A CURIOUS PROPERTY OF SERIES INVOLVING TERMS OF GENERALIZED SEQUENCES

Here we are concerned with series involving generalized Fibonacci numbers Un(p,q) and generalized Lucas numbers Vn(p,q). The aim of this paper is to find triples (p,q,r) for which the seriesUn(p,q)/rn and Vn(p,q)/rn (for r running from 0 to infinity) are unconcerned at the introduction of the factor n. The results established in this paper generalize the known fact that the series Fn/2n (Fn the nth Fibonacci number) and the series nFn/2n give the same result, namely −2/5.

1. Introduction.First, we define the sequences under study, then we explain the aim of our paper.
1.1.The generalized sequences.For a, b, p, q arbitrary real numbers (in particular, integers), in the notation of Horadam [3] write (a, b; p, q), (1.1) meaning that In particular, here we are concerned with the numbers U n (p, q) = U n := W n (0, 1; p, q) and V n (p, q) = V n := W n (2,p; p, q). ( Observe that U n (1, −1) = F n and V n (1, −1) = L n are the nth Fibonacci and Lucas number, respectively.The Binet forms for U n and V n are where ) are the roots, assumed distinct, of the equation x 2 − px + q = 0. Observe that (1.6) yields the two relations α + β = p and αβ = q (1.7) which will be widely used throughout this paper.As usual, we require that ∆ > 0, (1.8) so that α, β, and √ ∆ are real, where α ≠ β as assumed.We assume also that pq ≠ 0. (1.9) 1.2.Motivation and aim of the paper.We were amazed at the equality which is reported in [2, (4.20)] as a by-product result.One can immediately observe how the series on the left-hand side of (1.10) is, quite surprisingly, unconcerned at the introduction of the factor n. Our mathematical curiosity led us to seek analogs of (1.10) that involve the more general sequences U n and V n defined by (1.3).In fact, the aim of this paper is to find triples (p,q,r ) of real numbers (with r ≠ 0) for which the equality (here W stands for either U or V ) holds true.The cases where p and q are integers and r is rational (possibly an integer) are treated as particular instances.While the case W ≡ U (Section 3) is readily solved, some particular aspects of the case W ≡ V (Section 4) are worth investigating (Subsections 4.1 and 4.2).Several numerical examples illustrate the theoretical results whose proofs are given in full, except for the detailed discussions on certain inequalities which are omitted for the sake of brevity.

Preliminary results.
The following two lemmas will be needed in Sections 3 and 4, respectively.Lemma 2.1.If x and y are arbitrary complex numbers such that x ≠ y, then the equality is satisfied if and only if To prove Lemma 2.1, we show that (2.3) is nothing but an equivalent form of (2.2).The same technique will be used in the proofs of Lemma 2.2 and Theorems 3.1 and 4.1.
Proof of Lemma 2.1.Using the geometric series formula, (2.2) yields which, in turn, can be equivalently rewritten as (2.5) Since the first factor on the left-hand side of (2.5) cannot vanish as x ≠ y by hypothesis, let us equate to zero the second factor thus getting the following equation: which yields the desired result (2.3).
Lemma 2.2.If x and y are as in the statement of Lemma 2.1 (x = y allowed), then the equality is satisfied if and only if (2.8) Proof.Similar to the proof of Lemma 2.1, (2.7) yields which, in turn, can be equivalently rewritten as (2.10) whence (2.8) is immediately obtained.

Series involving the numbers U
r n for all nonzero p.
whence condition (3.1) appears to be sufficient.
Further, after replacing x (respectively y) by α/r (respectively β/r ) in conditions (2.1), and taking the value of r given by (3.1) into account, it becomes clear that we must have It is not hard to prove that the necessary and sufficient condition for inequalities (3.7) to be satisfied is that q/p 2 < −3/4.In other words, we must have The analog of (1.10) for the numbers G n (m) defined in [1] as Observe that (3.11) and (1.10) coincide for m = 1 whereas, for m = 2 (see [4]), (3.11) is the Jacobsthal-analog of (1.10).

Proof of Theorem
as expected.From (4.9) above, the quantity ∆ defined by (1.5) can be expressed as By (4.3), for a given integer p it is clear that q is not necessarily an integer.As a numerical example, if we let s = 3/2 (see the first part of Remark 4.1) and p = 14, then from (4.2) and (4.14) we get r = q = 21, so that Let us conclude this subsection by finding the analog of (1.10) for the Lucas numbers.If we put the Lucas parameters p = 1 and q = −1 in (4.3), then we get the cubic equation The values of r (= ps = 1•s = s) for which (4.4) holds are clearly given by all the (real) solutions of (4.16) that fulfill (4.1).It can be seen, with the aid of a computer, that only the root Question.For given p and q, how many distinct values of r do there exist for which (4.4) holds?
The answer is given in the following proposition.
Proposition.For given p and q, there exist at most two distinct values of r for which (4.4) holds.
Proof.By putting q/p 2 = a in (4.3), one gets the cubic equation in the unknown s after some simple manipulations.The discriminant δ of the second-degree factor above, given by We prove that, if s 0 is chosen in such a way that obtained by combining (4.1) and (4.23), then at most one of the roots s 1 and s 2 given by (4.24) satisfies (4.1).After a good deal of algebraic manipulation, one obtains the following: (a) s 1 and s 2 cannot be both greater than 1 because s 1 > 1 → s 0 < 1/3 and s 2 > 1 → s 0 > 1/3.Let us confine ourselves to sketch the proof only for the first implication.The second one can be proved in a similar way.From (4.24) we get the implication s 1 > 1 → s 0 < √ δ whose right-hand side is satisfied if s 0 < 0. If s 0 > 0, from (4.22) we can rewrite the implication above as  Finally, we observe that, if we replace V n (p, q) by H n (m) := V n (1, −m), m = 1, 2, 3,..., (see [1]) in (4.4), then this equality holds only for

A concluding remark.
It is worth noting that the results established in Section 2 allow us to obtain some interesting identities involving circular and hyperbolic functions.For example, we invite the interested reader to use Lemma 2.1 along with the

4 . 2 .
A question and its answer.By observing (4.4), a question arises quite naturally.
condition (3.2) is not satisfied whereas in all the previous examples it is.
If the real number s (see (4.1)) is a rational number s = m/d with gcd(m, d) = 1 (s an integer if d = 1), then (4.3) becomes 4.1.The integrality of q, and numerical examples.