Cahen's constant

In mathematics, Cahen's constant is defined as the value of an infinite series of unit fractions with alternating signs:
 * $$C = \sum_{i=0}^\infty \frac{(-1)^i}{s_i-1}=\frac11 - \frac12 + \frac16 - \frac1{42} + \frac1{1806} - \cdots\approx 0.643410546288...$$

Here $$(s_i)_{i \geq 0}$$ denotes Sylvester's sequence, which is defined recursively by
 * $$\begin{array}{l}

s_0 = 2; \\ s_{i+1} = 1 + \prod_{j=0}^i s_j \text{ 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:
 * $$C = \sum\frac{1}{s_{2i}}=\frac12+\frac17+\frac1{1807}+\frac1{10650056950807}+\cdots$$

This constant is named after Eugène Cahen (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 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 $$C = \left[a_0^2; a_1^2, a_2^2, a_3^2, a_4^2, \ldots\right] = [0;1,1,1,4,9,196,16641,\ldots]$$ where the sequence of coefficients

is defined by the recurrence relation $$a_0 = 0,~a_1 = 1,~a_{n+2} = a_n\left(1 + a_n a_{n+1}\right)~\forall~n\in\mathbb{Z}_{\geqslant 0}.$$ 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

$$1+a_{n+1}a_{n+2} = 1+a_{n+1} \cdot a_n(1+a_n a_{n+1})= 1+a_n a_{n+1} + (a_na_{n+1})^2 = s_{n-1} + (s_{n-1}-1)^2 = s_{n-1}^2-s_{n-1}+1 = s_n,$$where we used the recursion for $$(a_n)_{n \geq 0}$$ in the first step respectively the recursion for $$(s_n)_{n \geq 0}$$ in the final step. As a consequence, $$a_{n+2} = a_n \cdot s_{n-1}$$ holds for every $$n \geq 1$$, from which it is easy to conclude that

$$C = [0;1,1,1,s_0^2, s_1^2, (s_0s_2)^2, (s_1s_3)^2, (s_0s_2s_4)^2,\ldots]$$.

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 $$ 0 < \Big| C - \frac{p}{q} \Big| < \frac{K_1}{q^3} $$ has infinitely many solutions $$ (p,q) \in \mathbb{Z} \times \mathbb{N} $$, while the inequality $$ 0 < \Big| C - \frac{p}{q} \Big| < \frac{K_2}{q^3} $$ has at most finitely many solutions $$ (p,q) \in \mathbb{Z} \times \mathbb{N} $$. 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 $$(p_n/q_n)_{n \geq 0}$$ the sequence of convergents to Cahen's constant (that means, $$q_{n-1} = a_n \text{ for every } n \geq 1$$).

But now it follows from $$a_{n+2} = a_n \cdot s_{n-1}$$and the recursion for $$(s_n)_{n \geq 0}$$ that


 * $$\frac{a_{n+2}}{a_{n+1}^2} = \frac{a_{n} \cdot s_{n-1}}{a_{n-1}^2 \cdot s_{n-2}^2} = \frac{a_n}{a_{n-1}^2} \cdot \frac{s_{n-2}^2 - s_{n-2} + 1}{s_{n-1}^2} = \frac{a_n}{a_{n-1}^2} \cdot \Big( 1 - \frac{1}{s_{n-1}} + \frac{1}{s_{n-1}^2} \Big)$$

for every $$n \geq 1$$. As a consequence, the limits


 * $$\alpha := \lim_{n \to \infty} \frac{q_{2n+1}}{q_{2n}^2} = \prod_{n=0}^\infty \Big( 1 - \frac{1}{s_{2n}} + \frac{1}{s_{2n}^2}\Big)$$ and $$\beta := \lim_{n \to \infty} \frac{q_{2n+2}}{q_{2n+1}^2} = 2 \cdot \prod_{n=0}^\infty \Big( 1 - \frac{1}{s_{2n+1}} + \frac{1}{s_{2n+1}^2}\Big)$$

(recall that $$s_0 = 2$$) both exist by basic properties of infinite products, which is due to the absolute convergence of $$\sum_{n=0}^\infty \Big| \frac{1}{s_{n}} - \frac{1}{s_{n}^2} \Big|$$. Numerically, one can check that $$0 < \alpha < 1 < \beta < 2$$. Thus the well-known inequality


 * $$\frac{1}{q_n(q_n + q_{n+1})} \leq \Big| C - \frac{p_n}{q_n} \Big| \leq \frac{1}{q_nq_{n+1}}$$

yields


 * $$\Big| C - \frac{p_{2n+1}}{q_{2n+1}} \Big| \leq \frac{1}{q_{2n+1}q_{2n+2}} = \frac{1}{q_{2n+1}^3 \cdot

\frac{q_{2n+2}}{q_{2n+1}^2}} < \frac{1}{q_{2n+1}^3}$$ and $$\Big| C - \frac{p_n}{q_n} \Big| \geq \frac{1}{q_n(q_n + q_{n+1})} > \frac{1}{q_n(q_n + 2q_{n}^2)} \geq \frac{1}{3q_n^3}$$

for all sufficiently large $$n$$. Therefore $$C$$ has best approximation order 3 (with $$K_1 = 1 \text{ and } K_2 = 1/3$$), where we use that any solution $$ (p,q) \in \mathbb{Z} \times \mathbb{N} $$ to


 * $$0 < \Big| C - \frac{p}{q} \Big| < \frac{1}{3q^3}$$

is necessarily a convergent to Cahen's constant.