Let A a positive real number. Find all the possible values of ∞ ∑ x 2 j =0 j : = lim n

1.4.15. Let A a positive real number. Find all the possible values of ∞ ∑ x 2 j =0 j : = lim n

j =0 x j , where x 0 ,x 1 ,...,x n ,... are positive numbers satisfying ∑ j =0 x j =A .

Solution. We prove that the given expression may take any value in (0, A 2 ). We first observe that

= ∑ x j +2 ∑ x j x k .

j =0

j =0

0 ≤ j<k≤n

This implies that ∑ n j =0 x 2 j ≤A 2 − 2x 0 x 1 . Passing at the limit as n →∞ we obtain ∑ ∞ x 2 ≤A 2 − 2x

0 x 1 <A j 2 =0 j . Next, we prove that the expression can achieve any value in (0, A 2 ). Indeed, let (x

n ) n ≥0

be a geometric progression of ratio d. In this case, ∑ j =0 x j =x 0 /(1 − d) and

2 x 0 1 ∑ −d x

1 −d 2 1 +d ∑ j =0

j =0

If d grows from 0 to 1, then (1 − d)/(1 + d) decreases from 1 to 0. Consequently, if

(x ∞ n ) n ≥0 is a geometric progression of positive numbers satisfying ∑ j =0 x j = A,

then ∞ ∑ j =0 x 2 j takes values between 0 and A 2 . ⊓ ⊔

The terms of a sequence defined by a polynomial recurrence satisfy an interesting property. Our solution relies on a fundamental property of polynomials with integer coefficients.

1.4.16. Let f (x)

be a polynomial with integer coefficients. Define the sequence (a n ) n ≥0 by a 0 =0 and a n +1 = f (a n ) , for all n ≥0 . Prove that if there exists m ≥1

such that a m =0 , then either a 1 =0 or a 2 =0 .

Solution. We first recall that if f (x) is a polynomial with integer coefficients then m − n is a divisor of f (m) − f (n), for any different integers m and n. In particular, if b n =a n +1 −a n , then b n is a divisor of b n +1 , for any n. On the

other hand, a 0 =a m = 0, so a 1 =a m +1 , which implies b 0 =b m . If b 0 = 0, then

a 0 =a 1 = ··· = a m , and the proof is concluded. If not, then |b 0 | = |b 1 | = |b 2 | = ··· , so b n = ±b 0 , for all n. Using b 0 +...+b m −1 =a m −a 0 = 0 we deduce that half of the integers

b 0 ,...,b m −1 are positive. In particular, there exists an integer 0 < k < m such that

b k −1 = −b k , that is, a k −1 =a k +1 . It follows that a n =a n +2 , for all n ≥ k − 1. Taking m = n, we deduce that a 0 =a m =a m +2 = f ( f (a 0 )) = a 2 . ⊓ ⊔

For the next exercise the reader must be familiar with the properties of the scalar product of vectors.

such that the sequence a n = |x − ny| ( n ≥1 ) is (a) increasing; (b) decreasing.

1.4.17. Find all vectors x , y

1.4 Qualitative Results 41 Solution. We have x 2 2

2 (x · y) n = |x| − 2n(x· y) + n |y| = |y|

The nonconstant term of these quantities contains (n − c) 2 . This shows that the se- quence is never decreasing and the sequence is increasing if and only if c ≤ 3/2. Consequently, the sequence is increasing if and only if

An interesting convergent rearrangement is possible, provided a sequence of pos- itive real numbers has a finite nonzero accumulation point.

be a sequence of positive real numbers that has a finite nonzero accumulation point. Prove that (a n ) n ≥1 can be arranged in a sequence (x /n

1.4.18. Let (a n ) n ≥1

n ) n ≥1 so that 1 (x n ) n ≥1 is convergent. Solution. Let a ∈ (0,∞) be an accumulation point of (a n ) n ≥1 and choose α >0 such that α −1 <a< α . Let x 1 denote the a n of smallest subscript that lies in the interval ( α −1 , α ). Assume that x 1 ,x 2 ,...,x k −1 have been chosen. Then we define x k =a n , where n is the smallest subscript different from the k √ √ − 1 already cho- sen lying in the interval ( α − n , α n √ ). Such a subscript does exist, since a is an √ accumulation point. Moreover, by α n

→0 as n→∞, it follows that every a n belongs to a certain interval ( α − n , α n ) and will therefore become part √ √

→∞ and n α −

of the sequence √ (x n ) n ≥1 . For each n we have α − n √ √ <x n < α n and therefore (1/ α ) 1 / n

<x n < α 1 / n . Since lim n → ∞ α 1 / n = lim n → ∞ (1/ α ) 1 / n = 1, we con- clude that lim /n 1 n → ∞ x n = 1. ⊓ ⊔

We are now concerned with the maximum possible values of the terms of a se- quence satisfying a certain assumption.

1.4.19. Let (a n ) n ≥0

be a sequence of nonnegative numbers with a 0 =1 and sat- isfying ∞ ∑ k =n a k ≤ ca n , for all n ≥0 , where c >1 is a prescribed constant. Find the

maximum of a n in terms of c .

Solution. For any integer n ≥ 1, the maximum of a n is (c − 1) n /c n −1 . Indeed, since a n ≥ 0, we have

ca n −1 ≥ ∑ a k ≥a n −1 +a n ,

k =n−1

so a n −1 ≥a n /(c − 1). Similarly, we obtain

ca n −2 ≥a n −2 +a n −1 +a n ≥a n −2 + +1 a ,

c −1 n

42 1 Sequences so a n −2 ≥ ca n /(c − 1) 2 . With a similar argument we obtain, by induction,

n −k ≥c −1 a n /(c − 1) , for all 1 ≤ k ≤ n. For k = n we have 1 = a

n /(c − 1) , so a n ≤ (c − 1) /c −1 . For n ≥ 1 fixed, this upper bound is achieved by the sequence (a k ) k ≥0 defined by

, provided that 1 ≤ k < n. It follows that ∞ ∑ k =m a k ≤ ca m for m ≥ n, while for m < n we have

= ca m .⊓ ⊔ We know what monotone sequences are. What about monotone sequences start-

∑ a k =a n + ∑ k = c c n −1 −c c n −

k =m k =m

ing with a certain rank? More precisely, we are concerned with sequences of real numbers (x n ) n ≥0 such that (x n ) n ≥N is monotone for some positive integer N.