Dedekind Sums with Arguments near Certain Transcendental Numbers

Kurt Girstmair Institut für Mathematik Universität Innsbruck Technikerstr. 13/7 A-6020 Innsbruck Austria

               Kurt.Girstmair@uibk.ac.at

Abstract

We study the asymptotic behavior of the classical Dedekind sums s(s_{k}/t_{k}) for the sequence of convergents s_{k}/t_{k} k ≥ 0, of the transcendental number

∑_{j=0}^{∞} \frac{1}{b^{2^{j}}} , b ≥ 3.

In particular, we show that there are infinitely many open intervals of constant length such that the sequence s(s_{k}/t_{k}) has infinitely many transcendental cluster points in each interval.

1 Introduction and result

Dedekind sums have quite a number of interesting applications in analytic number theory (modular forms), algebraic number theory (class numbers), lattice point problems and algebraic geometry (for instance [1, 6, 7, 10]).

Let n be a positive integer and m ∈ ℤ, (m, n) = 1. The classical Dedekind sum s(m/n) is defined by

s(m/n) = ∑_{k=1}^{n}((k/n))((mk/n))

where ((⋯)) is the usual sawtooth function (for example, [7, p. 1]). In the present setting it is more natural to work with

S(m/n) = 12s(m/n)

instead.

In the previous paper [3] we used the Barkan-Hickerson-Knuth-formula to study the asymptotic behavior of S(s_{k}/t_{k}) for the convergents s_{k}/t_{k} of transcendental numbers like e or e^{2}. In this situation the limiting behavior of S(s_{k}/t_{k}) was fairly simple. It is much more complicated, however, for the transcendental number

x(b) = ∑_{j=0}^{∞} \frac{1}{b^{2^{j}}} , b ≥ 3.  (1)

In fact, we have no full description of what happens in this case. Its complexity is illustrated by the following theorem, which forms the main result of this paper.

Theorem 1. Let s_{k}/t_{k}, k ≥ 0, be the sequence of convergents of the number x(b) of (1). Then the sequence S(s_{k}/t_{k}), k ≥ 0, has infinitely many transcendental cluster points in each of the intervals

(b − 10 − 2i + \frac{1}{b}, b − 9 − 2i + \frac{1}{b − 1}) , i ≥ 0.

Note that each of the intervals of Theorem 1 has the length 1 + 1/(b(b − 1)), whereas the distance between two neighboring intervals is 1 − 1/(b(b − 1)).

2 The integer part

We start with the continued fraction expansion [a_{0}, a_{1}, a_{2}, …] of an arbitrary irrational number x. The numerators and denominators of its convergents

s_{k}/t_{k} = [a_{0}, a_{1}, …, a_{k}]  (2)

are defined by the recursion formulas

s_{−2} = 0, s_{−1} = 1, s_{k} = a_{k}s_{k−1} + s_{k−2} and
t_{−2} = 1, t_{−1} = 0, t_{k} = a_{k}t_{k−1} + t_{k−2}, for k ≥ 0.  (3)

Henceforth we will assume 0 < x < 1, so a_{0} = 0. Then the Barkan-Hickerson-Knuth formula says that for k ≥ 0

S(s_{k}/t_{k}) = ∑_{j=1}^{k}(−1)^{j−1}a_{j} + \begin{cases}(s_{k} + t_{k−1})/t_{k} − 3, & if k is odd; \\ (s_{k} − t_{k−1})/t_{k}, & if k is even; \end{cases}  (4)

see [2, 4, 5].

In the case of the number x = x(b), the continued fraction expansion has been given in [9]. It is defined recursively. To this end put

C(1) = C(1, b) = [0, b − 1, b + 2]

in the sense of (2) and (3). If C(j) = C(j, b) has been defined for j ≥ 1 and C(j) = [0, a_{1}, …, a_{n}] (where n = 2^{j}), then

C(j + 1) = C(j + 1, b) = [0, a_{1}, …, a_{n}, a_{n} − 2, a_{n−1}, a_{n−2}, …, a_{2}, a_{1} + 1].

Then x = lim_{j→∞} C(j). In particular, x = [0, a_{1}, a_{2}, …], where a_{k} is the corresponding partial denominator of each C(j) with 2^{j} ≥ k.

In view of formula (4) for x = x(b), it is natural to investigate

L(k) = L(k, b) = ∑_{j=1}^{k−1}(−1)^{j−1}a_{j}, k ≥ 0,

first. For the sake of simplicity we call L(k) the integer part of the Dedekind sum S(s_{k}/t_{k}). The following lemma comprises three easy observations.

Lemma 2. Let [0, a_{1}, a_{2}, …] be the continued fraction expansion of x = x(b) and n = 2^{j}, j ≥ 0.

(a) If n ≥ 4, then

a_{n+k} = a_{n−k+1} for 2 ≤ k ≤ n − 1.

(b) If n ≥ 8, then

a_{k} = a_{n−k+1} for 2 ≤ k ≤ n/2 − 1.

(c) If n ≥ 8, then

a_{k} = a_{n+k} for 2 ≤ k ≤ n/2 − 1.

Proof. Obviously, assertion (c) follows from (a) and (b). Assertion (a) is immediate from the definition of the continued fraction expansion of x(b). In order to deduce (b) from (a), we assume n ≥ 4 and put l = n − k + 1, 2 ≤ k ≤ n − 1. Then a_{l} = a_{n−k+1} = a_{n+k}, by (a). Since k = n − l + 1, this gives a_{l} = a_{n+(n−l+1)} = a_{2n−l+1}. So we have, for n ≥ 8 and 2 ≤ l ≤ n/2 − 1: a_{l} = a_{n−l+1}, which is (b).

Lemma 3. Let n = 2^{j}, n ≥ 4. For 1 ≤ k ≤ n − 1 we have

L(n + k) = −2 + L(n − k).

Proof. Since L(n + 1) = L(n − 1) + (−1)^{n−1}a_{n} + (−1)^{n}(a_{n} − 2) = L(n − 1) − 2, the assertion holds for k = 1. Let 2 ≤ k ≤ n − 1. Then

L(n + k) = L(n − 1) − 2 + ∑_{i=2}^{k}(−1)^{n+i−1}a_{n+i}.

By assertion (a) of Lemma 2, the sum on the right hand side equals

∑_{i=2}^{k}(−1)^{n+i−1}a_{n−i+1} = ∑_{i=2}^{k}(−1)^{i−1}a_{n−i+1} = ∑_{i=1}^{k−1}(−1)^{i}a_{n−i}.

We observe

∑_{i=1}^{k−1}(−1)^{i}a_{n−i} = ∑_{i=n−k+1}^{n−1}(−1)^{i}a_{i}.

This gives

L(n + k) = −2 + ∑_{i=1}^{n−1}(−1)^{i−1}a_{i} + ∑_{i=n−k+1}^{n−1}(−1)^{i}a_{i} = −2 + L(n − k).

Remark 4. By the construction of the sequence C(j), we have a_{n} = b for each n = 2^{j}, j ≥ 2. From Lemma 3 we obtain L(2n) = L(2n−1)+(−1)^{2n−1}a_{2n} = L(n+(n−1))−b = L(1)−2−b = b − 1 − 2 − b = −3.

Lemma 5. Let n = 2^{j}, n ≥ 8. For 2 ≤ k ≤ n/2 − 1,

L(n + k) = −4 + L(k).

In particular, L(n + k) = L(2n + k) = L(4n + k) = ⋯

Proof. We have L(n) = −3, by the remark. Hence L(n+1) = L(n)+(−1)^{n}a_{n+1} = −3+b−2 = b − 5. From Lemma 2, (c) we obtain

L(n + k) = b − 5 + (−1)^{n+1}a_{n+2} + ⋯ + (−1)^{n+k−1}a_{n+k} =
b − 5 + (−1)^{1}a_{2} + ⋯ + (−1)^{k−1}a_{k} = b − 5 + L(k) − a_{1} = −4 + L(k).

Let n = 2^{j}, n ≥ 8. We define a sequence k_{i}, i ≥ 0, in the following way:

k_{0} = n − 1.  (5)

If k_{i−1} has been defined, i ≥ 1, then

k_{i} = 2^{i}n − k_{i−1}.  (6)

Induction based on (5) and (6) gives

2 ≤ k_{i} ≤ 2^{i}n − 1,  (7)

and

k_{i} = \frac{2^{i+1} + (−1)^{i}}{3} n + (−1)^{i−1}  (8)

for all i ≥ 0. We have

L(k_{0}) = L(n − 1) = L(n) + a_{n} = −3 + b

from the remark. Further, Lemma 3 gives, by induction,

L(k_{i}) = −3 − 2i + b.

Indeed, if L(k_{i−1}) = −3 − 2(i − 1) + b, L(k_{i}) = L(2^{i}n − k_{i−1}) = L(2^{i−1}n + (2^{i−1}n − k_{i−1})) = −2 + L(k_{i−1}) = −3 − 2i + b. Altogether, we know the numbers k_{i} and the integer part of S(s_{k_{i}}/t_{k_{i}}) explicitly, namely

Lemma 6. Let n = 2^{j}, n ≥ 8. For i ≥ 0 let k_{i} be defined by (8). Then

L(k_{i}) = b − 3 − 2i.

Lemma 6 says that the integer part L(k_{i}) of S(s_{k_{i}}/t_{k_{i}}) is independent of n if n ≥ 8 is a power of 2. Suppose, therefore, that n_{l} = 2^{2+l}, l = 1, …, r. Fix i ≥ 0 for the time being and define

k_{i,l} = \frac{2^{i+1} + (−1)^{i}}{3} n_{l} + (−1)^{i−1}.  (9)

By (7),

k_{i,l} ≤ 2^{i}n_{l} − 1 ≤ 2^{i}n_{r} − 1 = 2^{i+r+2} − 1.

Suppose that n̂ is a power of 2, n̂ ≥ 2^{i+r+3}. Then we have

2 ≤ k_{i,l} ≤ \frac{n̂}{2} − 1

for all l = 1, …, r. Therefore, Lemma 5 and Lemma 6 give

Proposition 7. Let i ≥ 0 and r ≥ 1 be given and n_{l} = 2^{2+l}, l = 1, … r. Suppose that the numbers k_{i,l} are defined as in (9). If n̂ is a power of 2, n̂ ≥ 2^{i+r+3}, then

L(n̂ + k_{i,l}) = −4 + L(k_{i,l}) = b − 7 − 2i.

3 The fractional part

Note that the numbers k_{i,l} of the foregoing section are all odd. Hence Lemma 9 and the Barkan-Hickerson-Knuth formula give

S(s_{n̂+k_{i,l}}/t_{n̂+k_{i,l}}) = b − 7 − 2i + \frac{s_{n̂+k_{i,l}}}{t_{n̂+k_{i,l}}} + \frac{t_{n̂+k_{i,l−1}}}{t_{n̂+k_{i,l}}} − 3.  (10)

If n̂ tends to infinity s_{n̂+k_{i,l}}/t_{n̂+k_{i,l}} tends to x = x(b). Accordingly, we have to investigate the limiting behavior of t_{n̂+k_{i,l−1}}/t_{n̂+k_{i,l}} in order to understand the fractional part of formula (10).

To this end we suppose that n is a power of 2, n ≥ 8, and k is an integer, 2 ≤ k ≤ n/2−1. From (3) we have t_{n+k} = a_{n+k}t_{n+k−1} + t_{n+k−2}, hence

\frac{t_{n+k}}{t_{n+k−1}} = a_{n+k} + \frac{t_{n+k−2}}{t_{n+k−1}} = [a_{n+k}, \frac{t_{n+k−1}}{t_{n+k−2}} ].

When we repeat this procedure, we obtain the well-known fact

\frac{t_{n+k}}{t_{n+k−1}} = [a_{n+k}, a_{n+k−1}, \frac{t_{n+k−2}}{t_{n+k−3}} ] = [a_{n+k}, a_{n+k−1}, …, a_{1}].

From Lemma 2, (c), we infer

a_{n+k} = a_{k}, a_{n+k−1} = a_{k−1}, …, a_{n+2} = a_{2}.

Moreover, a_{n+1} = a_{n} − 2 = b − 2 and a_{n} = b. Finally, Lemma 2, (b) says

a_{n−1} = a_{2}, a_{n−2} = a_{3}, …, a_{n/2+2} = a_{n/2−1}.

Altogether,

\frac{t_{n+k}}{t_{n+k−1}} = [a_{k}, a_{k−1}, …, a_{2}, b − 2, b, a_{2}, a_{3}, … a_{n/2−1}, a_{n/2+1}, …, a_{1}].

The final terms a_{n/2+1}, a_{n/2}, …, a_{1} are not of interest. It suffices to write

\frac{t_{n+k}}{t_{n+k−1}} = [a_{k}, a_{k−1}, …, a_{2}, b − 2, b, a_{2}, a_{3}, …, a_{n/2−1}, c(n)]  (11)

for some c(n) ∈ ℚ. From [9, Theorem 8] we know that all numbers a_{1}, a_{2}, … are ≥ 1 and ≤ b + 2, hence we have

1 ≤ c(n) ≤ b + 3.

Proposition 8. Suppose that k remains fixed, 2 ≤ k ≤ n/2 − 1, but n = 2^{j} tends to infinity. Then t_{n+k}/t_{n+k−1} converges to

t(k) = t(k, b) = [a_{k}, a_{k−1}, …, a_{2}, b − 2, (x + 1)/x],

where x = x(b) is defined by (1).

Proof. We have x = lim_{i→∞} C(i) = [0, b − 1, y] with y = [a_{2}, a_{3}, …]. A short calculation shows

[b, y] = [b, a_{2}, a_{3}, …] = (x + 1)/x.

Let p_{i}/q_{i}, i = 0, 1, 2, … be the convergents of t_{n+k}/t_{n+k−1} (where the numbers p_{i}, q_{i} are defined in the same way as the numbers s_{i}, t_{i} in (3)). We have, by (11),

\frac{t_{n+k}}{t_{n+k−1}} = \frac{pc(n) + p′}{qc(n) + q′}

with p = p_{k+n/2−1}, p′ = p_{k+n/2−2}, q = q_{k+n/2−1}, q′ = q_{k+n/2−2}. We write

t(k) = [a_{k}, …, a_{2}, b − 2, b, a_{2}, …, a_{n/2−1}, z(n)],

where z(n) satisfies 1 ≤ z(n) ≤ b + 3 by the argument above. Accordingly,

t(k) = \frac{pz(n) + p′}{qz(n) + q′} .

This gives

t(k) − \frac{t_{n+k}}{t_{n+k−1}} = \frac{pz(n) + p′}{qz(n) + q′} − \frac{pc(n) + p′}{qc(n) + q′} .  (12)

The expression on the right hand side of (12) simplifies to

\frac{(pq′ − p′q)z(n) + (p′q − pq′)c(n)}{(qz(n) + q′)(qc(n) + q′)}.

However, it is well-known that pq′ − p′q = ±1. Observing 1 ≤ z(n), c(n) ≤ b + 3, we obtain

|t(k) − \frac{t_{n+k}}{t_{n+k−1}} | ≤ \frac{2b + 6}{(q + q′)^{2}} .

Since q and q′ tend to infinity for n → ∞, our proof is complete.

We conclude this section with two observations. Lemma 9. In the above setting, let 2 ≤ k < k′ be integers. Then t(k) ≠ t(k′). Proof. Suppose t(k) = t(k′), so

[a_{k′}, …, a_{k+1}, t(k)] = t(k).

An identity of this kind can only hold if t(k) is a quadratic irrationality. However, t(k) is a transcendental number since x is transcendental (see [8, p. 35, Satz 8]). Lemma 10. Let k ≥ 2 be an integer. Then x + 1/t(k) is a transcendental number. Proof. Suppose α = x + t(k) is algebraic. Since we may write

1/t(k) = [0, t(k)] = \frac{p(x + 1)/x + p′}{q(x + 1)/x + q′} = \frac{p(x + 1) + p′x}{q(x + 1) + q′x}

with integers p, p′, q, q′, q > 0, q′ ≥ 0, we obtain

x + \frac{p(x + 1) + p′x}{q(x + 1) + q′x} = α.

This, however, means that x satisfies a quadratic equation over the field ℚ(α). Accordingly, x is algebraic, a contradiction.

4 Proof of Theorem 1

As in the setting of Proposition 7, let i ≥ 0 and r ≥ 1 be given and n_{l} = 2^{2+l}, l = 1, …, r. Suppose that the numbers k_{i,l} are defined as in (9). Let n̂ be a power of 2, n̂ ≥ 2^{i+r+3}. By Proposition 7,

L(n̂ + k_{i,l}) = b − 7 − 2i.

If n̂ tends to infinity, Proposition 8 says that t_{n̂+k_{i,l}}/t_{n̂+k_{i,l}−1} tends to

t(k_{i,l}) = [a_{k_{i,l}}, a_{k_{i,l}−1}, …, a_{2}, b − 2, (x + 1)/x].

Therefore t_{n̂+k_{i,l}−1}/t_{n̂+k_{i,l}} tends to 1/t(k_{i,l}). Altogether, we have

S(s_{n̂+k_{i,l}}/t_{n̂+k_{i,l}}) → b − 10 − 2i + x + \frac{1}{t(k_{i,l})}.

For l < l′ ≤ r we obtain k_{i,l} < k_{i,l′} from (9). By Lemma 9, t(k_{i,l}) ≠ t(k_{i,l′}). Accordingly, the numbers 1/t(k_{i,l}) are pairwise different for 1 ≤ l ≤ r. Further, x + 1/t(k_{i,l}) is transcendental, by Lemma 10. The inequalities

1/b < x < 1/(b − 1) and 0 < 1/t(k_{i,l}) < 1

are obvious by (1) and x = [0, b − 1, …], 1/t(k_{i,l}) = [0, a_{k_{i,l}}, …]. Therefore, the sequence S(s_{j}/t_{j}), j ≥ 1, has r distinct transcendental cluster points in the interval

(b − 10 − 2i + \frac{1}{b}, b − 9 − 2i + \frac{1}{b − 1}) .

Since r can be chosen arbitrarily large, this proves Theorem 1.

References

[1] T. M. Apostol, Modular Functions and Dirichlet Series in Number Theory, Springer, 1976.

[2] Ph. Barkan, Sur les sommes de Dedekind et les fractions continues finies, C. R. Acad. Sci. Paris Sér. A-B 284 (1977) A923–A926.

[3] K. Girstmair, Dedekind sums with arguments near Euler’s number e, J. Integer Seq. 15 (2012), Article 12.5.8.

[4] D. Hickerson, Continued fractions and density results for Dedekind sums, J. Reine Angew. Math. 290 (1977), 113–116.

[5] D. E. Knuth, Notes on generalized Dedekind sums, Acta Arith. 33 (1977), 297–325.

[6] C. Meyer, Die Berechnung der Klassenzahl Abelscher Körper über Quadratischen Zahlkörpern, Akademie-Verlag, 1957.

[7] H. Rademacher and E. Grosswald, Dedekind Sums, Mathematical Association of America, 1972.

[8] Th. Schneider, Einführung in die Transzendenten Zahlen, Springer, 1957.

[9] J. Shallit, Simple continued fractions for some irrational numbers, J. Number Theory 11 (1979), 209–217.

[10] G. Urzúa, Arrangements of curves and algebraic surfaces, J. Algebraic Geom. 19 (2010), 335–365.

2010 Mathematics Subject Classification: Primary 11F20; Secondary 11A55. Keywords: asymptotic behavior of Dedekind sums, continued fraction expansion of transcendental numbers.

Received April 9 2013; revised version received June 20 2013. Published in Journal of Integer Sequences, July 29 2013.

Return to Journal of Integer Sequences home page.