Advances in Difference Equations
Volume 2009 (2009), Article ID 985161, 7 pages
doi:10.1155/2009/985161
Research Article

Convergence Results on a Second-Order Rational Difference Equation with Quadratic Terms

D. M. Chan, C. M. Kent, and N. L. Ortiz-Robinson

Department of Mathematics and Applied Mathematics, Virginia Commonwealth University, Harris Hall, 1015 Floyd Avenue, P.O. Box 842014, Richmond, VA 23284-2014, USA

Received 6 March 2009; Accepted 20 June 2009

Academic Editor: Martin J. Bohner

Copyright © 2009 D. M. Chan et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

We investigate the global behavior of the second-order difference equation 𝑥 𝑛 + 1 = 𝑥 𝑛 − 1 ( 𝛼 𝑥 𝑛 + 𝛽 𝑥 𝑛 − 1 ) / ( 𝐴 𝑥 𝑛 + 𝐵 𝑥 𝑛 − 1 ) , where initial conditions and all coefficients are positive. We find conditions on 𝐴 , 𝐵 , 𝛼 , 𝛽 under which the even and odd subsequences of a positive solution converge, one to zero and the other to a nonnegative number; as well as conditions where one of the subsequences diverges to infinity and the other either converges to a positive number or diverges to infinity. We also find initial conditions where the solution monotonically converges to zero and where it diverges to infinity.

1. Introduction and Preliminaries

There are a number of studies published on second-order rational difference equations (see, e.g., [1–9]). We investigate the global behavior of the second-order difference equation 𝑥 𝑛 + 1 = 𝑥 𝑛 − 1 𝛼 𝑥 𝑛 + 𝛽 𝑥 𝑛 − 1 𝐴 𝑥 𝑛 + 𝐵 𝑥 𝑛 − 1 , ( 1 . 1 ) where the numerator is quadratic and the denominator is linear with 𝐴 , 𝐵 , 𝛼 , 𝛽 ∈ ( 0 , ∞ ) . Under various hypotheses on the parameters, we establish the existence of different behaviors of even and odd subsequences of solutions of (1.1). Our results are summarized below.

(i)Let 𝛼 < 𝐴 and 𝛽 > 𝐵 , then we have the following. (a)There are infinitely many solutions, { 𝑥 𝑛 } ∞ 𝑛 = − 1 , such that for each, one of its subsequences, { 𝑥 2 𝑛 } ∞ 𝑛 = 0 , { 𝑥 2 𝑛 − 1 } ∞ 𝑛 = 0 , converges to zero and the other diverges to infinity.(b)There exist solutions, { 𝑥 𝑛 } ∞ 𝑛 = 0 , which(1)converge to zero if 𝐴 + 𝐵 > 𝛼 + 𝛽 ;(2)diverge to infinity if 𝐴 + 𝐵 < 𝛼 + 𝛽 ;(3)are constant if 𝐴 + 𝐵 = 𝛼 + 𝛽 .(i) Let 𝛼 = 𝐴 and 𝛽 > 𝐵 . Then for each positive solution { 𝑥 𝑛 } ∞ 𝑛 = − 1 , one of the subsequences, { 𝑥 2 𝑛 } ∞ 𝑛 = 0 , { 𝑥 2 𝑛 − 1 } ∞ 𝑛 = 0 , diverges to infinity and the other to a positive number that can be arbitrarily large depending on initial values. Further there, are positive initial values for which the corresponding solution, { 𝑥 𝑛 } ∞ 𝑛 = − 1 , increases monotonically to infinity.(ii)Let 𝛼 < 𝐴 and 𝛽 = 𝐵 . Then for each positive solution { 𝑥 𝑛 } ∞ 𝑛 = − 1 , one of the subsequences, { 𝑥 2 𝑛 } ∞ 𝑛 = 0 , { 𝑥 2 𝑛 − 1 } ∞ 𝑛 = 0 , converges to zero and the other to a nonnegative number. Further, there are positive initial values for which the corresponding solution, { 𝑥 𝑛 } ∞ 𝑛 = − 1 , decreases monotonically to zero.

We note that the following results address and solve the first five conjectures posed by Sedaghat in [10].

2. Results

In order to establish this first result, we reduce (1.1) to a first-order equation by means of the substitution 𝑟 𝑛 = 𝑥 𝑛 / 𝑥 𝑛 − 1 . This transforms (1.1) to 𝑟 𝑛 + 1 = 𝛼 𝑟 𝑛 + 𝛽 𝐴 𝑟 2 𝑛 + 𝐵 𝑟 𝑛 . ( 2 . 1 )

Theorem 2.1. Let 𝛼 < 𝐴 and 𝛽 > 𝐵 in (1.1). Then one has the following. (i)There are infinitely many solutions, { 𝑥 𝑛 } ∞ 𝑛 = − 1 , such that for each, one of its subsequences, { 𝑥 2 𝑛 } ∞ 𝑛 = 0 , { 𝑥 2 𝑛 − 1 } ∞ 𝑛 = 0 , converges to zero and the other to infinity.(ii)There exist solutions, { 𝑥 𝑛 } ∞ 𝑛 = − 1 , which(a)converge to zero if 𝐴 + 𝐵 > 𝛼 + 𝛽 ;(b)diverge to infinity if 𝐴 + 𝐵 < 𝛼 + 𝛽 ; (c)are constant if 𝐴 + 𝐵 = 𝛼 + 𝛽 .

Proof. Starting with (2.1), let the function 𝑔 ∶ ( 0 , ∞ ) → ( 0 , ∞ ) be defined as 𝑔 ( 𝑟 ) = ( 𝛼 𝑟 + 𝛽 ) / ( 𝐴 𝑟 2 + 𝐵 𝑟 ) . Note that for 𝑟 ∈ ( 0 , ∞ ) , 𝑔 ( 𝑟 ) is a decreasing function since 𝑔 ( 𝑟 ) = − ( 𝐴 𝛼 𝑟 2 + 2 𝐴 𝛽 𝑟 + 𝐵 𝛽 ) / ( 𝐴 𝑟 2 + 𝐵 𝑟 ) 2 < 0 . Also note that l i m 𝑟 → 0 + ( 𝑔 ( 𝑟 ) − 𝑟 ) = + ∞ and l i m 𝑟 → + ∞ ( 𝑔 ( 𝑟 ) − 𝑟 ) = − ∞ . Hence 𝑔 has a unique positive fixed point 𝑟 .
We next compute the expression 𝑔 2 ( 𝑟 ) − 𝑟 and simplify, it including canceling the common factor ( 𝐴 𝑟 + 𝐵 ) 𝑟 from the numerator and denominator, thereby obtaining the following:
𝑔 2 𝑎 ( 𝑟 ) − 𝑟 = 4 𝑟 4 + 𝑎 3 𝑟 3 + 𝑎 2 𝑟 2 + 𝑎 1 𝑟 𝑏 3 𝑟 3 + 𝑏 2 𝑟 2 + 𝑏 1 𝑟 + 𝑏 0 , ( 2 . 2 ) where 𝑎 1 = 𝛽 ( 𝐵 𝛼 − 𝐴 𝛽 ) , 𝑏 0 = 𝐴 𝛽 2 , 𝑎 2 = 𝛼 ( 𝐵 𝛼 − 𝐴 𝛽 ) , 𝑏 1 = 2 𝐴 𝛼 𝛽 + 𝐵 2 𝑎 𝛽 , 3 = 𝐵 ( 𝐴 𝛽 − 𝐵 𝛼 ) , 𝑏 2 = 𝐴 𝛼 2 + 𝐴 𝐵 𝛽 + 𝐵 2 𝑎 𝛼 , 4 = 𝐴 ( 𝐴 𝛽 − 𝐵 𝛼 ) , 𝑏 3 = 𝐴 𝐵 𝛼 . ( 2 . 3 ) Note that since 𝐴 𝛽 > 𝐵 𝛼 , 𝑎 1 , 𝑎 2 < 0 and 𝑎 3 , 𝑎 4 > 0 . Thus the numerator of 𝑔 2 ( 𝑟 ) − 𝑟 = 0 has one and only one sign change. Therefore, by Descartes' rule of signs, the numerator of 𝑔 2 ( 𝑟 ) − 𝑟 = 0 has exactly one positive root, 𝑟 .
In addition, we see that l i m 𝑟 → + ∞ [ 𝑔 2 ( 𝑟 ) − 𝑟 ] = + ∞ and so, given that 𝑟 is the only positive root of the numerator of 𝑔 2 ( 𝑟 ) − 𝑟 = 0 , we have 𝑔 2 ( 𝑟 ) − 𝑟 > 0 for 𝑟 > 𝑟 . Thus, since 𝑔 2 ( 0 ) = 0 and 𝑔 2 is continuous, we must have 𝑔 2 ( 𝑟 ) − 𝑟 < 0 for 𝑟 < 𝑟 . Therefore,
𝑔 2 ( 𝑟 ) − 𝑟 𝑟 − 𝑟 > 0 f o r 𝑟 ≠ 𝑟 . ( 2 . 4 ) We consider two cases depending on the initial value 𝑟 0 for (2.1).Case 1 ( 𝑟 0 ∈ ( 0 , 𝑟 ) ). Using induction and the fact that 𝑔 is a decreasing function so that 𝑔 2 is an increasing function, we have 0 < ⋯ < 𝑔 4 𝑟 0 < 𝑔 2 𝑟 0 < 𝑟 0 < 𝑟 𝑟 < 𝑔 0 < 𝑔 3 𝑟 0 < 𝑔 5 𝑟 0 ⋯ . ( 2 . 5 ) Thus, l i m 𝑛 → ∞ 𝑔 2 𝑛 ( 𝑟 0 ) ≥ 0 a n d l i m 𝑛 → ∞ 𝑔 2 𝑛 + 1 ( 𝑟 0 ) ≤ ∞ . Since 𝑟 is the only positive fixed point of 𝑔 2 , then we must have l i m 𝑛 → ∞ 𝑔 2 𝑛 ( 𝑟 0 ) = 0 and l i m 𝑛 → ∞ 𝑔 2 𝑛 + 1 ( 𝑟 0 ) = ∞ . Case 2 ( 𝑟 0 ∈ ( 𝑟 , ∞ ) ). The argument is similar to that in Case 1 in showing l i m 𝑛 → ∞ 𝑔 2 𝑛 ( 𝑟 0 ) = ∞ and l i m 𝑛 → ∞ 𝑔 2 𝑛 + 1 ( 𝑟 0 ) = 0 . In both cases, the solution, { 𝑟 𝑛 } ∞ 𝑛 = 0 , of (2.1) is divided into even and odd subsequences, { 𝑟 2 𝑛 } ∞ 𝑛 = 0 and { 𝑟 2 𝑛 + 1 } ∞ 𝑛 = 0 , where one subsequence converges monotonically to zero and the other to infinity.We now go back to (1.1) by inferring the behavior of 𝑥 𝑛 from 𝑟 𝑛 . To do this we first consider 𝑟 0 ≠ 𝑟 . Without loss of generality, we will assume that 0 < 𝑟 0 < 𝑟 and so l i m 𝑛 → ∞ 𝑔 2 𝑛 ( 𝑟 0 ) = ∞ and l i m 𝑛 → ∞ 𝑔 2 𝑛 + 1 ( 𝑟 0 ) = 0 .
Next, observe that
𝑥 2 𝑛 + 2 𝑥 2 𝑛 = 𝑥 2 𝑛 + 2 𝑥 2 𝑛 + 1 ⋅ 𝑥 2 𝑛 + 1 𝑥 2 𝑛 = 𝑟 2 𝑛 + 2 𝑟 2 𝑛 + 1 = 𝛼 𝑟 2 𝑛 + 1 + 𝛽 𝐴 𝑟 2 2 𝑛 + 1 + 𝐵 𝑟 2 𝑛 + 1 ⋅ 𝑟 2 𝑛 + 1 = 𝛼 𝑟 2 𝑛 + 1 + 𝛽 𝐴 𝑟 2 𝑛 + 1 + 𝐵 . ( 2 . 6 ) From this and our assumption with 𝑔 2 𝑛 + 1 , we have l i m 𝑛 → ∞ 𝑥 2 𝑛 + 2 𝑥 2 𝑛 = l i m 𝑛 → ∞ 𝛼 𝑟 2 𝑛 + 1 + 𝛽 𝐴 𝑟 2 𝑛 + 1 = 𝛽 + 𝐵 𝐵 > 1 . ( 2 . 7 ) Hence, for 0 < 𝜖 < 𝛽 / 𝐵 − 1 , there exists 𝑁 ≥ 0 such that 𝛽 1 < 𝐵 𝑥 − 𝜖 < 2 𝑛 + 2 𝑥 2 𝑛 < 𝛽 𝐵 + 𝜖 ( 2 . 8 ) for all 𝑛 ≥ 𝑁 . We then have 𝑥 2 ( 𝑁 + 1 ) > 𝛽 𝐵 − 𝜖 1 𝑥 2 𝑁 𝑥 2 ( 𝑁 + 2 ) > 𝛽 𝐵 − 𝜖 1 𝑥 2 ( 𝑁 + 1 ) > 𝛽 𝐵 − 𝜖 2 𝑥 2 𝑁 𝑥 2 ( 𝑁 + 3 ) > 𝛽 𝐵 − 𝜖 1 𝑥 2 ( 𝑁 + 2 ) > 𝛽 𝐵 − 𝜖 3 𝑥 2 𝑁 ( 2 . 9 ) and by induction, for 𝑚 ≥ 1 , 𝑥 2 ( 𝑁 + 𝑚 ) > 𝛽 𝐵 − 𝜖 𝑚 𝑥 2 𝑁 . ( 2 . 1 0 ) This, in turn, implies that l i m 𝑛 → ∞ 𝑥 2 𝑛 + 2 = ∞ . ( 2 . 1 1 ) The argument is similar in showing that l i m 𝑛 → ∞ 𝑥 2 𝑛 + 1 = 0 , since 𝑥 2 𝑛 + 1 𝑥 2 𝑛 − 1 = 𝑥 2 𝑛 + 1 𝑥 2 𝑛 ⋅ 𝑥 2 𝑛 𝑥 2 𝑛 − 1 = 𝑟 2 𝑛 + 1 𝑟 2 𝑛 = 𝛼 𝑟 2 𝑛 + 𝛽 𝐴 𝑟 2 2 𝑛 + 𝐵 𝑟 2 𝑛 ⋅ 𝑟 2 𝑛 = 𝛼 𝑟 2 𝑛 + 𝛽 𝐴 𝑟 2 𝑛 + 𝐵 . ( 2 . 1 2 ) Hence, result (i) is true.
Now consider 𝑟 0 = 𝑟 . Then 𝑟 𝑛 = 𝑟 for all 𝑛 ≥ 1 , and so 𝑥 𝑛 / 𝑥 𝑛 − 1 = 𝑟 for all 𝑛 ≥ 1 . Induction then gives us 𝑥 𝑛 = 𝑟 𝑛 + 1 𝑥 − 1 for all 𝑛 ≥ 1 . We thus have one of the following:
(1)If 𝑟 < 1 ( 𝐴 + 𝐵 > 𝛼 + 𝛽 ), then l i m 𝑛 → ∞ 𝑥 𝑛 = 0 . (2)If 𝑟 > 1 ( 𝐴 + 𝐵 < 𝛼 + 𝛽 ), then l i m 𝑛 → ∞ 𝑥 𝑛 = ∞ . (3)If 𝑟 = 1 ( 𝐴 + 𝐵 = 𝛼 + 𝛽 ), then { 𝑥 𝑛 } ∞ 𝑛 = − 1 is a constant solution 𝑥 − 1 = 𝑥 0 = 𝑥 1 = ⋯ . Thus the result (ii) is true and this completes the proof.

For the next couple of results we rewrite (1.1) in the form 𝑥 𝑛 + 1 𝑥 = 𝑓 𝑛 , 𝑥 𝑛 − 1 , 𝑛 = 0 , 1 , … . ( 2 . 1 3 ) Note that if either 𝛼 ≤ 𝐴 and 𝛽 < 𝐵 , or 𝛼 < 𝐴 and 𝛽 ≤ 𝐵 , then 𝑓 satisfies the following properties:

(P1) 𝑓 ∈ 𝐶 [ [ 0 , ∞ ) 2 − { 0 , 0 } , [ 0 , ∞ ) ] , with 𝑓 ( 𝑢 , 𝑣 ) undefined when 𝑢 = 𝑣 = 0 .(P2) 𝑓 ∈ 𝐶 [ [ 0 , ∞ ) × ( 0 , ∞ ) , ( 0 , ∞ ) ] (P3) 𝑓 ( 𝑢 , 𝑣 ) < 𝑣 if 𝑢 , 𝑣 ∈ ( 0 , ∞ ) .

If we consider the addition restriction that 𝛼 < 𝐴 and 𝛽 = 𝐵 , we also obtain

(P4)if 𝑓 ( 𝑢 , 𝑣 ) = 𝑣 , then 𝑢 = 0 , 𝑣 > 0 , or 𝑢 > 0 , 𝑣 = 0 .

Lemma 2.2. Let { 𝑥 𝑛 } ∞ 𝑛 = − 1 be a positive solution of (1.1) with 𝛼 < 𝐴 and 𝛽 = 𝐵 . Then there exist 𝐿 𝑜 ≥ 0 and 𝐿 𝑒 ≥ 0 such that the following statements are true: (1) 𝑥 2 𝑛 − 1 ↓ 𝐿 𝑜 as 𝑛 → ∞ ,(2) 𝑥 2 𝑛 ↓ 𝐿 𝑒 as 𝑛 → ∞ ,(3) 𝐿 𝑜 = 𝐿 𝑒 = 0 , and 𝑓 ( 𝐿 𝑜 , 𝐿 𝑒 ) and 𝑓 ( 𝐿 𝑒 , 𝐿 𝑜 ) are undefined; or if either 𝐿 𝑜 or 𝐿 𝑒 is not zero, then ( 𝐿 𝑜 , 𝐿 𝑒 , 𝐿 𝑜 , 𝐿 𝑒 , … ) is a solution of (1.1).(4) 𝐿 𝑜 ⋅ 𝐿 𝑒 = 0 .

Proof. Statements 1 and 2 follow from the fact that 0 < 𝑥 2 𝑛 + 1 𝑥 = 𝑓 2 𝑛 , 𝑥 2 𝑛 − 1 < 𝑥 2 𝑛 − 1 , 0 < 𝑥 2 𝑛 + 2 𝑥 = 𝑓 2 𝑛 + 1 , 𝑥 2 𝑛 < 𝑥 2 𝑛 ( 2 . 1 4 ) by properties (P2) and (P3). Statement 3 follows from the fact that either 𝐿 𝑜 = 𝐿 𝑒 = 0 , and so 𝑓 ( 𝐿 𝑜 , 𝐿 𝑒 ) and 𝑓 ( 𝐿 𝑒 , 𝐿 𝑜 ) are undefined by property (P1); or 𝐿 𝑜 ≠ 𝐿 𝑒 and 𝐿 𝑜 = l i m 𝑛 → ∞ 𝑥 2 𝑛 + 1 = l i m 𝑛 → ∞ 𝑓 𝑥 2 𝑛 , 𝑥 2 𝑛 − 1 𝐿 = 𝑓 𝑒 , 𝐿 𝑜 𝐿 𝑒 = l i m 𝑛 → ∞ 𝑥 2 𝑛 + 2 = l i m 𝑛 → ∞ 𝑓 𝑥 2 𝑛 + 1 , 𝑥 2 𝑛 𝐿 = 𝑓 𝑜 , 𝐿 𝑒 , ( 2 . 1 5 ) where Statements 1 and 2 and the continuity of 𝑓 (Property (P1) hold. Finally, Statement 4 follows immediately from Statement 3 and Property (P4).

In the first three results, we characterize the convergence of the odd and even subsequences of solutions of (1.1).

Theorem 2.3. Let 𝛼 < 𝐴 and 𝛽 = 𝐵 in (1.1). Then for each positive solution, { 𝑥 𝑛 } ∞ 𝑛 = − 1 , one of the subsequences, { 𝑥 2 𝑛 } ∞ 𝑛 = 0 , { 𝑥 2 𝑛 − 1 } ∞ 𝑛 = 0 , converges to zero and the other to a nonnegative number.

Proof. Consider (1.1) with 𝛼 < 𝐴 , 𝛽 = 𝐵 , and 𝑓 ( 𝑢 , 𝑣 ) = 𝑣 ( ( 𝛼 𝑢 + 𝛽 𝑣 ) / ( 𝐴 𝑢 + 𝐵 𝑣 ) ) . Then it follows from Lemma 2.2 that for each positive solution of (1.1), { 𝑥 𝑛 } ∞ 𝑛 = − 1 , one of the subsequences, { 𝑥 2 𝑛 } ∞ 𝑛 = 0 , { 𝑥 2 𝑛 − 1 } ∞ 𝑛 = 0 , converges to zero and the other to a nonnegative number.

Theorem 2.4. Let 𝛼 = 𝐴 and 𝛽 > 𝐵 in (1.1). Then for each positive solution { 𝑥 𝑛 } ∞ 𝑛 = − 1 , one of the subsequences, { 𝑥 2 𝑛 } ∞ 𝑛 = 0 , { 𝑥 2 𝑛 − 1 } ∞ 𝑛 = 0 , diverges to infinity and the other to a positive number or diverges to infinity.

Proof. Consider (1.1) with 𝛼 = 𝐴 and 𝛽 > 𝐵 . Using the transformation 𝑦 𝑛 = 1 / 𝑥 𝑛 , convert (1.1) to the equation 𝑦 𝑛 + 1 = 𝑦 𝑛 − 1 𝐵 𝑦 𝑛 + 𝐴 𝑦 𝑛 − 1 𝛽 𝑦 𝑛 + 𝛼 𝑦 𝑛 − 1 . ( 2 . 1 6 ) Then 𝑓 ( 𝑢 , 𝑣 ) = 𝑣 ( ( 𝐴 𝑣 + 𝐵 𝑢 ) / ( 𝛼 𝑣 + 𝛽 𝑢 ) ) , and so it follows from Lemma 2.2 that for each positive solution of (2.16), { 𝑦 𝑛 } ∞ 𝑛 = − 1 , one of the subsequences, { 𝑦 2 𝑛 } ∞ 𝑛 = 0 , { 𝑦 2 𝑛 − 1 } ∞ 𝑛 = 0 , converges to zero and the other to a nonnegative number. Hence, for each positive solution of (1.1), { 𝑥 𝑛 } ∞ 𝑛 = − 1 , one of the subsequences, { 𝑥 2 𝑛 } ∞ 𝑛 = 0 , { 𝑥 2 𝑛 − 1 } ∞ 𝑛 = 0 , diverges to infinity and the other to a positive number or diverges to infinity.

In the following results, we show the existence of monotonic solutions for (1.1). As with Theorem 2.1 we use the substitution 𝑟 𝑛 = 𝑥 𝑛 / 𝑥 𝑛 − 1 .

Theorem 2.5. Let 𝛼 < 𝐴 and 𝛽 = 𝐵 in (1.1). Then there are positive initial values for which the corresponding solutions, { 𝑥 𝑛 } ∞ 𝑛 = − 1 , decrease monotonically to zero.

Proof. Note that an equilibrium equation for (2.1) satisfies, 𝐴 𝑟 3 + 𝐵 𝑟 2 − 𝛼 𝑟 − 𝛽 = 0 . ( 2 . 1 7 ) Set 𝑝 ( 𝑟 ) = 𝐴 𝑟 3 + 𝐵 𝑟 2 − 𝛼 𝑟 − 𝛽 . Given Descartes' rule of signs, we have that there exists a unique positive equilibrium, 𝑟 < 1 , where 𝑝 ( 0 ) < 0 and 𝑝 ( 1 ) > 0 . Recall that 𝑟 𝑛 = 𝑥 𝑛 / 𝑥 𝑛 − 1 , and let 𝑟 𝑛 = 𝑟 for all 𝑛 ≥ 0 . Then 𝑥 𝑛 / 𝑥 𝑛 − 1 = 𝑟 for all 𝑛 ≥ 0 . It follows from induction that 𝑥 𝑛 = 𝑟 𝑛 + 1 𝑥 − 1 for all 𝑛 ≥ 0 . Since 𝑟 < 1 , { 𝑥 𝑛 } ∞ 𝑛 = − 1 , with 𝑥 0 = 𝑟 𝑥 − 1 , decreases monotonically to zero.

Theorem 2.6. Let 𝛼 = 𝐴 and 𝛽 > 𝐵 in (1.1). Then there are positive initial values for which the corresponding solution, { 𝑥 𝑛 } ∞ 𝑛 = − 1 , increases monotonically to infinity.

Proof. As in the previous proof, an equilibrium equation for (2.1) satisfies (2.17). Setting 𝑝 ( 𝑟 ) = 𝐴 𝑟 3 + 𝐵 𝑟 2 − 𝛼 𝑟 − 𝛽 , we obtain from Descartes' rule of signs, a unique positive equilibrium, 𝑟 > 1 , where 𝑝 ( 0 ) < 0 and l i m 𝑟 → ∞ 𝑝 ( 𝑟 ) > 0 . Recall that 𝑟 𝑛 = 𝑥 𝑛 / 𝑥 𝑛 − 1 , and let 𝑟 𝑛 = 𝑟 for all 𝑛 ≥ 0 . Then 𝑥 𝑛 / 𝑥 𝑛 − 1 = 𝑟 for all 𝑛 ≥ 0 . It follows from induction that 𝑥 𝑛 = 𝑟 𝑛 + 1 𝑥 − 1 for all 𝑛 ≥ 0 . Since 𝑟 > 1 , { 𝑥 𝑛 } ∞ 𝑛 = − 1 , with 𝑥 0 = 𝑟 𝑥 − 1 , increases monotonically to infinity.

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