Infinite Products

2.4 Infinite Products

I recognize the lion by his paw. [After reading an anonymous solution to a problem that he realized was Newton’s solution.]

Jakob Bernoulli (1654–1705) If a real sequence (x n ) n ∈N satisfies the Coriolis conditions, then necessarily not

only ∞ ∏ ∞ n =1 (1 + x n ), but also ∏ n =1 (1 + cx n ) for every c ∈ R is convergent. It turns out that this, or even the convergence of ∏ ∞ n =1 (1 + c 1 x n ), ∏ ∞ n =1 (1 + c 2 x n ) for any

two different nonzero real numbers c 1 ,c 2 , is equivalent to the Coriolis conditions. There remains a “pathological” special case of convergence of a real product that is characterized by the following properties: the product ∏ ∞ (1 + x n ) con- verges, while the series ∑ ∞

n =1

n =1 x n =∑x 2 n diverges. In this case the balance of factors is destroyed by any scaling of the deviations from unity: ∏ ∞ n =1 (1 + cx n ) diverges for

c ∈ R \ {0,1}. The following exercise summarizes properties of this type.

be a sequence of real numbers. (i) If any two of the four expressions

2.4.1. Let (x n ) n ≥1

are convergent, then this holds also for the remaining two. (ii) If ∞

n =1 x n is convergent and ∑ n =1 x n is divergent, then ∏ n =1 (1 + x n ) diverges to zero. (iii) If ∞

∑ ∞ n =1 x 2 n is convergent and ∑ n =1 x n is divergent, then

∏ (1 + x k )/exp ∑ x k k =1

k =1

tends to a finite limit for n →∞ . (iv) If ∏ ∞

∞ n =1 (1 + x n ) is convergent and ∑ n =1 x n is divergent, then ∑ n =1 x n =∞ (v) If . ∏ n =1 (1 + cx n ) is convergent for two different values c ∈ R \ {0} , then the

product is convergent for every c ∈R .

Solution. (i) If any two of the four expressions are convergent, then there is a positive integer n 0 such that |x n | ≤ 1/2 for n ≥ n 0 . Thus, for n 0 ≤n 1 ≤n 2 ,

(1 + x )= x 2 v log ˜ x n = x

=n n ∑ 1 n =n 1 2 n =n ∑

2 1 ∑ n =n 1

where v n , ˜v ∈ 4

2.4 Infinite Products 87 Thus, by Cauchy’s criterion for series, if any two of the three expressions

∏ ∞ n =1 (1 + x n ), ∑ n =1 x n , ∑ n =1 x 2 n are convergent, then so is the third one. Since the convergence of ∞ ∑ n =1 x

2 n and ∑ n =1 x n is not affected by changing the sign of each x

n , there remains only to be shown that the convergence of ∏ n =1 (1 + x n ∞ ) and ∞ ∏

n =1 (1 − x ) implies that of ∑

But if ∞ n

∞ n =1 n

∏ n =1 (1 + x n ) and ∏ n =1 (1 − x n ) are convergent, then so is the product

∏ 2 (1 + x

n )(1 − x n )= ∏ (1 − x n ),

n =1

n =1

and hence ∞ ∑ n =1 x 2 n converges. The assertions (ii), (iii), and (iv) follow directly from relation (2.4).

(v) Without loss of generality we assume that ∏ ∞ n =1 (1 + x n ) and ∏ ∞ n =1 (1 + c 0 x n ) are convergent, where c 0 ∈ R \ {0,1}. Then with |x n |, |c 0 x n | < 1 for n ≥ n 0 ,

∏ (1 + x n ) = ∏ (1 + x ) c n 0 .

n =n 0 n =n 0

Thus, the infinite product

∞ (1 + x

n =n 0 1 +c 0 x n

converges. But

(1 + x n ) c 0 2 c 0 (c 0 − 1)

2 (n ≥ n 0 ) as ε

=1+x n ·

(1 + ε n )

1 +c 0 x n

n → 0 for n → ∞. This implies the convergence of ∑ n =1 x 2 n , and hence by (i), the conclusion follows.

Instructive examples are furnished by the products

, c ∈ R, and

∏ 1 +c·

for α

for α , c ∈ R. (2.6)

In order to discuss (2.5), the convergence properties of the series

need to be investigated. To do this it is most convenient to use the Gauss test. Using this criterion and the fact that |( α n )| ց 0 as n → ∞ for −1 < α < 0, whereas

|( α n )| → ∞ as n → ∞ for α < −1, we deduce the following properties:

88 2 Series

⋄ if α ≥ 0, then the product (2.5) is absolutely convergent for every c ∈ R; ⋄ if −1/2 < α < 0, then the product (2.5) is conditionally convergent for c ∈ R \

{0}; ⋄ if −1 < α ≤ −1/2, then the product (2.5) diverges to zero for c ∈ R \ {0}; √ ⋄ if α = −1, then the product (2.5) diverges to zero for 0 < |c| < √

2, and is indef- initely divergent for |c| ≥ 2; ⋄ if α < −1, then the product (2.5) is indefinitely divergent for c ∈ R \ {0}. We now discuss the infinite product (2.6). ⋄ If | α | < 1, then the product (2.6) obviously is absolutely convergent; ⋄ if | α | > 1, then the product (2.6) is divergent for c ∈ R \ {0}; ⋄ if α = 1, then the product (2.6) is convergent for c = 1, since

for all n ∈ N. Thus, from ∑x 2 n = ∞ and the above exercise, we conclude that for every c ∈ R \ {0,1}, the infinite product (2.6) is divergent. ⊓ ⊔

2.4.2. Establish the following formula of Vi`ete:

2 2 2 2 · 2 2 2 2 2 ··· . Solution. Let t

sint lim cos cos n →∞

2 n Hence

sint /2 n But

2 2 n sint

From (2.7) we obtain, for t = π /2,

4 8 ··· cos 2 n +1 . Since

= lim cos cos

π n →∞

cos = √

and cos =

+ cos θ ,

our conclusion follows. ⊓ ⊔

2.5 Qualitative Results 89