Find all positive real numbers a and b such that the limit of every sequence (x n ) n ≥1 satisfying lim n → ∞ (ax n +1 − bx n )=0 equals zero.

1.4.23. Find all positive real numbers a and b such that the limit of every sequence (x n ) n ≥1 satisfying lim n → ∞ (ax n +1 − bx n )=0 equals zero.

Solution. We prove that this property holds if and only if a > b. Indeed, if a < b then the sequence x n = (b/a) n satisfies the hypothesis but does not converge to 0. If a = b, a counterexample is offered by the sequence x n = 1 + 1/2 + ···+ 1/n.

Next, we assume that a > b. Let ℓ − (resp., ℓ + ) be the limit inferior (resp., limit superior) of the given sequence. Now we pass to the “lim sup” in our hypothesis, using the relation lim sup n → ∞ (a n +b n ) ≥ limsup n → ∞ a n + lim sup n → ∞ b n .It follows

44 1 Sequences that

0 = lim

∞ (ax n +1 − bx n ) ≥ limsup (ax n +1 ) + lim sup (−bx n → ) n → ∞ n → ∞ = aℓ + − liminf

(bx n ) = aℓ + − bℓ − .

Thus, ℓ + ≤ (b/a)ℓ − . Since ℓ − ≤ℓ + , it follows that ℓ + ≤ (b/a)ℓ + , and so ℓ − ≤ ℓ + ≤ 0. Similarly, passing at the “liminf,” we obtain ℓ − ≥ (b/a)ℓ + . Since ℓ + ≥ℓ − , we deduce that ℓ − ≥ (b/a)ℓ − . Thus, ℓ + ≥ℓ − ≥ 0. Therefore ℓ + =ℓ − = 0, and the sequence converges to 0. ⊓ ⊔

1.4.24. Let (a n ) n ≥0

be a sequence of nonnegative numbers such that a 2k −a 2k +1 ≤

a 2 k , a 2k +1 −a 2k +2 ≤a k a k +1 for any k ≥0 and lim sup n → ∞ na n < 1/4 . Prove that lim sup

√ n n → ∞ a n <1 .

Solution. Let c l = sup n ≥2 l (n + 1)a n for l ≥ 0. We show that c l +1 ≤ 4c 2 l . Indeed, for any integer n

≥2 +1 l there exists an integer k ≥2 l such that n = 2k or n = 2k + 1. In the first case we have

+1 2k +2 while in the second case we obtain

(k + 1)(k + 2)

2k

Hence a sequence n − 4c 2 l (n + 1) −1 n ≥2 l +1 is nondecreasing and its terms are nonpositive, since it converges to 0. Therefore a n ≤ 4c 2 l l +1 /n + 1 for n ≥ 2 , meaning that c 2 l +1 ≤ 4c 2 l . This implies that a sequence ((4c l ) 2 −l ) l ≥0 is nonincreasing and therefore bounded from above by some number q ∈ (0,1), since all its terms except finitely many are less than 1. Hence c l ≤q 2 l for l large enough. For any n between 2 l

and 2 l +1 there is a n ≤c l /n + 1 ≤ q 2 l

√ ≤( q ) n , yielding lim sup

√ n ≤ q < 1,

which ends the proof. ⊓ ⊔ Leopold Kronecker (1823–1891) was a German mathematician with deep contri-

butions in number theory, algebra, and analysis. One of his famous quotations is the following: “God made the integers; all else is the work of man.” The next density property has a crucial importance for the understanding of the set of real numbers.

1.4.25. (Kronecker’s Theorem). Let α be an irrational real number. Prove that the set A = {m + n α ;m , n ∈ Z} is dense in R .

Solution. We recall that a set A ⊂ R is dense in R if for all ε > 0 and for any x ∈ R, there exists a ∈ A such that |x − a| < ε . In other words, the set A ⊂ R is dense in R if for all x ∈ R there exists a sequence (a n ) n ≥1 in A such that a n →x as n→∞.

We first prove that there exists a sequence (a n ) n ≥1 in A such that a n →0 as n→∞. For this purpose, consider the numbers { α }, {2 α },..., {(n + 1) α }, where n ≥ 1 is

1.5 Hardy’s and Carleman’s Inequalities 45

an arbitrary integer. Since α ∈ R \ Q, it follows that { j α α }, for any 1 ≤ j < k ≤ n + 1. Next, we observe that

n n ∪ ··· ∪ n So, by the pigeonhole principle 1 (Dirichlet’s principle), there exist two elements,

{{ α },{2 α },...,{(n + 1) α }} ⊂ [0,1] = 0 ,

say {p α } and {q α } (1 ≤ p < q ≤ n + 1), that belong to the same interval [k/n,(k +

1 )/n). This means that |{p α } − {q α }| < 1/n. Now setting

a n = {p α } − {q α } = (p − q) α + [q α ] − [p α ] ∈ A,

we deduce that a n →0 as n→∞. So, since a n n ;k ∈ Z ,n∈N ∗ } is dense in R. Since B ⊂ A, this implies that A is dense in R, too. ⊓ ⊔

We have proved in Exercise 1.1.15. that the sequence (sin n) n ≥1 is divergent. The next result yields much more information about the behavior of this sequence.