Let E denote the set of all real sequences A = (A n ) n ≥0 satisfying A 0 =1 ,

1.5.2. Let E denote the set of all real sequences A = (A n ) n ≥0 satisfying A 0 =1 ,

A n ≥1 and (A n ) 2 ≤A n +1 A n −1 , f or all n ≥1 .

48 1 Sequences

(a) Check that the sequence A defined by A n = n! is an element of E . Fix A ∈E . Define the sequence ( λ n ) n ≥0 and ( µ n ) n ≥0 by λ 0 = µ 0 =1 , λ n =

A n −1 /A n and µ n = (A n ) −1/n for all integers n ≥1 . (b) Prove that the sequence ( λ n ) is decreasing and ( λ n ) n ≤ λ 1 λ 2 ··· λ n for any integer

n ≥0 . (c) Show that the sequence ( µ n ) is decreasing. (d) Prove that for all n ∈N , and for any j ∈ [0,n] ∩ N , A n +1 /A n +1− j ≥A n /A n −j . Deduce that for all n ∈N and for any j ∈ [0,n] ∩ N , A j A n −j ≤A n . (e) Establish the inequality λ n ≤ µ n .

Let (a n ) and (c n )

be sequences of positive numbers such that the sequence de- fined by s n =∑ n k =1 a k is convergent. Let u n = (a 1 a 2 ··· a n ) 1 /n and b n = (c 1 c 2 ···

c n ) −1/n for all integers n ≥1 . (f) Show that u

n ≤ n ∑ n k =1 a k c k .

(g) We assume, additionally, that ∑ n b k k =1 k is convergent and define, for any k ≥1 ,

p =k p . Prove that ∑ p =1 u p ≤∑ k =1 B k c k a k . (h) (Carleman’s inequality). Let c n = (n + 1) n /n n −1 . Deduce that the sequence (∑ n u ) is convergent and ∑ ∞

B k = lim n → ∞ ∑ n b p

∞ k =1 k n

n =1 (a 1 a 2 ...a n )

<e∑ n =1 a n . Solution. (a) Set A = (n!). Then A 2 n /A n +1 A n −1 = n/n + 1 ≤ 1, so A ∈ E.

1 /n

(b) From A 2 n ≥A n +1 A n −1 we deduce that λ n +1 ≤ λ n . Fix n ≥ 1. We have 0 < λ n ≤ λ k for all k = 1, . . . , n. Therefore

0 < λ n n ≤ λ 1 λ 2 ... λ n .

(c) We may write

µ n (n+1)

n +1 λ n .

The inequality proved in (b) can be rewritten as A n λ n n ≤ 1, since λ 1 λ 2 ··· λ n = 1/A n . Hence

µ n /(n+1)

n +1

µ n +1

which means that the sequence ( µ n ) is decreasing. (d) Let j and n be integers such that 0 ≤ j ≤ n. The sequence ( λ m ) is decreasing, so

λ n +1 ≤ λ n − j+1 , that is, A n /A n +1 ≤A n −j /A n − j+1 . It follows that A n +1 /A n − j+1 ≥

A n /A n −j . The sequence (A n /A n −j ) n ≥j is increasing, so A n /A n −j ≥A j /A 0 . Since A 0 = 1, it

follows that A n ≥A n −j A j . (e) Using (b), we have λ n n ≤ λ 1 λ 2 ... λ n = 1/A n = µ n n , that is, λ n ≤ µ n .

1.5 Hardy’s and Carleman’s Inequalities 49 (f) Let n

≥ 1. We have u /n n /b n =( α 1 α 2 ... α n ) 1 , where α i =a i c i for all 1 ≤ i ≤ n. By the AM–GM inequality we deduce that

≤ (a 1 c 1 +a 2 c 2 + ··· + a n c n ).

(g) Observe that b n /n = B n −B n +1 and set S n =a 1 c 1 + ··· + a n c n . We have

∑ u k ≤ ∑ (B k −B k +1 )S k .

k =1

k =1

Applying Abel’s inequality, we obtain

n +1

∑ (B k −B k +1 )S k = ∑ B k S k − ∑ B k S k −1 = ∑ B k a k c k −B n +1 S n .

Since B n +1 S n ≥ 0, we have

k =1

k =1

(h) Set c n = (n + 1) n /n n −1 . Thus c n = γ n +1 / γ n , where γ n =n n −1 . It follows that

c 1 c 2 ··· c n = (n + 1) n . Thus we see that b n = 1/(n + 1) satisfies

n → ∞ ∑ < +∞ k =1 k

n = 1/n satisfies B n c n = (1 + 1/n) < e. Using (g), it follows that ∑ n

→∞, for all n ≥ 1. Thus we see that B n

k =1 u k ≤ e∑ k =1 a k . We deduce that the sequence defined by ∑ n k =1 u k is convergent and, moreover,

lim

n → ∞ ∑ (a 1 a 2 ···a k ) < e lim

Carleman proved that e is the best constant in (1.16), that is, Carleman’s inequal- ity is no longer valid if we replace e with a smaller constant. We give in what follows the following alternative proof to Carleman’s inequality, based on the arithmetic–

50 1 Sequences geometric means inequality. We first observe that for all positive integers k,

∑ a n = ∑ na n ∑

∑ na n

Strict inequality holds, since we cannot have equality at the same time in all terms of the inequality. This can occur only if a

k = c/k for some c > 0 and all k ≥ 1. Such

a choice is not possible, since ∑ k =1 a k is convergent.

Using Hardy’s inequality we can give the following proof of Carleman’s inequal- ity. We first apply Hardy’s inequality for a k replaced by a 1 k /p . Hence

Since a b =e b ln a (for a , b > 0), it follows that

1 + ··· + a n ) − lnn √ → n n a 1 1 ··· a n , (1.18)

= exp

/p

as p →∞. We point out that we have used above a basic result that asserts that ln (a x 1 + ··· + a x n ) − lnn

1 ··· a n ) 1 x as x →0, where n ≥ 1 is a fixed integer (see Chapter 3 for a proof of this result).

tends to ln

(a /n

Since (p/p − 1) p increases to e as p →∞, we deduce from (1.17) and (1.18) that

a 1 + a 1 a 2 + 3 a 1 a 2 a 3 + ··· + n a 1 a 2 ··· a n + ··· < e(a 1 +a 2 + ··· + a n + ···), which is just Carleman’s inequality.

The following alternative proof of Carleman’s inequality is due to George P´olya (1887–1985) (see [109]). We first observe that the arithmetic–geometric means ine- quality implies

1.6 Independent Study Problems 51 √ n

a a 1 a 1 α 2 a 2 + ··· + α n a 1 n 2 ···a n ≤

α 1 α 2 ··· α n

for all positive numbers α 1 ,..., α n . Choosing α n = (n + 1) n /n n −1 and observing that √ n α 1 α 2 ··· α n = n + 1, we obtain

1 (k + 1) k

a 1 a 2 ···a n ≤

(n + 1) k ∑ k k

α m a m n (n+1)

It follows that

∑ a 1 a 2 ···a n ≤ ∑

n =1

n =1 n (n + 1) m =1

∑ x m ,n = =1 ∑ =1 ∑ x m ,n

m = m ∑ <e

m =1

m =1

since α m < em, for all m ≥ 1. This concludes the proof of Carleman’s inequality. ⊓ ⊔