For any real number a and for any positive integer n define the sequence (x k ) k ≥0 by x 0 =0 , x 1 =1 , and

1.3.5. For any real number a and for any positive integer n define the sequence (x k ) k ≥0 by x 0 =0 , x 1 =1 , and

= k − (n − k)x k k +2 , for all k ≥ 0.

cx +1

k +1

Fix n and let c be the largest real number such that x n +1 =0 . Find x k in terms of n and k , 1 ≤k≤n .

Solution. We first observe that x n +1 is a polynomial of degree n in c. Thus, it is enough to find n values of c such that x n +1 = 0. We will prove that these values are

c = n − 1 − 2r, for r = 0,1,...,n − 1. In this case, x k is the coefficient of t k −1 in the polynomial f

(t) = (1 −t) r (1 + t) n −1−r . This property follows after observing that f satisfies the identity

f ′ (t)

n −1−r

that is, (1 − t 2 )f ′ (t) = f (t)[(n − 1 − r)(1 − t) − r(1 +t)] = f (t)[(n − 1 − 2r) − (n − 1)t].

Identifying the coefficients of t k in both sides, we obtain

(k + 1)x k +2 − (k − 1)x k = (n − 1 − 2r)x k +1 − (n − 1)x k .

In particular, the largest c is n − 1, and x k =C k −1 n −1 , for k = 1, 2, . . . , n. ⊓ ⊔ The next problem is devoted to the study of the normalized logistic equation,

which is a successful model of many phenomena arising in genetics and mathemati- cal biology. In its simplest form, the logistic equation is a formula for approximating the evolution of an animal population over time. The unknown a n in the following recurrent sequence represents the number of animals after the nth year. It is easy to observe that the sequence (a n ) n ≥1 converges to zero. The interesting part of the problem is to deduce the first-and second-order decay terms of this sequence.

1.3.6. Consider the sequence (a n ) n ≥1 satisfying a 1 ∈ (0,1) and a n +1 =a n (1−a n ) , for all n ≥1 . Show that lim n → ∞ na n =1 and lim n → ∞ n (1 − na n )/ ln n = 1 .

Solution. We first prove that a n < 1/(n + 1), for all n ≥ 2. For this purpose we use the fact that the mapping f (x) = x(1 −x) is increasing on (0,1/2] and decreasing on [1/2, 1). Thus, a 2 =a 1 (1 −a 1 ) ≤ 1/4 < 1/3. Let us now assume that a n < 1/(n + 1). Hence

a n +1 =a n (1 − a n )≤

n +1

(n + 1) 2

n +2

22 1 Sequences We observe that the sequence (a n ) n ≥1 is decreasing, because a n +1 −a n =−a 2 n < 0.

Moreover, (a n ) n ≥1 is bounded and lim n → ∞ a n = 0. Let c n = 1/a n . We compute

lim n → ∞ c n /n = lim n → ∞ 1 /(na n ). We have

n →∞ a n a n (1 − a n ) n →∞ 1 −a n So, by the Stolz–Ces`aro lemma, lim n → ∞ c n /n = 1. Therefore lim n → ∞ na n = 1.

We provide the following alternative proof to this result. Let us first observe that (n + 1)a +1 = na +a

n − (n + 1)a n = na n +a n (1 − (n + 1)a n ). (1.4) To see that (na n ) n ≥1 is increasing, we need to show that 1 − (n + 1)a n ≥ 0. From the

graph of y = x(1 −x) we note that a 2 ≤ 1/4 and a n ≤ a ≤ 1/2 imply a n +1 ≤ a(1−a). So, by induction,

1 1 1 (n + 1)a n ≤ (n + 1)

≤ 1. Furthermore, na n < (n + 1)a n ≤ 1, and so (na n ) n ≥1 is bounded above by 1. Thus

na n converges to a limit ℓ with 0 < na n < ℓ ≤ 1. Now summing (1.4) from 2 to n, we obtain

1 ≥ (n+1)a n +1 = 2a 2 +a 2 (1 −3a 2 )+a 3 (1 −4a 3 ) + ···+a n (1 −(n+1)a n ) . (1.5) If n ) ≥ (1 − ℓ)/2 for all large n, and thus relation (1.5)

shows that the series ∑ ∞ n =1 a n is convergent. However, na

n ≥a 1 and so ∑

n =1 a n ≥

a 1 ∞ ∑ n =1 1 /n. But ∑ n =1 1 /n is divergent. This contradiction shows that ℓ = 1. For the second part, set b n = 1/a n . Then

Using this recurrence relation, a standard induction argument implies that b n ≥ n, for all n ≥ 0. Thus, by (1.6),

≤b 1 +n+ ∑ ≤ n + lnn +C ,

k =1 k

where C is a real constant [we have used here the fact that 1 + 1/2 + ···+ 1/n − lnn tends to a finite limit]. This in turns proves the existence of a constant C 0 such that for all integer n ≥ 1,

≥b 1 +n+ ∑

≥ n + lnn +C 0 . (1.8)

k =1 k + ln(k + 1) + C

1.3 Recurrent Sequences 23 It follows from (1.7) and (1.8) that

= 1 + o(1) as n→∞ ln n

= 1 + o(1) as n→∞. ln n

b n ln n

(n + lnn + C) lnn

This concludes the proof. ⊓ ⊔ Independent Study. Find a sequence (x n ) n ≥1 such that

The following problem is difficult, and it offers a curious comparison property with respect to the Fibonacci sequence.

1.3.7. Fix x 1 ∈ [0,1) and define the sequence (x n ) n ≥1 by x n +1 =0 if x n =0 and x n +1 = 1 x n − [1/x n ] . Prove that for all integers n ≥1 ,

k =1

k =1 F k +1

where (F n ) n ≥1 is the Fibonacci sequence. USA, Proposed to the 33rd International Mathematical Olympiad, 1992 Solution. Consider the function f (x) = (x + 1) −1 , and for any positive integer n,

define the mapping g n (x) = x+ f (x)+ f 2 (x)+ ···+ f n (x), where f n is the nth iterate of f , that is, f n = f ◦ ··· ◦ f (n times). We easily check the following properties:

(i) for all 0 ≤ x < y ≤ 1 we have 0 < f (x) − f (y) < y − x; (ii) g n is increasing in [0, 1]; (iii) F 1 /F 2 = 1, f (F n /F n +1 )=F n +1 /F n +2 and g n −1 (1) = ∑ n k =1 F k /F k +1 , for all pos- itive integers n.

We just remark that for proving (i) we use

g n (x) − g n (y) = (x − y) + [f (x) − f (y)] + ··· + [ f n (x) − f n (y)] combined with the fact that any difference is less in modulus than and of opposite

sign with respect to the previous one [see (i)]. If for some 2 ≤ k < n we have x k = 0, then by definition, x n = 0, and we conclude the proof by induction with respect to the first n − 1 terms of the sequence. If this

24 1 Sequences does not occur, then for all 2 ≤ k ≤ n, we may write x k −1 = (a k +x k ) −1 , where

a k > 0 is the integer part of x −1 k −1 . Hence

1 1 ···an−1+ an+xn

We prove by induction with respect to k that for any fixed x n ∈ [0,1), the right- hand side of the above equality is maximum if and only if a k = 1, for all k. We first observe that a 2 must be 1, since a 2 appears only in the last expression. Next, let us assume that k > 2 and that for any values of x n ,a n ,a n −1 ,...,a k +1 , the above expression attains its maximum for a k −1 =a k −2 = ··· = a 2 = 1. Observe that only the last k − 1 terms on the right-hand side contain a k , and moreover, their sum is

g k −2 ⎜

...an−1+ 1 an+xn

By (ii), the function g k −2 is increasing, so its maximum is achieved for a k = 1. Using now (ii) and (iii), we obtain

n −1 (x) < g n −1 (1) = ∑

k =1 F k +1

This concludes our proof. ⊔ ⊓

A simple linear recurrence generates a sequence of perfect squares, as shown below.