INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 7 (2007), #A31

ON THE IRRATIONALITY OF A DIVISOR FUNCTION SERIES

John B. Friedlander^{1}

Department of Mathematics, University of Toronto, Toronto, Ontario M5S 3G3, Canada

                          frdlndr@math.toronto.edu

Florian Luca^{2}

Instituto de Matemáticas, Universidad Nacional Autónoma de México, C.P. 58089, Morelia, Michoacán,

México

                           fluca@matmor.unam.mx

Mihai Stoiciu

Department of Mathematics and Statistics, Williams College, Williamstown, MA 01267, USA

                           mstoiciu@williams.edu

Received: 12/8/06, Revised: 3/20/07, Accepted: 6/12/07, Published: 7/3/07

Abstract

Here, we show, unconditionally for k = 3, and on the prime k-tuples conjecture for k ≥ 4, that ∑_{n=1}^{∞} \frac{σ_{k}(n)}{n!} is irrational, where σ_{k}(n) denotes the sum of the kth powers of the divisors of n.

1. Introduction

For a positive integer n put

σ_{k}(n) = ∑_{d | n} d^{k}

for the sum of the kth powers of the (positive) divisors of n. In [4], Erdős and Kac showed that the series

∑_{n=1}^{∞} \frac{σ_{k}(n)}{n!}

is irrational for both k = 1 and k = 2. The problem is mentioned also in [5], wherein it is stated that the method does not seem to extend to k ≥ 3, and it appears as B14 in [6]. Let us quickly give proofs of the irrationality of these series when k = 0, 1, 2. Assume that the given series is A/B, where A and B are positive integers. Multiplying by (n − 1)!, where n > B, we get that

1Research of J. F.was supported in part by NSERC grant A5123.

2Research of F. L. was supported in part by grants SEP-CONACYT 46755 and a Guggenheim Fellowship.

∑_{j=1}^{n−1} σ_{k}(j)\frac{(n − 1)!}{j!} + \frac{σ_{k}(n)}{n} + \frac{σ_{k}(n + 1)}{n(n + 1)} + ∑_{j≥2} \frac{σ_{k}(n + j)}{n(n + 1) ⋯ (n + j)} = \frac{A(n − 1)!}{B}.

The first of the above sums is an integer and the right hand side is also an integer. The second sum is positive and, for k ≤ 2, its size is

≤ \frac{σ_{2}(n + 2)}{n(n + 1)(n + 2)} + ∑_{j≥3} \frac{σ_{2}(n + j)}{n(n + 1) ⋯ (n + j)} ≪ \frac{1}{n} + ∑_{m≥n} \frac{1}{m^{2}} ≪ \frac{1}{n},

so that this term belongs to the interval [0, c/n], where c is an absolute constant. We shall take n to be a prime n ≡ 1 (mod 4) such that (n + 1)/2 has no prime factor < y, where y is a large positive real number. Such primes exist by Dirichlet’s theorem on primes in arithmetical progressions. Then

\frac{σ_{1}(n + 1)}{n(n + 1)} ≤ \frac{c_{1} log log n}{n}

for some constant c_{1}, and

\frac{5}{4} ≤ \frac{σ_{2}(n + 1)}{n(n + 1)} ≤ \frac{5(n + 1)}{4n} ∏_{q≥y} (1 + \frac{1}{q^{2}} + ⋯)

≤ \frac{5(n + 1)}{4n} exp (∑_{q≥y} \frac{1}{q(q − 1)}) = \frac{5(n + 1)}{4n}e^{O(1/y)}

= \frac{5}{4} + O (\frac{1}{y})  (1)

whenever n > y. Thus, when k = 1, we get that σ_{1}(n)/n = 1 + 1/n, and we conclude that there is an integer in the interval [1/n, (c + 1 + c_{1} log log n)/n], while when k = 2 we get that σ_{2}(n)/n = n+1/n, and so there exists an integer in the interval [1/4, 1/4+c/n+c_{2}/y], where c_{2} is the constant implied by the O symbol in (1). Choosing n (and y) to be sufficiently large, we get a contradiction in the case k = 1 (and k = 2), which completes the proof.

In this paper, we make a modest contribution to the problem, proving two results about the irrationality of the above series for k ≥ 3.

Theorem 1. The sum of the series

∑_{n=1}^{∞} \frac{σ_{3}(n)}{n!}  (2)

is irrational.

We recall the statement of the Prime k-tuples Conjecture (see [3, 7, 9]), which is due to Dickson.

Conjecture 1. For any k ≥ 2, let a_{1}, …, a_{k} and b_{1}, …, b_{k} be integers with a_{i} > 0 for each i = 1, …, k. Suppose that for every prime number p there exists an integer n such that ∏_{i=1}^{k}(a_{i}n + b_{i}) is not a multiple of p. Then there exist infinitely many positive integers n such that p_{i} = a_{i}n + b_{i} is prime for all i = 1, …, k.

We have the following result for any positive integer k.

Theorem 2. The Prime k-tuples Conjecture implies that

∑_{n=1}^{∞} \frac{σ_{k}(n)}{n!}

is irrational.

Throughout this paper, we use the Landau symbols O and o as well as Vinogradov’s symbols ≫, ≪ and ≍ with their regular meaning. The constants implied by these symbols depend at most on k. We use p and q to denote prime numbers.

Acknowledgements. We thank the referee for a careful reading of the manuscript and for suggestions which improved the quality of this paper. Most of this paper was written during a very enjoyable visit by the first two authors to Williams College; these authors wish to express their thanks to that institution for the hospitality and support.

2. Proof of Theorem 1

Our main tool is the following well-known theorem of Chen (see [1], [2]).

Theorem 3. Let a be an integer. There exists x_{a} such that for x > x_{a}, the interval [x/2, x] contains ≫ xa/φ(a)^{2}(log x)^{2} prime numbers p ≡ 1 ( mod a) such that q = (p + 1)/2 is either a prime or a product of two primes each one of which exceeds x^{1/10}.

A proof containing the basic ideas for this theorem appears, for example, in Chapter 11 of [8]. Chen actually proved that for large even integers N there are ≫ N/(log N)^{2} primes p such that N − p is either a prime or a product of two large primes. Easy and well-known modifications of Chen’s argument yield the above theorem.

To prove our Theorem 1 we assume that the sum of the series shown at (2) is rational and deduce a contradiction. We write it, as we did for k = 0, 1 and 2 as A/B, multiply across by (n − 1)! for some large n (with n > B), and obtain

\frac{σ_{3}(n)}{n} + \frac{σ_{3}(n + 1)}{n(n + 1)} + \frac{σ_{3}(n + 2)}{n(n + 1)(n + 2)} + ∑_{j≥3} \frac{σ_{3}(n + j)}{n(n + 1) ⋯ (n + k)} = (n − 1)! (\frac{A}{B} − ∑_{j=1}^{n−1} \frac{σ_{k}(j)}{j!}) , (3)

where the right hand side is an integer. In what follows, we shall exploit the above relation. Since σ_{3}(n) ≪ n^{3}, we have

∑_{j≥3} \frac{σ_{3}(n + j)}{n(n + 1) ⋯ (n + j)} ≪ \frac{1}{n} + ∑_{m≥n} \frac{1}{m^{2}} ≪ \frac{1}{n}.

Furthermore, the sum appearing on the left hand side of formula (3) above is a positive integer. We now let y be a large positive real number which we shall fix later, and take a = 72 ∏_{5≤p≤y} p. Note that, by Chebyshev’s bounds, this means that y ≍ log a. We let x be large compared to a and m ∈ [x/2a, x/a] be such that p = am + 1 is prime and q = (p + 1)/2 = (am + 2)/2 is either a prime or a product of two primes q_{1}q_{2} each exceeding x^{1/10}. Choose n to be the prime n = p. We then note that

\frac{σ_{3}(n)}{n} = n^{2} + \frac{1}{n}.

Furthermore, writing n + 2 = am + 3 = 3t, we have that t is coprime to all primes q ≤ y, and so obtain the inequalities

\frac{28}{27} < \frac{σ_{3}(n + 2)}{n(n + 1)(n + 2)} = \frac{28}{27} \frac{σ_{3}(t)}{t^{3}} (1 + O (\frac{1}{n}))
< \frac{28}{27} ∏_{p>y} (1 − \frac{1}{p^{3}} )^{−1} (1 + O (\frac{1}{n}))

< 1 + \frac{1}{27} + O (\frac{1}{y^{2}} ) ,

since y < \sqrt{x}. Combining this with our estimate for the tail of the series we have, for n a prime as above,

\frac{σ_{3}(n)}{n} + \frac{σ_{3}(n + 2)}{n(n + 1)(n + 2)} + ∑_{j≥3} \frac{σ_{3}(n + j)}{n(n + 1) ⋯ (n + j)} = A_{n} + \frac{1}{27} + O (\frac{1}{y^{2}} ) ,  (4)

where A_{n} is a positive integer.

We now consider the remaining term σ_{3}(n + 1)/(n(n + 1)).

Assume first that for some large x there is a prime p = am + 1 ∈ [x/2, x] such that q = (p + 1)/2 = (a/2)m + 1 is prime. Then

\begin{aligned}\frac{σ_{3}(n + 1)}{n(n + 1)} & = \frac{σ_{3}(2q)}{2q(2q − 1)} = \frac{9(q^{3} + 1)}{2q(2q − 1)} \\ & = \frac{9}{4}q + \frac{9}{8} + \frac{9q − 4}{8q(2q − 1)} = B_{n} + \frac{3}{8} + O (\frac{1}{n}) \end{aligned}

with some integer B_{n}, where we have used the fact that q = (a/2)m + 1 ≡ 1 (mod 4), because 8 | a. Summing up everything, we find that

\frac{1}{27} + \frac{3}{8} + O (\frac{1}{y^{2}} ) ∈ ℤ,

which is impossible if y is chosen to be sufficiently large.

So, we are left with the more interesting part of the problem where q = q_{1}q_{2}, where q_{1} > q_{2} > x^{1/10} holds for all the ≫ xa/φ(a)^{2}(log x)^{2} choices of primes p = am + 1 ∈ [x/2, x] guaranteed by Chen’s Theorem 3.

We put

M = ⌊\frac{σ_{3}(n + 1)}{n(n + 1)}⌋ = ⌊\frac{9σ_{3}(q)}{2q(2q − 1)}⌋ .

Note that since q_{2} ≤ q^{1/2}, we have that

\frac{σ_{3}(q)}{2q(2q − 1)} = \frac{q^{3} + q_{1}^{3} + q_{2}^{3} + 1}{2q(2q − 1)} = \frac{q^{3} + q_{1}^{3} + O(q^{3/2})}{2q(2q − 1)} = \frac{q^{3} + q_{1}^{3}}{2q(2q − 1)} + O (\frac{1}{q^{1/2}} ) .

Thus, using also the previous calculations of fractional parts (see (4)), we find that for large x,

M + \frac{26}{27} − \frac{c_{0}}{y^{2}} ≤ \frac{9q^{3} + 9q_{1}^{3}}{2q(2q − 1)} ≤ M + \frac{26}{27} + \frac{c_{0}}{y^{2}}

holds for all large x with some constant c_{0} > 0, an inequality which can be rewritten as

q^{2} (4M + 3 − 9q + \frac{23}{27} − \frac{2q(m + O(1))}{q^{2}} − \frac{4c_{0}}{y^{2}} ) ≤ 9q_{1}^{3}

≤ q^{2} (4M + 3 − 9q + \frac{23}{27} − \frac{2q(m + O(1))}{q^{2}} + \frac{4c_{0}}{y^{2}} ) .

From this inequality and the fact that q_{1} < qx^{−1/10}, we find that

\frac{M}{q} = \frac{9(q^{3} + q_{1}^{3})}{2q^{2}(2q − 1)} + O (\frac{1}{q}) = \frac{9}{4} + O (\frac{1}{x^{3/10}} )

so

−\frac{2q(M + O(1))}{q^{2}} = −\frac{9}{2} + O (\frac{1}{x^{3/10}} ) .

Hence, we deduce that

4M + 3 − 9q + \frac{23}{27} − \frac{2q(M + O(1))}{q^{2}} = L + \frac{19}{54} + O (\frac{1}{x^{3/10}} )

holds with some non-negative integer L. Thus, for large x, provided that y < x^{3/20}, we get that

\frac{q^{2}}{9} (L + \frac{19}{54} − \frac{5c_{0}}{y^{2}} ) < q_{1}^{3} < \frac{q^{2}}{9} (L + \frac{19}{54} + \frac{5c_{0}}{y^{2}} ) ,

which is equivalent to

\frac{q_{2}^{2}}{9} (L + \frac{19}{54} − \frac{5c_{0}}{y^{2}} ) < q_{1} < \frac{q_{2}^{2}}{9} (L + \frac{19}{54} + \frac{5c_{0}}{y^{2}} ) .  (5)

The above inequalities certainly tell us that L ≥ 0, provided that y is sufficiently large. Further, the left hand side of the above inequality is > q_{2}^{2}/27 if y^{2} > 220c_{0}, so if y satisfies the above inequality and x is large, then q_{2}^{2}/27 ≤ q_{1} ≤ x/q_{2}, and therefore q_{2} ≤ 3x^{1/3}. Now fix q_{2}. Then the left hand side of the above inequality is ≥ (L + 1/3)q_{2}^{2}/27 and the middle term satisfies q_{1} ≤ x/q_{2}. Thus, L + 1/3 ≤ 27x/q_{2}^{3}. Since L ≥ 0 is an integer, it follows that the number of possibilities for L is

1 + ⌊\frac{27x}{q_{2}^{3}}⌋ ≤ \frac{81x}{q_{2}^{3}}.  (6)

We now fix also L ≥ 0. Then q_{1} is a prime in the interval shown at (5) above such that q_{1}q_{2} ≡ 1 (mod a) and 2q_{1}q_{2} − 1 = p is a prime. By the Brun sieve, the number of such primes is

≪ \frac{c_{0}q_{2}^{2}}{y^{2}} · (\frac{a}{φ(a)^{2}} ) · \frac{1}{(log(10c_{0}q_{2}^{2}/(9y^{2}a)))^{2}} .

Since q_{2} > x^{1/10}, it follows that for large x we have 10c_{0}q_{2}^{2}/(9y^{2}a) > x^{1/6}. Thus, the number of possibilities for q_{1} once q_{2} and L are fixed is

≪ \frac{1}{y^{2}} · \frac{a}{φ(a)^{2}} · \frac{q_{2}^{2}}{(log x)^{2}} .

Summing up over all the possibilities for L shown at (6), we get that the total number of possibilities for q_{1} when q_{2} is fixed is

≪ \frac{1}{y^{2}} · \frac{a}{φ(a)^{2}} · \frac{x}{(log x)^{2}} · \frac{1}{q_{2}} ,

and now summing up the above bound over all primes q_{2} ∈ [x^{1/10}, 3x^{1/3}], we get that the total number of possibilities for p is

≪ \frac{1}{y^{2}} \frac{a}{φ(a)^{2}} \frac{x}{(log x)^{2}} ∑_{x^{1/10}≤q_{2}≤3x^{1/3}} \frac{1}{q_{2}} ≪ \frac{1}{y^{2}} \frac{a}{φ(a)^{2}} \frac{x}{(log x)^{2}} ,  (7)

where the fact that the last sum above is O(1) follows from Mertens’s estimate for the summatory function of the reciprocal of the primes. Let c_{1} be the constant implied in the Vinogradov symbol from Chen’s Theorem 3 and c_{2} be the constant implied in the last Vinogradov symbol in (7). Comparing the estimates for the number of primes p under scrutiny we get

\frac{c_{1}ax}{φ(a)^{2}(log x)^{2}} ≤ \frac{c_{2}ax}{φ(a)^{2}y^{2}(log x)^{2}}

leading to y^{2} < c_{3}, where c_{3} = c_{2}/c_{1}. Choosing y larger than this we complete the proof of Theorem 1.

3. Proof of Theorem 2

Let k ≥ 4. For i = 1, …, k we let Q_{i}(X), R_{i}(X) ∈ ℤ[X] be the polynomials given by the division algorithm:

(X + i)^{k} = Q_{i}(X)(X + 1) ⋯ (X + i) + R_{i}(X) i = 1, …, k,

where deg Q_{i}(X) = k − i and deg R_{i}(X) ≤ i − 1. Note that when i = k we have that Q_{k}(X) = 1. For each of those i = 1, …, k for which Q_{i}(−i) ≠ 0 choose distinct primes p_{i} > k such that p_{i} ∤ σ_{k}(i)Q_{i}(−i). (For any other i, for notational purposes, take p_{i} = 1.) As we have seen, Q_{k}(−k) = 1 ≠ 0, so that p_{k} ∤ σ_{k}(k)Q_{k}(−k) can be arranged simply by choosing p_{k} > σ_{k}(k).

Let P = ∏_{i=1}^{k} p_{i} and let m be a positive integer such that

\frac{(k!)^{2}}{i} m + 1 ≡ p_{i} (mod p_{i}^{2}) i = 1, …, k.

By the Chinese Remainder Theorem, there are infinitely many such positive integers m and they form the arithmetic progression m_{0} (mod P^{2}), where m_{0} is the first positive integer in this progression. Given such an m, we choose n = (k!)^{2}m and write m = m_{0} + ℓP^{2} with some nonnegative integer ℓ. Then

n + i = i (\frac{(k!)^{2}}{i} m + 1) = ip_{i} (\frac{(k!)^{2}}{i} \frac{P^{2}}{p_{i}} ℓ + \frac{(k!)^{2}m_{0}/i + 1}{p_{i}}) .

Put

A_{i} = \frac{(k!)^{2}}{i} \frac{P^{2}}{p_{i}} and B_{i} = \frac{(k!)^{2}m_{0}/i + 1}{p_{i}}

for i = 1, …, k.

One checks easily that we can apply the Prime k-tuples Conjecture 1 to the linear polynomials A_{i}ℓ + B_{i}. Indeed, the prime numbers dividing A_{i} are exactly the primes p ≤ k together with the primes p_{i}. Since B_{i} | (k!)^{2}m_{0}/i + 1, we see that B_{i} is coprime to all primes p ≤ k. Further, B_{i} ≡ 1 (mod p_{i}) from the way m_{0} was chosen. To see that B_{i} is coprime to p_{j} if j ≠ i assume otherwise. Then p_{j} | ((k!)^{2}/j)m_{0} + 1 and p_{j} | B_{i} | ((k!)^{2}/i)m_{0} + 1. Hence, p_{j} | (k!)^{2}m_{0} + j and p_{j} | (k!)^{2}m_{0} + i, so p_{j} | (j − i), but this is false because p_{j} > k. Thus, the conditions for the applicability of the Prime k-tuples Conjecture 1 are fulfilled and we can let ℓ be large and such that A_{i}ℓ + B_{i} = q_{i} are all primes for i = 1, …, k. Put a_{i} = ip_{i} and note that if m = m_{0} + ℓP^{2}, then n + i = a_{i}q_{i} for i = 1, …, k.

Now assume that the series

∑_{j=1}^{∞} \frac{σ_{k}(j)}{j!} = \frac{A}{B}

with A and B coprime integers. Suppose n is sufficiently large, and in particular n > B. Multiplying across by n!, we get

∑_{j≤n} σ_{k}(j)\frac{n!}{j!} + ∑_{i=1}^{k} \frac{σ_{k}(n + i)}{(n + 1) ⋯ (n + i)} + ∑_{j≥k+1} \frac{σ_{k}(n + j)}{(n + 1) ⋯ (n + j)} ∈ ℤ.  (8)

The first sum above is an integer, while the last sum is positive and, since σ_{k}(n) ≪ n^{k},

∑_{j≥k+1} \frac{σ_{k}(n + j)}{(n + 1) ⋯ (n + j)} = \frac{σ_{k}(n + k + 1)}{(n + 1) ⋯ (n + k + 1)} + ∑_{j≥k+2} \frac{σ_{k}(n + j)}{(n + 1) ⋯ (n + j)}
≪ \frac{1}{n} + ∑_{j≥k+2} \frac{1}{(n + j)^{2}} ≪ \frac{1}{n}.

As for the intermediate terms, since q_{i} is prime we have

\frac{σ_{k}(n + i)}{(n + 1) ⋯ (n + i)} = \frac{σ_{k}(a_{i})(q_{i}^{k} + 1)}{(n + 1) ⋯ (n + i)}
= \frac{σ_{k}(a_{i})}{a_{i}^{k}} (\frac{(n + i)^{k} + a_{i}^{k}}{(n + 1) ⋯ (n + i)})
= \frac{σ_{k}(a_{i})}{a_{i}^{k}} (Q_{i}(n) + \frac{R_{i}(n) + a_{i}^{k}}{(n + 1) ⋯ (n + i)})
= \frac{σ_{k}(a_{i})}{a_{i}^{k}}Q_{i}(n) + O (\frac{1}{n}) ,

where for the above error term we used the fact that deg R_{i} ≤ i−1. Since n ≡ −i (mod p_{i}), we get that Q_{i}(n) ≡ Q_{i}(−i) (mod p_{i}) so that Q_{i}(n) = Q_{i}(−i) + p_{i}ℓ_{i}(n), for some integers ℓ_{i}(n). Thus,

\frac{σ_{k}(n + i)}{(n + 1) ⋯ (n + i)} = \frac{σ_{k}(ip_{i})}{(ip_{i})^{k}} (Q_{i}(−i) + ℓ_{i}(n)p_{i}) + O (\frac{1}{n}) .

We add everything together, multiply formula (8) by (k!)^{k}P^{k−1} and get

∑_{i=1}^{k} P_{i}^{k−1}σ_{k}(ip_{i})\frac{(k!)^{k}}{i^{k}} \frac{1}{p_{i}} (Q_{i}(−i) + p_{i}ℓ_{i}(n)) + O (\frac{1}{n}) ∈ ℤ,

where

P_{i} = P p_{i}^{−1} = ∏_{\begin{matrix}1≤j≤k \\ j≠i \end{matrix}} p_{j} .

From this it follows that

∑_{i=1}^{k} P_{i}^{k−1} \frac{(k!)^{k}}{i^{k}} \frac{σ_{k}(ip_{i})Q_{i}(−i)}{p_{i}} + O (\frac{1}{n}) ∈ ℤ.

At this moment, we observe that the sum on the left does not depend on n and so

∑_{1≤i≤k} P_{i}^{k−1} \frac{(k!)^{k}}{i^{k}} \frac{σ_{k}(ip_{i})}{p_{i}}Q_{i}(−i) ∈ ℤ.

Using also the fact that σ_{k}(ip_{i}) = σ_{k}(i)σ_{k}(p_{i}) = σ_{k}(i)(p_{i}^{k} + 1), we get that the above relation implies

∑_{1≤i≤k} P_{i}^{k−1} \frac{(k!)^{k}}{i^{k}} \frac{σ_{k}(i)}{p_{i}}Q_{i}(−i) ∈ ℤ.

This is a non-empty (since Q_{k}(−k) = 1) sum of non-zero rational numbers whose reduced denominators are distinct primes. Thus, the above sum cannot be an integer. The proof of Theorem 2 is complete. Note Added: March 2007. Shortly after this paper was submitted, we learned about the recent appearance of the paper [10], where the same two results as the present ones have been obtained. The proof of the case k = 3 in [10] also uses sieve methods but is rather different, whereas the conditional proof for larger k’s is similar to ours. We thank Professor Igor Shparlinski for pointing out this reference to us.

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[10] J.-C. Schlage-Puchta, ‘The irrationality of a number theoretical series’, The Ramanujan J. 12 (2006), 455–460.