Summation of Certain Infinite Lucas-Related Series

Transcription

1 J. Integer Sequences 22 (209) Article Summation of Certain Infinite Lucas-Related Series arxiv: v [math.nt] Jan 209 Bakir Farhi Laboratoire de Mathématiques appliquées Faculté des Sciences Exactes Université de Bejaia Bejaia Algeria Abstract In this paper we find the sums in closed form of certain type of Lucas-related convergent series. More precisely we generalize the results already obtained by the author in his arxiv paper entitled: Summation of certain infinite Fibonacci related series. Introduction Throughout this paper we let N denote the set N \ {0} of positive integers. We let Φ denote the golden ratio (Φ : + 5) and Φ its conjugate in the quadratic field Q( 5); that 2 is Φ : 5. Further we let sg(x) denote the sign of a nonzero real number x; that 2 Φ is sg(x) if x > 0 and sg(x) if x < 0. Let PQ R be fixed such that : P 2 4Q > 0 and consider L(PQ) the R-vectorial space of linear sequences (w n ) n Z satisfying w n+2 Pw n+ Qw n ( n Z). (.) The Lucas sequences of thefirst andsecond kind arethe sequences of L(PQ)corresponding respectively to the initial values (w 0 w ) (0) and (w 0 w ) (2P). Those sequences are respectively denoted by ( ) n and (V n ) n. Let α and β be the roots of the quadratic equation: X 2 PX +Q 0 such that α > β (note that α β because by hypothesis P 0 and P 2 4Q > 0). It is well known that for all n Z we have αn β n α β and V n α n +β n. (.2)

2 The connections and likenesses between the Lucas sequences of the first and second kind are numerous; among them we will use the following which are easy to check V n + Q (.3) U 2n V n (.4) β n U m β m Q m m (.5) U m+r U m +r Q m U r m (.6) which holdforanynmr Z. Notethatifwe take(pq) ( ) weobtainforthelucas sequence of the first kind the classical Fibonacci sequence (F n ) n (referenced by A in the OEIS [9]) and for the Lucas sequence of the second kind the classical Lucas sequence (L n ) n (referenced by A in the OEIS). Further if (PQ) (2 ) we obtain the so called Pell sequences: (+ 2) n ( 2) n 2 V n (+ 2) n +( 2) n 2 (respectively referenced by A00029 and A in the OEIS). Next if (PQ) (32) then the sequences obtained are simply: 2 n V n 2 n + (respectively referenced by A and A00005 in the OEIS). The exact evaluation of infinite Lucas-related series is an old and fascinating subject of study with many still open questions. The particular case dealing with Fibonacci numbers (considered as the most important) has been the subject of several researches that the reader can consult in [ ]. For the general case we just refer the reader to the papers [6] and [] that are close enough to the present work. This paper is devoted to generalize the results already obtained by the author in[8] which only concerns the classical Fibonacci sequence to the larger class of the Lucas sequences of the first kind. We investigate two types of infinite Lucas-related series. The first one consists of the series of one of the two forms: Q an U a n+k a n or ( ) n Q an U a n+k a n where k is a positive integer and (a n ) n is a sequence of positive integers tending to infinity with n. The second type of series which we consider consists of the series of one of the two forms: ( ) nβan U an or β an U an where (a n ) n is an increasing sequence of positive integers. We show in particular that if P and Q are integers then some of such series can be transformed on series with rational terms. 2

3 2 The first type of series We begin with the following general result. Theorem 2.. Let (a n ) n be a sequence of positive integers tending to infinity with n and let k be a positive integer. Then we have Q an U a n+k a n To prove this theorem we need the following lemma. β an U an. (2.) Lemma 2.2. Let (x n ) n be a convergent real sequence and let x R be its limit. Then for all k N we have (x n+k x n ) kx x n. Proof. Let k N be fixed. For any positive integer N we have N N (x n+k x n ) (x n+i x n+i ) i i N (x n+i x n+i ) (x N+i x i ) i x N+i i x i. The formula of the lemma immediately follows by tending N to infinity. Proof of Theorem 2.. Following an idea of Bruckman and Good [6] let us apply Lemma 2.2 for x n : βan ( n ) which converges to 0 (since α > β ). For any nk N by α an β an using (.2) and the fact that αβ Q we get x n+k x n β a n+k α a n+k β a n+k i β an α an β an α an β a n+k α a n+kβ a n (α an β an )(α a n+k β n+k ) (αβ)an (β a n+k a n α a n+k a n ) (α an β an )(α a n+k β a n+k) Q an (β α)u an+k a n (α β)u an (α β)u an+k α β Qan U a n+k a n. 3

4 So Lemma 2.2 gives Thus ( U ) a n+k a n α β Qan as required. The theorem is proved. x n Q an U a n+k a n β an U an β an α an β an. From Theorem 2. we immediately deduce the following corollary which is already pointed out by Bruckman and Good [6] and also by Hu et al. []: Corollary 2.3. Let (a n ) n be a sequence of positive integers tending to infinity with n. Then we have Q U an a n+ a n βa. (2.2) U an U an+ U a If (a n ) n is an arithmetic sequence of positive integers Theorem 2. gives the following result (already pointed out by Hu et al. [] for the case k ): Corollary 2.4. Let (a n ) n be an increasing arithmetic sequence of positive integers and let r be its common difference. Then for any positive integer k we have In particular we have Q r(n ) Q a U kr β an U an. (2.3) Q r(n ) U an U an+ (β/q)a U a U r. (2.4) Proof. To obtain Formula (2.3) it suffices to apply Formula (2.) of Theorem 2. and use that a n r(n ) + a ( n ). Then to obtain Formula (2.4) we simply set k in formula (2.3). This completes the proof. Before continuing with general results let us give some applications of the preceding results for the usual Fibonacci sequence; so we must fix (PQ) ( ). An immediate application of Formula (2.3) of Corollary 2.4 gives the following wellknown formulas (see for example the survey paper of Duverney and Shiokawa [7]): 5 F 2n F 2n+ 2 ( ) n F n F n F 2n F 2n ( ) n F n F n (2.5)

5 Let k be a positive integer and a 2 be an integer. By taking in Formula (2.2) of Corollary 2.3: a n ka n ( n ) we obtain the following formula: F (a )ka n F ka nf ka n+ F ka Φka. (2.6) By taking in addition a 2 in (2.6) we derive the following formula: F n0 k2 n + + F k F 2k F 2k Φ2k (2.7) which is already pointed out by Hoggatt and Bicknell [9]. By taking again k in (2.7) we derive the following remarkable formula discovered since the 870 s by Lucas [2]: 7 5. (2.8) F 2 n 2 n0 From Formula (2.7) we deduce that + n0 Q( F k2 5) (for any positive integer k). n But except the geometric sequences with common ratio 2 we don t know any other regular sequence (a n ) n N ofpositiveintegers satisfyingthepropertythat + n0 Q( 5). More precisely we propose the following open question: Open question. Is there any linear recurrence sequence (a n ) n N of positive integers which is not a geometric sequence with common ratio 2 and which satisfies the property that n0 F an Q( 5)? F an Next by taking a 3 in Formula (2.6) we deduce (according to Formula (.4)) the following formula of Bruckman and Good [6]: L k3 n F k3 n+ F 3k Φ3k. (2.9) By taking k in Formula (2.9) we deduce (after some calculations) the formula: L 3 n F n0 3 n+ 5 2 (also already pointed out by Bruckman and Good [6]). (2.0) By taking in Formula (2.) of Theorem 2.: a n F n ( n ) and k we obtain the following: F ( ) Fn Fn 5 (2.) F Fn F Fn+ 2 5

6 (also already pointed out by Bruckman and Good [6]). Next by taking in Formula (2.) of Theorem 2.: a n F n ( n ) and k 2 we obtain the following: F Fn+ ( ) Fn 5. (2.2) F Fn F Fn+2 Now with the same context as Theorem 2. the following corollary gives the sums in closed form of the series ( ) n Q U an a n+k a n when k is chosen even. Corollary 2.5. Let (a n ) n be a sequence of positive integers tending to infinity with n and let k be a positive integer. Then we have ( ) n Q an U a n+2k a n U an U an+2k Q a 2n U a 2n a 2n U a2n U a2n. (2.3) Proof. By applying Formula (2.) of Theorem 2. for the sequence (a 2n ) n (instead of (a n ) n ) we obtain Q a U a 2n 2n+2k a 2n β a 2n. (2.4) U a2n U a2n+2k U a2n Next by applying Formula (2.) of Theorem 2. for the sequence (a 2n ) n (instead of (a n ) n ) we obtain Q a U a 2n 2n+2k a 2n β a 2n. (2.5) U a2n U a2n+2k U a2n Now by subtracting (2.5) from (2.4) we get which we can write as: ( Q a U a 2n 2n+2k a 2n Q a U ) a 2n 2n+2k a 2n U a2n U a2n+2k U a2n U a2n+2k ( ) n Q an U a n+2k a n U an U an+2k ( β a 2n U a2n βa2n U a2n β a 2n U a2n β a 2n U a2n U a2n U a2n. Finally since for any n Z we have β a 2n U a2n β a 2n U a2n Q a 2n U a2n a 2n (according to Formula (.5)) we conclude that: as required. The proof is achieved. ( ) n Q an U a n+2k a n U an U an+2k Q a 2n U a 2n a 2n U a2n U a2n ) 6

7 If (a n ) n is an arithmetic sequence of positive integers then Corollary 2.5 reduces to the following corollary: Corollary 2.6. Let (a n ) n be an increasing arithmetic sequence of positive integers and let r be its common difference. Then for any positive integer k we have ( ) n Q r(n ) U an U an+2k U r U 2kr Q 2r(n ) U a2n U a2n. (2.6) In particular we have ( ) n Q r(n ) U an U an+2 U a U a2 V r. (2.7) Proof. To establish Formula (2.6) it suffices to apply Corollary 2.5 together with the formula u n r(n )+u ( n ). To establish Formula (2.7) we take k in (2.6) and we use in addition Formula (.4). Remark 2.7. Let (a n ) n N be an increasing arithmetic sequence of natural numbers and let r be its common difference. By Corollary 2.4 we know a closed form of the sum Q r(n ) (k N ) and by Corollary 2.6 we know a closed form of the sum ( ) n Q r(n ) (2.8) when k is an even positive integer. But if k is an odd positive integer the closed form of the sum (2.8) is still unknown even in the particular case a n n. However we shall prove in what follows that if a 0 0 there is a relationship between the sums (2.8) where k lies in the set of the odd positive integers. We have the following: Theorem 2.8. Let S rk : ( ) n Q r(n ) U rn U r(n+k) ( rk N ). Then for any positive integer r and any odd positive integer k we have S rk U (k )/2 r S r +Q r Q 2r(n ). (2.9) U rk U 2nr U (2n+)r 7

8 Proof. Let r and k be positive integers and suppose that k is odd. Because the formula of the theorem is trivial for k we can assume that k 3. We have S r U rk U r S rk ( ) n Q r(n ) U rn U r(n+) U rk U r ( ) n Q r(n ) U rn U r(n+k) ( ) n Q r(n )U rku r(n+) U r U r(n+k) U r U rn U r(n+) U r(n+k). But according to Formula (.6) ( applied to the triplet (rkrrn) instead of (nmr) ) we have U rk U r(n+) U r U r(n+k) Q r U rn U r(k ). Using this it follows that: S r U rk U r S rk U r(k ) U r Q ru r(k ) U r ( ) n Q rn U r(n+) U r(n+k) ( ) n Q r(n ) U bn U bn+k where we have put b n : r(n+) ( n ). Next because (k ) is even (since k is supposed odd) it follows by Corollary 2.6 that: S r U rk S rk Q ru r(k ) U (k )/2 r U r U r U (k )r The formula of the theorem follows. (k )/2 Q r Q 2r(n ) U 2nr U (2n+)r. Q 2r(n ) U b2n U b2n Remark 2.9. The particular case of Formula (2.9) corresponding to (P Q) ( ) and r is already established by Rabinowitz [8]. Remark 2.0. In [] Hu et al. established an expression of S r in terms of the values of the Lambert series; and before them Jeannin [4] obtained the same expression in the particular case when Q and r is odd. 3 The second type of series We begin this section by dealing with series of the form n ( )n β an /U an where (a n ) n is an increasing sequence of positive integers. By grouping terms we transform such series to another type of series whose terms are rational numbers when PQ Z. As we will specify later some results of this section can be deduced from the results of the previous one by tending the parameter k to infinity. We have the following: 8

9 Theorem 3.. Let (a n ) n be an increasing sequence of positive integers. Then we have ( ) nβan U an Q a 2n U a 2n a 2n U a2n U a2n. (3.) Proof. The increase of (a n ) n ensures the convergence of the two series in (3.). By grouping terms we have ( ) ( ) nβan ( ) 2nβa 2n +( ) 2n βa2n U an U a2n ( β a 2n U a2n βa2n U a2n ) β a 2n U a2n β a 2n U a2n U a2n U a2n as required. This achieves the proof of the theorem. U a2n Q a 2n U a2n a 2n U a2n U a2n (according to Formula (.5)) Remark 3.2. We can also prove Theorem 3. by tending k to infinity in Formulas (2.3) of Corollary 2.5. To do so we must previously remark that for any positive integer n we have U an+2k a lim n k + U an+2k (according to (.2)) and that the series of functions α an ( ) n Q an U a n+2k a n U an U an+2k (from N to R) converges uniformly on N as we can see for example by observing that: U an+2k a n U an+2k α a n+2k a n ( ) β an+2k a n α α a n+2k ( 2 ) β an+2k ( nk N β ). α an α α According to the closed-form formulas of and V n (see (.2)) it is immediate that V lim n n + α β sg(p). So the approximations sg(p) Vn (n ) are increasingly better when n increases. When supposing P > 0 and Q < 0 we derive from Theorem 3. a curious formula in which the sum of the errors of the all approximations sg(p) V n Vn (n ) is transformed to a series whose terms are rational numbers when PQ Z. We have the following: Corollary 3.3. Suppose that P > 0 and Q < 0. Then we have V + n 2 β n Q 2n 2. (3.2) U n U 2n U 2n 9

10 Proof. From the hypothesis P > 0 and Q < 0 we deduce that α P+ and β P 2 2 (since α > β ). Thus α > 0 β < 0 and α β. In addition (P > 0 and Q < 0) implies that > 0 ( n N ). Using all these facts together with (.2) we have for any positive integer n: V n (α β) V n (α β) V n (αn β n ) (α n +β n ) 2 β n which gives the first equality of (3.2). The second equality of (3.2) follows from Theorem 3. by taking a n n ( n ). The proof is complete. Before continuing with general results we will give some important applications of the two preceding results for the usual Fibonacci and Lucas sequences. By taking in Corollary 3.3 (PQ) ( ) which corresponds to ( V n ) (F n L n ) we immediately deduce the following: Corollary 3.4. We have L + 5 n 2 F n F n Φ 2 + n F 2n F 2n. Next the application of Corollary 3.3 for (P Q) (4 ) can be announced in the following form: Corollary 3.5. We have 5 r 2 r Λ F 3n Φ 4 + (3.3) 3n F 6n F 6n 3 where Λ denotes the set of the regular continued fraction convergents of the number 5. Proof. The continued fraction expansion of the number 5 is known to be equal to: 5 [2;444...] (see e.g. [5 page 6]). From this we derive that for all n N the nth-order convergent of 5 is rn p n /q n where (p n ) n N and (q n ) n N are the sequences of positive integers satisfying the recurrence relations: p n+2 4p n+ +p n q n+2 4q n+ +q n ( n N) with initial values: p 0 2 q 0 p 9 q 4. Using those recurrence relations the sequences (p n ) n and(q n ) n canbeextended to negativeindices n. Doingso we obtainp and q 0. It follows from this that ( V n ) (q n 2p n ) (n N) is exactly the couple of Lucas sequences corresponding to (PQ) (4 ). Since (PQ) (4 ) gives 20 0

11 and (αβ) ( ) (Φ 3 Φ 3 ) we deduce by applying Formulas (.2) that we have for all n N: (Φ 3 ) n (Φ 3) n q n Φ 3 Φ 3 Φ3n 3n Φ F 3n 2p n ( ( Φ 3)n + Φ 3) n Φ 3n +Φ 3n L 3n. Thus (p n q n ) ( 2 L 3n+3 ) 2 F 3n+3 By applying then Corollary 3.3 for (PQ) (4 ) we get that is + 2p n 20 4 n0 q n + 5 r n 2 as required. The corollary is proved. F 3n Φ 8 + 3n F 3n Φ 4 + 3n ( n N). F 6n F 6n 3 F 6n F 6n 3 Further by applying Theorem 3. for (PQ) ( ) and taking successively: a n 2n a n 2n and then a n 2n+ we respectively obtain the three following formulas: ( ) n F 2n Φ + 2n F 4n F 4n 2 ( ) n F 2n Φ + 2n ( ) n F 2n+ Φ + 2n+ F 4n F 4n 3 F 4n+ F 4n. (3.4) Remarkthattheaddition(sidetoside) ofthetwo lastformulasof (3.4)givesthefirst formula of (2.5). Now by applying another technique of grouping terms we are going to show that some series of the form n 0 βan /U an (where (a n ) n is an arithmetic sequence of a particular type) can be also transformed to series whose terms are rational numbers when PQ Z. We have the following: Theorem 3.6. For any positive integer r we have n0 β 2r n+2 r U 2 r n+2 r Q 2r n. (3.5) U 2 r n Proof. Let r be a positive integer. From the trivial equality of sets: { 2 r n+2 r n N } { 2 r n n N } \{2 r n n N }

12 we get n0 β 2r n+2 r U 2 r n+2 r β 2r n U 2 r n ( β 2r n U 2 r n β 2r n U 2 r n β2r n U 2 r n ) β 2r n V 2 r n β 2r n (since U 2 r n U 2 r nv 2 r n according to (.4)). But since β 2r n V 2 r n β 2r n ( ) β 2r n α 2r n +β 2r n β 2rn (αβ) 2r n Q 2r n (according to (.2)) we conclude that: β 2r n+2 r Q 2r n U n0 2 r n+2 r U 2 r n as required. The theorem is proved. To finish let us see what Theorem 3.6 gives for the usual Fibonacci sequence. By taking in Theorem 3.6: (PQ) ( ) and r we obtain the following formula: n0 F 2n+ Φ + 2n+ U 2 r n ( ) n F 2n. (3.6) Next by taking in Theorem 3.6: (PQ) ( ) and r 2 we obtain the formula: n0 F 2 r n+2 r Φ2r n+2 r F 2 r n ( r 2). (3.7) Note that the particular case corresponding to r 2 of the last formula; that is the formula n0 F 4n+2 Φ + 4n+2 F 4n was already pointed out by Melham and Shannon [4] who proved it by summing both sides of Formula (2.7) over k lying in the set of the odd positive integers. Remark 3.7. Transforming a series of real terms into a series of rational terms could serve for example to show the irrationality of the sum of such a series. For example for (w n ) n L(PQ) André-Jeannin [3] proved the irrationality of the series + tn /w n when P Z Q ± t Z and t < α ; thus including the series in (3.5) of Theorem 3.6 (if we assume P Z and Q ±). Other similar and related results can also be found in [ 2 3 7]. 4 Acknowledgments The author would like to thank the editor and the anonymous referee for their helpful comments. 2

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