Cahen's constant explained

In mathematics, Cahen's constant is defined as the value of an infinite series of unit fractions with alternating signs:

C = \sum ^{inf ty} _{i =0} (( -1 ) ^{i})/(= s _{i-1}) (11)/(-) (12)/(+) (16)/(-) (1 { 42 })/(+) (1 { 1806 })/(-) … ≈ 0 . 643410546288 ... Here (s _{i)} _{i} denotes Sylvester's sequence, which is defined recursively by\begin{array}{l} s _{0~~~} = 2 ; \\ s _{i+1} = 1 + \prod ^{i} _{j =0} s _{j} for i \geq 0 . \end{array}

Combining these fractions in pairs leads to an alternative expansion of Cahen's constant as a series of positive unit fractions formed from the terms in even positions of Sylvester's sequence. This series for Cahen's constant forms its greedy Egyptian expansion:

C = \sum (1)/(= s _{2i}) (12+ (17+ (1 { 1807 } + (1 { 10650056950807 } + …)/())/())/())/() This constant is named after (also known for the Cahen–Mellin integral), who was the first to introduce it and prove its irrationality.

Continued fraction expansion

The majority of naturally occurring^[1] mathematical constants have no known simple patterns in their continued fraction expansions. Nevertheless, the complete continued fraction expansion of Cahen's constant

C is known: it is$$C\; =\; \backslash left[a\_0^2;\; a\_1^2,\; a\_2^2,\; a\_3^2,\; a\_4^2,\; \backslash ldots\backslash right]\; =\; [0;1,1,1,4,9,196,16641,\backslash ldots]$$where the sequence of coefficientsis defined by the recurrence relation$$a\_0\; =\; 0,~a\_1\; =\; 1,~a\_\; =\; a\_n\backslash left(1\; +\; a\_n\; a\_\backslash right)~\backslash forall~n\backslash in\backslash mathbb\_.$$All the partial quotients of this expansion are squares of integers. Davison and Shallit made use of the continued fraction expansion to prove that C is transcendental.

Alternatively, one may express the partial quotients in the continued fraction expansion of Cahen's constant through the terms of Sylvester's sequence: To see this, we prove by induction on

n \geq 1 that 1+ a _{n} a _{n+1} = s _{n-1} . Indeed, we have 1+ a _{1} a _{2} = 2 = s _{0} , and if 1+ a _{n} a _{n+1} = s _{n-1} holds for some n \geq 1 , then 1+ a _{n+1} a _{n+2} = 1+ a _{n+1} ⋅ a _{n(1+a} _{n} a _{n+1} ) = 1+ a _{n} a _{n+1} + (a _{na} _{n+1} ) ^{2} = s _{n-1} + (s _{n-1} -1 ) ^{2} = s ^{2- s} _{n -1} +1 = s _{n,} where we used the recursion for (a _{n)} _{n} in the first step respectively the recursion for (s _{n)} _{n} in the final step. As a consequence, a _{n+2} = a _{n} ⋅ s _{n-1} holds for every n \geq 1 , from which it is easy to conclude that C = [ 0 ; 1 , 1 , 1 ,s ^{2 ,} _{0} s ^{2 ,} _{1} (s _{0s} ^{2 ,} ^{2 )} (s _{1s} ^{2 ,} ^{3 )} (s _{0s} _{2s} ^{2 ,\ldots]} ^{4 )} .

Best approximation order

Cahen's constant

C has best approximation order q ^{-3} . That means, there exist constants K _{1,} K _{2} > 0 such that the inequality 0 < | C - (p)/(q) | < (K _{1})/(q ^{3}) has infinitely many solutions (p,q) \in Z x N , while the inequality 0 < | C - (p)/(q) | < (K _{2})/(q ^{3}) has at most finitely many solutions (p,q) \in Z x N .This implies (but is not equivalent to) the fact that C has irrationality measure 3, which was first observed by .

To give a proof, denote by

(p _{n/q} _{n)} _{n} the sequence of convergents to Cahen's constant (that means, q _{n-1} = a _{n} forevery n \geq 1 ).^[2]

But now it follows from

a _{n+2} = a _{n} ⋅ s _{n-1} and the recursion for (s _{n)} _{n} that (a _{n+2})/(a ^{2} _{n +1}) = (a _{n} ⋅ s _{n-1})/(a ^{2} _{n -1} ⋅ s ^{2} _{n -2}) = (a _{n})/(a ^{2} _{n -1}) ⋅ (s ^{2} _{n -2} - s _{n-2} + 1)/(s ^{2} _{n -1}) = (a _{n})/(a ^{2} _{n -1}) ⋅ ( 1 - (1)/(s _{n-1}) + (1)/(s ^{2} _{n -1}) )

for every

n \geq 1 . As a consequence, the limits \alpha : = \lim _{n} (q _{2n+1})/(q ^{2} _{2 n}) = \prod ^{inf ty} _{n =0} ( 1 - (1)/(s _{2n}) + (1)/(s ^{2} _{2 n}) ) and \beta : = \lim _{n} (q _{2n+2})/(q ^{2} _{2 n +1}) = 2 ⋅ \prod ^{inf ty} _{n =0} ( 1 - (1)/(s _{2n+1}) + (1)/(s ^{2} _{2 n +1}) )

(recall that

s _{0} = 2 ) both exist by basic properties of infinite products, which is due to the absolute convergence of \sum ^{inf ty} _{n =0} | (1)/(s _{n}) - (1)/(s ^{2} _{n}) | . Numerically, one can check that 0 < \alpha < 1 < \beta < 2 . Thus the well-known inequality (1)/(q _{n(q} _{n} + q _{n+1} )) \leq | C - (p _{n})/(q _{n}) | \leq (1)/(q _{nq} _{n+1})

yields

| C - (p _{2n+1})/(q _{2n+1}) | \leq (1)/(q _{2n+1} q _{2n+2}) = (1)/(q ^{3} _{2 n +1} ⋅ (q _{2n+2})/(q ^{2} _{2 n +1})) < (1)/(q ^{3} _{2 n +1}) and | C - (p _{n})/(q _{n}) | \geq (1)/(q _{n(q} _{n} + q _{n+1} )) > (1)/(q _{n(q} _{n} + 2 q ^{2 )} _{n}) \geq (1)/(3 q ^{3} _{n})

for all sufficiently large

n . Therefore C has best approximation order 3 (with K _{1} = 1 and K _{2} = 1 / 3 ), where we use that any solution (p,q) \in Z x N to 0 < | C - (p)/(q) | < (1)/(3 q ^{3})

is necessarily a convergent to Cahen's constant.

Notes and References

1. A number is said to be naturally occurring if it is *not* defined through its decimal or continued fraction expansion. In this sense, e.g., Euler's number e = \lim _{n} ( 1+ (1)/(n) ) ^{n} is naturally occurring.
2. cs2.