1 at s = 1. The Riemann hypothesis states that the nonreal zeros of the Riemann zeta function ζ(s) all lie on the line ℜ(s) = 1 2 . One reason for the great interest in the Riemann hypothesis, regarded by many as the most important unsolved problem in pure mathematics, is its connection with the distribution of prime numbers, described below. More significantly, the Riemann hypothesis is a special case of questions concerning generalizations of the zeta function (L-functions) and their connections with problems in number theory, algebraic geometry, topology, representation theory and perhaps even physics, see Berry and Keating [3], Katz and Sarnak [8] and Murty [11].
The connection of the Riemann hypothesis with prime numbers was the original question studied by Riemann [17]. Let π(x) count the number of primes p with 1 < p ≤ x. C. F. Gauss noted empirically that π(x) is well approximated by the logarithmic integral which itself satisfies The Riemann hypothesis is equivalent to the assertion that for each ǫ > 0 there is a positive constant C ǫ such that for all x ≥ 2, see Edwards [5, p. 90]. The force of the Riemann hypothesis lies in the small size of the error term. The strongest form known of the Prime Number Theorem with error term asserts that for any positive ǫ, for certain positive constants C 1 and C 2 depending on ǫ; this result is due to Vinogradov and Korobov in 1958. Problem E encodes a modification of a criterion of Guy Robin [18] for the Riemann hypothesis. Robin's criterion states that the Riemann hypothesis is true if and only if σ(n) < e γ n log log n for all n ≥ 5041, (1.2) where γ ≈ 0.57721 is Euler's constant. This criterion is related to the density of primes, as explained in §2. Our aim was to obtain a problem statement as elementary as possible, containing no undefined constants. However the hard work underlying the equivalence resides in the results of Robin stated in §3, where we give a proof of Theorem 1.1. Before coming to the proof, in the next section we describe how the Riemann hypothesis is related to the sum of divisors function. The connection traces back to Ramanujan's work on highly composite numbers, and involves several results of Erdős with coauthors.
so that n attains the maximum value of σ(k) k 1+ǫ over all k. The set of colossally abundant numbers is infinite. They form a subset of the superabundant numbers, a fact which can be deduced from the definition. Table 1 gives the colossally abundant numbers up to 10 18 , as given in [1]. Robin showed that if the Riemann hypothesis is false, there will necessarily exist a counterexample to the inequality (1.2) which is a colossally abundant number, cf. [18, Proposition 1 of Section 3]; the same property can be established for counterexamples to (1.1). (There could potentially exist other counterexamples as well.) Superabundant and colossally abundant numbers were studied 1 in detail by Alaoglu and Erdős [1] in 1944. As evidenced in the table, colossally abundant numbers consist of a product of all the small primes up to some bound, with exponents which are nonincreasing as the prime increases. The values of these exponents have a characteristic smooth shape which can be almost completely described (see (2.3) below). In fact these classes of numbers had been studied earlier, by Ramanujan, in his 1915 work on highly composite numbers [14]. The notes in Ramanjuan's Collected Papers report: "The London Mathematical Society was in some financial difficulty at the time, and Ramanujan suppressed part of what he had written in order to save expense." [14, p. 339]. Only the first 52 of 75 sections were printed. The manuscript of the unpublished part was eventually rediscovered among the papers of G. N. Watson, in "Ramanujan's Lost Notebook" (see Andrews [2] 1 Alaoglu and Erdős [1] use a slightly stronger definition of colossally abundant number; they impose the additional requirement that σ(n) n 1+ǫ > σ(k) k 1+ǫ must hold for 1 ≤ k < n. With their definition the colossally abundant numbers are exactly those given by (2.3) below. and Ramanujan [15, pp. 280-308]) and finally published in 1997, in [16]. In it superabundant and colossally abundant numbers are considered in Section 59, as special cases of the concepts of generalised highly composite numbers and superior generalised highly composite numbers, for the parameter value s = 1. Ramanujan derived upper and lower bounds for the maximal order of generalised highly composite numbers in Section 71, assuming the Riemann hypothesis. (The Riemann hypothesis was assumed from Section 40 onward in his paper.) His bounds imply, assuming the Riemann hypothesis, that (1.2) holds for all sufficiently large n.
The results of Alaoglu and Erdős in their 1944 paper are unconditional, and mainly concerned the exponents of primes occurring in highly composite and superabundant numbers. In considering the exponents of primes appearing in colossally abundant numbers, they raised the following question, which is still unsolved.
Conjecture. (Alaoglu and Erdős) If p and q are both primes, is it true that p x and q x are both rational only if x is an integer?
This conjecture would follow as a consequence 2 of the four exponentials conjecture in transcendental number theory, which asserts that if a 1 , a 2 form a pair of complex numbers, linearly independent over the rationals Q, and b 1 , b 2 is another such pair, also linearly independent over the rationals, then at least one of the four exponentials {e a i b j : i, j = 1 or 2} is transcendental, see Lang [10, pp. 8-11]. Alaoglu and Erdős [1,Theorem 10] showed that for "generic" values of ǫ there is exactly one maximizing integer n , and the exponent a p (ǫ) of each prime p in it is furthermore for all ǫ > 0 this value of n is maximizing. However for a discrete set of ǫ there will be more than one maximizing integer. Erdős and Nicolas [6, p. 70] later showed that for a given value of ǫ there will be exactly one, two or four integers n that attain the maximum value of σ(k) k 1+ǫ . If the conjecture of Alaoglu and Erdős above has a positive answer, then no values of ǫ will have four extremal integers. This would imply that the ratio of two consecutive colossally abundant numbers, the larger divided by the smaller, will always be a prime, and that every colossally abundant number has a factorization (2.3) for some nonempty open interval of "generic" values of ǫ. Now we can explain the relevance of the Riemann hypothesis to the extremal size of σ(n). Colossally abundant numbers are products of all the small primes raised to powers that are a smoothly decreasing function of their size. Fluctuations in the distribution of primes will be reflected in fluctuations in the growth rate of σ(n) n taken over the set of colossally abundant numbers. Recall that the Riemann hypothesis is equivalent to the assertion that for each ǫ > 0, holds for all sufficiently large x. It is also known that if the Riemann hypothesis is false, then there will exist a specific positive constant δ such that the one-sided inequality is true for an infinite set of values with x → ∞. Heuristically, if we choose a value of x where such an excess of primes over Li(x) occurs, and then take a product of all primes up to this bound, choosing appropriate exponents, one may hope to construct numbers n with σ(n) exceeding e γ n log log n by a small amount. If the Riemann hypothesis holds, there is a smaller upper bound for excess of the number of primes above Li(x), and from this one can deduce a slighly better upper bound for σ(n). The analysis of Robin [18] gives a quantitative version of this, as formulated in Propositions 3.1 and 3.2 below.
One can prove unconditionally that the inequality (1.1) holds for nearly all integers. Even if the Riemann hypothesis is false, the set of exceptions to the inequality (1.1) will form a very sparse set. Furthermore, if there exists any counterexample to (1.1), the value of n will be very large.

Proofs
The proof of Theorem 1.1 is based on the following two results of Robin [18]. where γ is Euler's constant.
Proof. This is Theorem 1 of Robin [18]. holds for infinitely many n.
Proof. This appears in Proposition 1 of Section 4 of Robin [18]. The constant β can be chosen to take any value 1 − b < β < 1 2 , where b = ℜ(ρ) for some zero ρ of ζ(s) with ℜ(ρ) > 1 2 , and C > 0 must be chosen sufficiently small, depending on ρ. The proof uses ideas from a result of Nicolas [12], [13,Proposition 3], which itself uses a method of Landau.
Proof of Theorem 1.1 ⇐ Suppose the Riemann hypothesis is true. Then Proposition 3.1 and Lemma 3.1 together give, for n ≥ 5041, d|n d ≤ e γ n log log n < H n + exp(H n ) log H n .
For 1 ≤ n ≤ 5040 one verifies (1.1) directly by computer, the only case of equality being n = 1.
⇒ Suppose (1.1) holds for all n. We argue by contradiction, and suppose that the Riemann hypothesis is false. Then Proposition 3.2 applies. However its lower bound, valid for infinitely many n, contradicts the upper bound of Lemma 3.2 for sufficiently large n. We conclude that the Riemann hypothesis must be true.