Convergence of a linear recursive sequence

Transcription

1 int. j. math. educ. sci. technol., 2004 vol. 35, no. 1, Convergence of a linear recursive sequence E. G. TAY*, T. L. TOH, F. M. DONG and T. Y. LEE Mathematics and Mathematics Education, National Institute of Education, Nanyang Technological University, Singapore egtay@nie.edu.sg. (Received 4 November 2002) A necessary and sufficient condition is found for a linear recursive sequence to be convergent, no matter what initial values are given. Its limit is also obtained when the sequence is convergent. Methods from various areas of mathematics are used to obtain the results. 1. Introduction Let k be a positive integer and p 1, p 2,, p k be real numbers. Define a sequence fa n g 1 n¼1 as follows: a nþkþ1 ¼ Xk p i a nþi, n ¼ 0, 1, 2, ð1þ In this paper, we shall study the convergence of fa n g 1 n¼1 under the condition: p i 5 0 for i ¼ 1, 2,, k. Hence in this paper, we always assume that p i 5 0 for i ¼ 1, 2,, k. The convergence of the above sequence was first studied by Zhao et al. [3]. A partial result for k ¼ 3 was obtained. The convergence of the sequence fa n g 1 n¼1 sometimes depends not only on p 1, p 2,, p k, but also on the initial values a 1, a 2,, a k. For example, if k >1, p 1 ¼ 1 and p i ¼ 0 for 2 4 i 4 k, then {a n } is divergent, unless a 1 ¼ a 2 ¼¼a k. But if p k ¼ 1 and p i ¼ 0fori ¼ 1, 2,, k 1, then fa n g 1 n¼1 is always convergent, no matter what the initial values a 1, a 2,, a k are. In this paper, we do not consider special initial values a 1, a 2,, a k. Our purpose is to find a necessary and sufficient condition on p 1, p 2,, p k such that fa n g 1 n¼1 is always convergent, no matter what the initial values a 1, a 2,, a k are. Integers a 1, a 2,, a n, where n 5 2, are called coprime if there is no integer q with q 5 2 such that a i is divisible by q for all i ¼ 1, 2,, n. Our goal in this paper is to prove that this sequence is convergent for all initial values a 1, a 2,, a k if and only if either (i) p 1 þ p 2 þþp k < 1, or (ii) p 1 þ p 2 þþp k ¼ 1 and the integers in the following set are coprime: ¼fkþ1 i : p i 6¼ 0, i ¼ 1, 2,, kg * The author to whom correspondence should be addressed. International Journal of Mathematical Education in Science and Technology ISSN X print/issn online # 2004 Taylor & Francis Ltd DOI: /

2 52 E. G. Tay et al. We also find its limit if the sequence is convergent, P k P i lim a p j a i n ¼ P n!1 k ðk þ 1 iþp i If k ¼ 1, then a n ¼ a 1 p n 1 1 for n 5 1. So the above result is obvious when k ¼ 1. Thus in the following, we always assume that k 5 2. In section 2, we use an analysis method to prove that if p 1 þ p 2 þþp k ¼ 1 and the integers in are coprime, then fa n g 1 n¼1 is always convergent, and its limit is also obtained. In section 3, we prove that fa n g 1 n¼1 is not always convergent if p 1 þ p 2 þþ p k ¼ 1 and the integers in are not coprime. In section 4, we use a matrix method to prove fa n g 1 n¼1 is convergent under the condition that p 1 þ p 2 þþp k ¼ 1 and the integers in are coprime. Its limit is found by another method. In section 5, we finally settle the last two cases: p 1 þ p 2 þþp k < 1 and p 1 þ p 2 þþp k > 1. Thus we more than extend the results of Zhao et al.: we are also motivated by the pedagogical implications of using methods from different areas of mathematics, as stated in [3]. In the teaching of mathematics, an example demonstrating the application of methods or techniques in one branch of mathematics to solve a problem formulated in another field of mathematics can not only help students to see the intrinsic links between different areas of mathematics but also motivate them to develop a passion and interest in their learning of mathematics. 2. The integers in are coprime and p 1 þ p 2 þþp k ¼ 1 Recall that k 5 2 and p i 5 0 for i ¼ 1, 2,, k. In this section, we always assume that p 1 þ p 2 þþp k ¼ 1. We shall show that the sequence fa n g 1 n¼1 is always convergent if all integers in are coprime. The limit of this sequence is also obtained if it is convergent. Define another two sequences fm n g 1 n¼1 and fm ng 1 n¼1 : and M n ¼ maxfa nþi : 0 4 i 4 k 1g n ¼ 1, 2, ð2þ m n ¼ minfa nþi : 0 4 i 4 k 1g n ¼ 1, 2, ð3þ Lemma 2.1. fm n g 1 n¼1 is a descending sequence. Proof: For any n 5 1, by definition, a nþk ¼ Xk p i a n 1þi 4 Xk p i M n ¼ M n Hence M nþ1 ¼ maxfa nþi : 1 4 i 4 kg 4 maxfm n, a nþk g¼m n

3 Convergence of a linear recursive sequence 53 Similarly we can prove the following result. Lemma 2.2. fm n g 1 n¼1 is an ascending sequence. Since M 1 5 M n 5 m n 5 m 1 for all n 5 1, by Lemmas 2.1 and 2.2, the two sequences fm n g 1 n¼1 and fm ng 1 n¼1 are convergent. Thus if the following sequence tends to zero: M 1 m 1, M 2 m 2, M 3 m 3, then fa n g 1 n¼1 is convergent. Let ¼fkþ1 i : p i > 0, 1 4 i 4 kg Let t ¼jj (i.e. the cardinality of ) and ¼fr 1, r 2,, r t g where 1 4 r 1 < r 2 < < r t 4 k. Define ( ) ¼ Xt w i r i : w i is a non-negative integer for i ¼ 1, 2,, t Lemma 2.3. The integers in are coprime if and only if r, r þ 1 2 for some integer r. Proof. (Necessity). Assume that r 1, r 2,, r t are coprime. Then there exist integers w 1, w 2,, w t such that X t w i r i ¼ 1 Let Then r ¼ X ð w i Þ r i 1 4 i 4 t w i <0 r þ 1 ¼ X w i r i 1 4 i 4 t w i >0 Hence r, r þ 1 2. (Sufficiency) Assume that r, r þ 1 2 for some r. Let r ¼ Xt w i r i and r þ 1 ¼ Xt w 0 i r i

4 54 E. G. Tay et al. Then 1 ¼ Xt ðw 0 i w iþr i implying that r 1, r 2,, r t are coprime. Lemma 2.4. The integers in are coprime if and only if fg, g þ 1, g þ 2, g for some positive integer g. Proof. By Lemma 2.3, we just need to prove the necessity of this lemma. Assume that the integers in are coprime. By Lemma 2.3, r, r þ 1 2 for some r. Thus fðr 1 1Þr þ i : i ¼ 0, 1,, r 1 1g since ðr 1 1Þr þ i ¼ iðr þ 1Þþðr 1 i 1Þr 2. Let g ¼ðr 1 1Þr. For any integer n > g, we have n ¼ g þ wr 1 þ i for some non-negative integers w and i with 0 4 i 4 r 1 1. Observe that g þ i 2. Hence n 2. This proves the necessity. By definition, for any integer n, there exist f i (n) for i ¼ 1, 2,, k such that a n ¼ f 1 ðnþa 1 þ f 2 ðnþa 2 þþf k ðnþa k ð4þ where f i (n) is a polynomial of p 1, p 2,, p k with non-negative integer coefficients, and f i (n) is independent of a 1, a 2,, a k. Lemma 2.5. In (4), f i ðnþ 5 0 and f i (n) is unique for all i ¼ 1, 2,, k and n 5 1. Further, if p 1 þ p 2 þþp k ¼ 1, then f 1 ðnþþf 2 ðnþþþf k ðnþ ¼1 Proof. As p j 5 0 for all j with 1 4 j 4 k and f i (n) is a polynomial of p 1, p 2,, p k with non-negative integer coefficients, we have f i ðnþ 5 0. Assume that there exists another expression for a n : a n ¼ f1 0 ðnþa 1 þ f2 0 ðnþa 2 þþfk 0 ðnþa k Then ð f i ðnþ f 0 i ðnþþa i ¼ 0 Letting a i ¼ 1 and a j ¼ 0 for j 6¼ i, we have f i ðnþ ¼fi 0ðnÞ. Hence f iðnþ ¼fi 0 ðnþ for i ¼ 1, 2,, k. Thus the lemma holds. Assume that p 1 þ p 2 þþp k ¼ 1. If a 1 ¼ a 2 ¼¼a k ¼ 1, then we have a n ¼ 1 for all n 5 1 by definition. Hence f 1 ðnþþf 2 ðnþþþf k ðnþ ¼1. Lemma 2.5 shows that f i (n) is well defined for any integers n and i with n 5 1 and 1 4 i 4 k. In the following, we show some properties on f i (n), and these results will be used in the proof of the main result.

5 Convergence of a linear recursive sequence 55 Lemma 2.6. For any integers n and i: n 5 k þ 1 and 1 4 i 4 k, f i ðnþ > 0 if and only if f i ðn r j Þ > 0 for some j with 1 4 j 4 t. Proof: By definition, a n ¼ Xk p s a n kþs 1 s¼1 ¼ Xk s¼1 p s f i ðn k þ s 1Þa i So ¼ Xk s¼1 p s f i ðn k þ s 1Þa i f i ðnþ ¼ Xk s¼1 p s f i ðn k þ s 1Þ ¼ Xt p kþ1 rj f i ðn r j Þ Since p kþ1 rj > 0 and f i ðn r j Þ 5 0 for j ¼ 1, 2,, t, the result follows immediately. Corollary 2.1. If n 5 k and f i ðnþ > 0, then f i ðn þ uþ > 0 for any u 2. Proof. Assume that u > 0. Since n þ r j > k for j ¼ 1, 2,, t, the result follows immediately by Lemma 2.6 repeatedly. Corollary 2.2. Let i and j be any integers with 1 4 i 4 k and 1 4 j 4 t. If i þ r j 5 k þ 1, then f i ði þ u þ r j Þ > 0 for any u 2. Proof. Note that f i ðiþ ¼1 > 0. Since i þ r j > k, we have f i ði þ r j Þ > 0 by Lemma 2.6. Thus the result follows from Corollary 2.1. If the integers in are coprime, then by Lemma 2.4, there exists g 2 such that n 2 if n 5 g. Lemma 2.7. Assume that p 1 > 0 and the integers in are coprime. Let g be the minimum integer such that u 2 for every integer u 5 g. Then f i ðnþ > 0 for i ¼ 1, 2,, k and n 5 g þ 2k. Proof. Since p 1 > 0, we have r t ¼ k. Thus i þ r t ¼ i þ k > k. Let n 0 ¼ n i k 5 g. Then by Corollary 2.2, we have f i ðnþ ¼f i ðn 0 þ i þ r t Þ > 0

6 56 E. G. Tay et al. Define a matrix B ¼ðb i, j Þ kk : (i) b i, iþ1 ¼ 1for, 2,, k 1; (ii) b k, j ¼ p j for j ¼ 1, 2,, k; (iii) b i, j ¼ 0ifi6¼ k or j 6¼ i þ 1. Let X n ¼ða n, a nþ1,, a nþk 1 Þ be a column vector. Then X nþ1 ¼ BX n, n ¼ 1, 2, Thus X nþ1 ¼ B n X 1, n ¼ 1, 2, Lemma 2.8. For any positive integers n and j, Proof. a nþjk ¼ f 1 ðnþa jkþ1 þ f 2 ðnþa jkþ2 þþf k ðnþa ðjþ1þk We just need to show that it holds when j ¼ 1, i.e. a nþk ¼ f 1 ðnþa kþ1 þ f 2 ðnþa kþ2 þþf k ðnþa 2k Observe that X n ¼ B n 1 X 1 and X nþk ¼ B n 1 þ1. By Lemma 2.5, the matrix B n 1 is unique in the expression X n ¼ B n 1 X 1. Hence the result holds. Assume that p 1 > 0 and the integers in are coprime. Let g be the minimum integer such that u 2 for every integer u 5 g. Let f ¼ minf f i ðuþ : 1 4 i 4 k, g þ 2k 4 u 4 g þ 3k 1g ð5þ By Lemma 2.7 and the result P k f iðnþ ¼1, we have 0 < f < 1. Lemma 2.9. Assume that p 1 > 0 and the integers in are coprime. If n 5 g þ 2k, then a n 4 fm jkþ1 þð1 fþm jkþ1 where j ¼bðn gþ=kc 2. Proof. Observe that g þðj þ 2Þk 4 n < g þðj þ 3Þk. Let n 0 ¼ n jk. Then g þ 2k 4 n 0 4 g þ 3k 1. By Lemma 2.8, a n ¼ a n0 þjk ¼ Xk f i ðn 0 Þa jkþi By definition, m jkþ1 ¼ a jkþs for some s with 1 4 s 4 k. Thus a n ¼ f s ðn 0 Þa jkþs þ X f i ðn 0 Þa jkþi 1 4 i 4 k i6¼s 4 f s ðn 0 Þm jkþ1 þ X 1 4 i 4 k i6¼s f i ðn 0 ÞM jkþ1 ¼ f s ðn 0 Þm jkþ1 þð1 f s ðn 0 ÞÞM jkþ1 4 fm jkþ1 þð1 fþm jkþ1 where the last inequality follows from the fact that f 4 f s ðn 0 Þ and m jkþ1 4 M jkþ1. The proof is thus completed.

7 Now we are going to establish the main result in this section. Theorem 2.1. If p 1 þ p 2 þþp k ¼ 1 and the integers in are coprime, then the sequence fa n g 1 n¼1 is always convergent. Proof. There are two cases: p 1 >0 or p 1 ¼ 0. Case 1. p 1 >0. We first prove that for any non-negative integer j 5 0, M gþðjþ2þk m gþðjþ2þk 4 ð1 fþðm jkþ1 m jkþ1 Þ By definition, M gþðjþ2þk ¼ a n for some n with g þðj þ 2Þk 4 n < g þðj þ 3Þk. Then by Lemma 2.9, M gþð jþ2þk ¼ a n 4 fm jkþ1 þð1 fþm jkþ1 So M gþð jþ2þk m gþð jþ2þk 4 fm jkþ1 þð1 fþm jkþ1 m jkþ1 ¼ð1 fþðm jkþ1 m jkþ1 Þ Let c ¼dðg 1Þ=ke. Then g þðj þ 2Þk 4 ðc þ j þ 2Þk þ 1 and M ðcþjþ2þkþ1 m ðcþjþ2þkþ1 4 M gþðjþ2þk m gþðjþ2þk Hence Convergence of a linear recursive sequence 57 4 ð1 fþðm jkþ1 m jkþ1 Þ M n m n 4 ð1 fþ bðn 1Þ=ðcþ2Þkc ðm 1 m 1 Þ As 0 < f < 1, lim ðm n m n Þ¼0 n!1 Therefore the sequence fa n g 1 n¼1 is convergent by the definition of M n and m n. Case 2. p 1 ¼ 0. Let j ¼ minfi : p i 6¼ 0, i ¼ 1, 2,, kg. So24j4k. Then a nþkþ1 ¼ Xk p i a n kþi for any integer n 5 0. Define another sequence {b n }, where b n ¼ a nþj 1 n ¼ 1, 2, 3,. Let k 0 ¼ k j þ 1 and p 0 i ¼ p iþj 1 for i ¼ 1, 2,, k 0. Then i¼j for b nþk0 þ1 ¼ a nþk0 þ1þj 1 ¼ a nþkþ1 ¼ Xk p i a n kþi ¼ Xk0 p 0 i b n k 0 þi i¼j for any integer n 5 0. Observe that p 0 i 5 0 for i ¼ 1, 2,, k0, p 0 1 > 0 and p 0 1 þ p0 2 þþp0 k ¼ 1. It is easy to verify that 0 0 ¼fk 0 i þ 1 : p 0 i > 0, i ¼ 1, 2,, k0 g¼ since p 0 i ¼ p iþj 1 and k 0 ¼ k j þ 1. Thus the integers in 0 are coprime. By the result in case 1, the sequence b 1, b 2, is convergent and thus fa n g 1 n¼1 is convergent.

8 58 E. G. Tay et al. Now we assume that the sequence fa n g 1 n¼1 is convergent and we are going to determine d ¼ lim n!1 a n. Theorem 2.2. to d, then If p 1 þ p 2 þþp k ¼ 1 and the sequence fa n g 1 n¼1 is convergent d ¼ lim n!1 a n ¼ P k P i p j a i P k ðk i þ 1Þp i Proof. It is easy to verify that for any n 5 0, X i p j!ða nþiþ1 a nþi Þ¼0 Consider the sum of the above expression for n ¼ 0, 1, 2,, t. Using this result repeatedly, we have X i p j!ða tþiþ1 a i Þ¼0 When t tends to infinity,! X i 0 ¼ lim p j ða tþiþ1 a i Þ¼ Xk t!1 Hence! d Xk X i p j ¼ Xk X i X i p j!ð lim t!1 a tþiþ1 a i Þ p j!a i Observe that So X i p j! ¼ Xk ðk i þ 1Þ p i P k P i p j a i d ¼ P k ðk i þ 1Þp i 3. The integers in are not coprime and p 1 þ p 2 þþp k ¼ 1 For any u 2 and 1 4 i 4 k, define 8 >< 1, if u ¼ 0 Jði, uþ ¼ 1, if i þ r j > k for some j with w j > 0 >: 0, otherwise where u ¼ P t w jr j.

9 Convergence of a linear recursive sequence 59 Lemma 3.1. For any integers n and i: n 5 1 and 1 4 i 4 k, f i ðnþ > 0 if and only if n ¼ u þ i for some u 2 with Jði, uþ ¼1. Proof: (Necessity) Assume that f i ðnþ > 0. If 1 4 n 4 k, then f i ðnþ > 0 if and only if n ¼ i and so the necessity holds. Now let n > k. By Lemma 2.6, f i ðn r j Þ > 0 for some j. By induction, n r j ¼ i þ u 0 for some u 0 2 with Jði, u 0 Þ¼1. Let u ¼ u 0 þ r j. It is easy to verify that Jði, uþ ¼1. Hence the necessity holds. (Sufficiency) Assume that n ¼ u þ i for some u 2 with Jði, uþ ¼1. It is obvious that fðnþ > 0ifu¼0. If u 6¼ 0, then u ¼ P t w jr j and w j > 0 for some j such that i þ r j > k. Let u 0 ¼ u r j. Then n ¼ u 0 þ r j þ i and u 0 2. By Corollary 2.1, f i ðr j þ iþ > 0. By Lemma 2.6 repeatedly, we have f i ðr j þ i þ u 0 Þ > 0, i.e. f i ðnþ > 0. The sufficiency holds. Let q be the largest common factor of integers in. Lemma 3.2. For 1 4 i 4 k, ifn i is not divisible by q, then f i ðnþ ¼0. Proof. Assume that n i is not divisible by q. By Lemma 3.1, we just need to prove that n i 62. Otherwise, n i ¼ Xt for some non-negative integers w j, j ¼ 1, 2,, t. As q is a common factor of r 1, r 2,, r t, n i is divisible by q, a contradiction. By Lemma 3.2, for any integer l 5 0 and s ¼ 1, 2,, q, w j r j a sþlq ¼ bðk sþ=qc X c¼0 f sþcq ðs þ lqþa sþcq ð6þ So the following result is obtained immediately. Theorem 3.1. If p 1 þ p 2 þþp k ¼ 1 and the integers in are not coprime, then the sequence fa n g 1 n¼1 is not always convergent. Proof. Choose special initial values: for 1 4 i 4 k, ifi 1 is divisible by q, then a i ¼ 1, and otherwise, a i ¼ 0. Then by (6), for any n 5 1, if ðn 1Þ is divisible by q, then a n ¼ 1 and otherwise, a n ¼ 0. This shows that fa n g 1 n¼1 is divergent for some initial values. Hence the theorem holds. 4. By algebraic approach In the following, another method is used to prove the convergence of the sequence fa n g 1 n¼1 if p 1 þ p 2 þþp k ¼ 1 and the integers in are coprime. Recall the column vector X n and a matrix B defined in section 2. We have X nþ1 ¼ BX n, n ¼ 1, 2,

10 60 E. G. Tay et al. and thus X nþ1 ¼ B n X 1, n ¼ 1, 2, Observe that B is a stochastic matrix, since each entry is non-negative and the sum of entries in each row of B is 1. We shall prove that if p 1 þ p 2 þþp k ¼ 1 and the integers in are coprime, then all entries in B s are positive for some s, and thus B is a regular stochastic matrix. Construct another k k matrix A ¼ða i, j Þ kk : a i, j ¼!ðb i, j Þ, i ¼ 1, 2,, k, j ¼ 1, 2,, k where!ðxþ ¼1ifx > 0 and!ðxþ ¼0 otherwise. Lemma 4.1. For any integer s 5 1, the entry of B s at (i, j ) is positive if and only if the entry of A s at (i, j ) is positive. Proof. Observe that all entries in both A s and B s are non-negative for all s. By the definition of A, the lemma can be proved by induction on s. Let D be the digraph with vertex set VðDÞ ¼fx 1, x 2,, x k g, and arc set fx i x j : a i, j ¼ 1, i, j ¼ 1, 2,, kg The following is the digraph D when k ¼ 6 and p 1 > 0, p 3 > 0 and p 4 >0. Then A is an adjacency matrix of D. Let A s ¼ða i, j ðsþþ kk. We refer to the following theorem in Graph Theory (see, for example, [1]). Theorem 4.1. For any s 5 1, a i, j ðsþ is the number of walks in D from x i to x j of length s. Lemma 4.2. If p 1 > 0 and all integers in are coprime, then there exists an integer s > 0 such that all entries in B s are positive. Proof. By Lemma 2.4, n 2 for all n 5 g. We shall show that for any i, j with 1 4 i 4 k and 1 4 j 4 k, D has a walk from x i to x j of length g þ 2k. Then the lemma follows from Lemma 4.1 and Theorem 4.1. We first prove that D has a walk from x k to x j of length g þ k for j ¼ 1, 2,, k. For any i ¼ 1, 2,, t, D has a basic circle of length r i, denoted by C i : x k! x k ri þ1! x k ri þ2!!x k For any j with 1 4 j 4 k, we have g þ k j 5 g. Thus g k þ j ¼ Xt v¼1 w v r v

11 for some non-negative integers w v. Then the following walk from x k to x j is of length g þ k: start from x k ; go along circle C v, w v times, for v ¼ 1, 2,, t; finally take the walk x k! x 1! x 2!!x j : We then prove that for any s > g þ k, D has a walk from x k to x j of length s for j ¼ 1, 2,, k. This can be done by induction. By induction, D has a walk P s 1 ðx k, x j 1 Þ from x k to x j 1 of length s 1, where x 0 ¼ x k. So P s 1 ðx k, x j 1 Þ followed by x j 1 x j is a walk from x k to x j of length s. Now let s ¼ g þ 2k and i, j be integers with 1 4 i 4 k and 1 4 j 4 k. By the above result, D has a walk P from x k to x j of length s k þ i, since s k þ i 5 g þ k. Thus the walk x i! x iþ1!!x k followed by P is a walk from x i to x j of length s. By Theorem 4.1, all entries of A gþ2k and hence all entries of B gþ2k are positive. By Lemmas 4.1 and 4.2, B is a regular stochastic matrix. By a theorem on regular stochastic matrices [2], we have (a) B has a unique fixed probability vector t, i.e., t is a row vector satisfying tb ¼ t (b) the sequence B, B 2, B 3, approaches the matrix T whose rows are each the fixed probability vector t. By (b), the sequence fa n g 1 n¼1 approaches tx 1, where t is the unique probability vector satisfying tb ¼ t. Now we are going to determine the probability vector t satisfying tb ¼ t. Let t ¼ðt 1, t 2,, t k Þ. Then this system of equations is t 1 ¼ p 1 t k t iþ1 ¼ t i þ p iþ1 t k, i ¼ 1, 2,, k 1 It is easy to show that Convergence of a linear recursive sequence 61 t i ¼ðp 1 þ p 2 þþp i Þt k for i ¼ 1, 2,, k 1. As t is a probability vector, we have P k t i ¼ 1. Also notice that p 1 þ p 2 þþp k ¼ 1. Thus ð p 1 þþp i Þt k ¼ 1 Hence t k t k ¼ ðk i þ 1Þp i ¼ 1 1 P k ðk i þ 1Þp i

12 62 E. G. Tay et al. implying that for i ¼ 1, 2,, k. Therefore t i ¼ p 1 þ p 2 þþp i P k ðk i þ 1Þp i lim n!1 a n ¼ðt 1, t 2,, t k ÞX 1 ¼ P k P i p j a i P k ðk i þ 1Þp i 5. p 1 þ p 2 þþp k 6¼ 1 Let U ¼ p 1 þ p 2 þþp k. In sections 2, 3 and 4, we study the convergence of fa n g 1 n¼1 under the condition that U ¼ 1. In this section, we consider the other two cases: U > 1 and U<1. Lemma 5.1. For any n 5 1, M n 4 U bðn 1Þ=kc M 1 and m n 5 U bðn 1Þ=kc m 1 Proof. By definition, for any n 5 1, a nþk ¼ Xk p i a nþi 1 4 Xk p i M n ¼ UM n Hence a nþkþi 4 UM nþi 4 UM n for any i 5 0. Thus M nþk ¼ maxfa nþkþi 1 : i ¼ 1, 2,, kg 4 UM n Therefore M n 4 U bðn 1Þ=kc M 1. Similarly, we have m n 5 U bðn 1Þ=kc m 1. Theorem 5.1. If U<1, then the sequence fa n g 1 n¼1 always tends to zero. If U > 1 and a i > 0 for i ¼ 1, 2,, k, then the sequence fa n g 1 n¼1 is divergent. Proof. (1) Assume that U<1. By Lemma 5.1, for any integer n 5 1, U bðn 1Þ=kc m 1 4 m n 4 M n 4 U bðn 1Þ=kc M 1 By Lemmas 2.1 and 2.2, the two sequences {M n } and {m n } are convergent. Hence lim M n ¼ lim m n ¼ 0 n!1 n!1 implying that lim a n ¼ 0. n!1 (2) Assume that U > 1 and a i > 0 for i ¼ 1, 2,, k. By Lemma 5.1, m n 5 U bðn 1Þ=kc m 1 Since m 1 > 0 and U >1, {m n } tends to infinity. Hence the sequence fa n g 1 n¼1 is divergent.

13 Convergence of a linear recursive sequence 63 Remarks (i) It is shown in this paper that fa n g 1 n¼1 is not always convergent if p 1 þ p 2 þþp k ¼ 1 and the integers in are not coprime. A problem arises naturally: what initial values a 1, a 2,, a k can be chosen such that fa n g 1 n¼1 is convergent if fa ng 1 n¼1 is not always convergent? (ii) In this paper, we study the convergence of the sequence fa n g 1 n¼1 under the condition that a i 5 0 for all i ¼ 1, 2,, k. Ifp i <0 for some i, the problem appears quite difficult. We have verified a few cases by Excel, and no patterns were found. We think it is worth studying the problem: find conditions on p 1, p 2,, p k, where p 1 þ p 2 þþp k 5 0 and p i <0 for some i, such that fa n g 1 n¼1 is always convergent. Acknowledgment We thank the referees for their valuable comments. References [1] HARARY, F., 1969, Graph Theory (Reading, MA: Addison-Welsey). [2] LIPSCHUTZ, S., 1983, Schaum s Outline of Theory and Problems of Finite Mathematics (New York: McGraw-Hill). [3] DONGSHENG, Z., YEONG, L. T., SENG, L. C., and FWE, Y. S., 2002, Int. J. Math. Educ. Sci. Technol., 33, 123.