Continued Fractions, Convergence Theory. Vol. 1, 2nd Editions. Loretzen, Waadeland. Atlantis Press. 2008

8 Chapter 1: Introductory examples
Remark: For a given continued fraction b 0 + K(a n /b n ), we shall use the notation
s n , S n , A n , B n , f n , f (n) and Δ n throughout the book.
In Lemma 1.1 we saw that the canonical numerators {A n } and denominators {B n }
for a given continued fraction b 0 + K(a n /b n ) are uniquely determined. In fact,
Daniel Bernoulli ([Berno75]) proved that also the converse holds true if all Δ n ≠0:
✬
Theorem 1.2. The given sequences {A n } ∞ n=0 and {B n} ∞ n=0 of complex
numbers are the canonical numerators and denominators of some continued
fraction b 0 + K(a n /b n ) if and only if Δ n ≠0for n ≥ 1 and B 0 =1.
Then b 0 + K(a n /b n ) is uniquely determined by
✩
a n := − Δ n
Δ
,
n−1
✫
b 0 := A 0 , b 1 := B 1 , a 1 := A 1 − A 0 B 1 ,
b n := A n−2B n − B n−2 A n
Δ n−1
for n ≥ 2 .
(1.2.15)
✪
Proof : Let {A n } and {B n } be given with all Δ n ≠ 0 and B 0 = 1. Then the
elements a n and b n must be solutions of the system
b n A n−1 + a n A n−2 = A n ,
b n B n−1 + a n B n−2 = B n
(1.2.16)
of linear equations. The determinant of this system is −Δ n−1 ≠ 0. Hence the
solution is given by (1.2.15), and it is unique. □
This theorem allows us to construct as many continued fraction identities
as we may possibly want.
f = K(a n /b n )
Example 2. We shall find the continued fraction b 0 +K(a n /b n ) for which A n := n 2
and B n := n 2 + 1 for n =0, 1, 2,.... In this case
Δ n = A n−1 B n − B n−1 A n =1− 2n
and
so by Theorem 1.2
A n−2 B n − B n−2 A n =4− 4n,
b 0 =0, b 1 =2, a 1 =1,
a n = − 2n − 1
2n − 3 , b n = 4n − 4
2n − 3
for n =2, 3, 4,....