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. ⊓ ⊔