On the length of the continued fraction for values of quotients of power sums

Generalizing a result of Pourchet, we show that, if α, β are power sums over Q satisfying suitable necessary assumptions, the length of the continued fraction for α(n)/β(n) tends to infinity as n→∞. This will be derived from a uniform Thue-type inequality for the rational approximations to the rational numbers α(n)/β(n), n ∈ N. Introduction. The features of the continued fraction of a positive real number are usually extraordinarily difficult to predict. However, if the numbers in question run through certain parametrized families, some regularity occasionally appears. For instance, we may recall remarkable results by Schinzel [S1], [S2], on the lengths of the periods of the continued fractions for √ f(n), where f is a polynomial with rational coefficients and n varies through N. In a similar direction, but in a much more simpler way, one can deal with the continued fractions of the rational numbers r(n), n ∈ N, where now r ∈ Q(X) is a rational function; for example, we may now look at the length of the fraction: in [S2, Lemma 2], Schinzel proves that this length is bounded as a function of n (see also [M] for a more precise result). Substantially, the reason for this is that r(X) has a finite expansion as a simple continued fraction with partial quotients in Q[X]. More generally, one may inquire about the length of the continued fraction for the rational values of other arithmetical functions. In general, due to the lack of an Euclid’s algorithm, we have not a suitable “functional Manuscrit reçu le 4 février 2004. 1Observe however that such continued fraction specialized at X = n may not coincide with the continued fraction for r(n). 738 Pietro Corvaja, Umberto Zannier continued fraction”; this fact makes the situation different from the case of Q(X) and generally speaking considerably more difficult. For certain exponential functions, we have an appealing question of Mendès-France (see [M2, p. 214]), stating that, for coprime integers a, b > 1, the length of the continued fraction for (a/b)n tends to infinity as n→∞, in marked contrast to the rational function case. This was answered affirmatively by Pourchet (unpublished) and independently by Choquet (in a weaker form); we refer to the paper [M3] for a discussion and several references. We also mention a paper by Grisel [G] solving a function-field analogue. This theorem of Pourchet may be derived from the (deep) lower bound |(a/b)n − (p/q)| q−2 exp(− n) (any positive ), where p, q are integers with 0 < q < bn (which follows e.g. from Ridout generalization of Roth’s Theorem [R]). The estimate is in fact amply sufficient, since it implies that all the partial quotients of the continued fraction in question are exp( n), and the statement about the length follows at once. (See also [Z, Ex. II.6].) In this direction, one may consider more general power sums in place of an, bn. Namely, in analogy to [CZ1] (where the notation is slightly different) we consider the ring (actually a domain) E made up of the functions on N of the form α(n) = c1a1 + . . .+ cra n r , (1) where the coefficients ci are rational numbers, the ai are positive integers and the number r of summands is unrestricted. (We shall often normalize (1) so that the ci are nonzero and the ai are distinct; in this case the ai are called the “roots” of α.) We then consider the ratio α/β of nonzero elements of E and ask about the length of the continued fraction for the values α(n)/β(n) (provided β(n) 6= 0, which is the case for all large n). It turns out (Corollary 2 below) that the length of the relevant continued fraction does not tend to infinity if and only if α/β admits an expansion as a (finite) continued fraction over E . Note that E is not euclidean, so in practice this condition “rarely” holds. This result plainly generalizes Pourchet’s, but now the Ridout Theorem seems no longer sufficient for a proof. In fact, we shall need the full force of the Schmidt Subspace Theorem, similarly to [CZ1], where (among others) the related question of the integrality of the values α(n)/β(n) was investigated. Similarly to the above sketched argument for Pourchet’s Theorem, the conclusions about the continued fraction will be derived from a “Thue-type” inequality for the rational approximations to the values in question, which is the content of the Theorem below. Its proof follows [CZ1] in the main On the length of the continued fraction 739 lines, except for a few technical points; however the present new simple applications of those principles are perhaps not entirely free of interest. Preliminary to the statements, we introduce a little more notation. We let EQ be the domain of the functions of type (1), but allowing the ai to be positive rationals. If α ∈ EQ is written in the form (1), with nonzero ci and distinct ai, we set `(α) = max(a1, . . . , ar), agreeing that `(0) = 0. It is immediate to check that `(αβ) = `(α)`(β), `(α+ β) ≤ max(`(α), `(β)) and that `(α)n |α(n)| `(α)n for n ∈ N tending to infinity (so in particular, α(n) cannot vanish infinitely often if α 6= 0). Theorem. Let α, β ∈ E be nonzero and assume that for all ζ ∈ E , `(α − ζβ) ≥ `(β). Then there exist k = k(α, β) > 0 and Q = Q(α, β) > 1 with the following properties. Fix > 0; then, for all but finitely many n ∈ N and for integers p, q, 0 < q < Qn, we have ∣∣∣∣α(n) β(n) − pq ∣∣∣∣ ≥ 1 qk exp(− n). We tacitly mean that the alluded finite set of exceptional integers includes those n ∈ N with β(n) = 0; this finite set may depend on . Remarks. (i) The assumption about α, β expresses the lack of an “Euclid division” in E for α : β and cannot be omitted. In fact, suppose that `(α − ζβ) < `(β) for some ζ ∈ E . Then, the values ζ(n) have bounded denominator and verify |(α(n)/β(n))− ζ(n)| exp(− 0n) for large n and some positive 0 independent of n, against the conclusion of the Theorem (with p/q = pn/qn = ζ(n)). Note that the assumption is automatic if `(α) ≥ `(β) and `(β) does not divide `(α). (ii) By the short argument after Lemma 1 below, given α, β, one can test effectively whether a ζ ∈ E such that `(α−ζβ) < `(β) actually exists. Also, the proof will show that a suitable Q and “exponent” k may be computed. On the contrary, the finitely many exceptional n’s cannot be computed with the present method of proof. (iii) Naturally, the lower bound is not significant for q larger than `(β)n/(k−1). So, some upper bound Qn for q is not really restrictive. (Also, since the dependence of k on α, β is unspecified, and since `(β)1/(k−1) → 1 as k →∞, it is immaterial here to specify a suitable Q in terms of k.) 740 Pietro Corvaja, Umberto Zannier Note also that a lower bound Qn for the denominator of α(n)/β(n) follows, sharpening [CZ1, Thm. 1]; in this direction, see [BCZ], [CZ3, Remark 1] and [Z, Thm. IV.3] for stronger conclusions in certain special cases. (iv) Like for the proofs in [CZ1], the method yields analogous results for functions of the form (1), but where ai are any algebraic numbers subject to the sole (but crucial) restriction that there exists a unique maximum among the absolute values |ai|. Using the (somewhat complicated) method of [CZ2], one may relax this condition, assuming only that not all the |ai| are equal. For the sake of simplicity, we do not give the proofs of these results, which do not involve new ideas compared to [CZ1], [CZ2] and the present paper, but only complication of detail. (v) The Theorem can be seen as a Thue-type inequality with moving targets (similarly to [C] or [V]). It seems an interesting, but difficult, problem, to obtain (under suitable necessary assumptions) the “Roth’s exponent” k = 2, or even some exponent independent of α, β. (As recalled above, this holds in the special cases of Pourchet’s Theorem.) Corollary 1. Let α, β be as in the Theorem. Then the length of the continued fraction for α(n)/β(n) tends to infinity as n→∞. This corollary is in fact a lemma for the following result, which gives a more precise description. Corollary 2. Let α, β ∈ E be nonzero. Then the length of the continued fraction for α(n)/β(n) is bounded for infinitely many n ∈ N if and only if there exist power sums ζ0, . . . , ζk ∈ E such that we have the identical continued fraction expansion α(n) β(n) = [ζ0(n), ζ1(n), . . . , ζk(n)]. If this is the case, the mentioned length is uniformly bounded for all n ∈ N. It will be pointed out how the condition on α/β may be checked effectively. The special case α(n) = an− 1, β(n) = bn− 1 appears as [Z, Ex. IV.12]. For completeness we give here this application. Corollary 3. Let a, b be multiplicatively independent positive integers. Then the length of the continued fraction for (an − 1)/(bn − 1) tends to infinity with n. On the length of the continued fraction 741 Proofs. We start with the following very simple: Lemma 1. Let α, β ∈ E be nonzero and let t be any positive number. Then there exists η ∈ EQ such that `(α − ηβ) < t. Such an η may be computed in terms of α, β, t. Proof. Write β(n) = cbn(1 − δ(n)), where c ∈ Q∗, b = `(β) and where δ ∈ EQ satisfies u := `(δ) < 1. In particular, we have |δ(n)| un, so for a fixed integer R we have an approximation, for n→∞, α(n) β(n) = α(n) cbn ( R ∑