# Cahen's constant

In mathematics, Cahen's constant is defined as the value of an infinite series of unit fractions with alternating signs:
    $C=\sum _{i=0}^{\infty }{\frac {(-1)^{i}}{s_{i}-1}}={\frac {1}{1}}-{\frac {1}{2}}+{\frac {1}{6}}-{\frac {1}{42}}+{\frac {1}{1806}}-\cdots \approx 0.643410546288...$ (sequence A118227 in the OEIS)
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:
    $C=\sum {\frac {1}{s_{2i}}}={\frac {1}{2}}+{\frac {1}{7}}+{\frac {1}{1807}}+{\frac {1}{10650056950807}}+\cdots$
This constant is named after Eugène Cahen [fr] (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
$1+a_{n+1}a_{n+2}=1+a_{n+1}\cdot a_{n}(1+a_{n}a_{n+1})=1+a_{n}a_{n+1}+(a_{n}a_{n+1})^{2}=s_{n-1}+(s_{n-1}-1)^{2}=s_{n-1}^{2}-s_{n-1}+1=s_{n},$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
$C=[0;1,1,1,s_{0}^{2},s_{1}^{2},(s_{0}s_{2})^{2},(s_{1}s_{3})^{2},(s_{0}s_{2}s_{4})^{2},\ldots ]$.

## 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 Duverney & Shiokawa (2020).
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
    ${\frac {a_{n+2}}{a_{n+1}^{2}}}={\frac {a_{n}\cdot s_{n-1}}{a_{n-1}^{2}\cdot s_{n-2}^{2}}}={\frac {a_{n}}{a_{n-1}^{2}}}\cdot {\frac {s_{n-2}^{2}-s_{n-2}+1}{s_{n-1}^{2}}}={\frac {a_{n}}{a_{n-1}^{2}}}\cdot {\Big1-{\frac {1}{s_{n-1}}}+{\frac {1}{s_{n-1}^{2}}}{\Big )}$
for every $n\geq 1$. As a consequence, the limits
    $\alpha :=\lim _{n\to \infty }{\frac {q_{2n+1}}{q_{2n}^{2}}}=\prod _{n=0}^{\infty }{\Big1-{\frac {1}{s_{2n}}}+{\frac {1}{s_{2n}^{2}}}{\Big )}$ and $\beta :=\lim _{n\to \infty }{\frac {q_{2n+2}}{q_{2n+1}^{2}}}=2\cdot \prod _{n=0}^{\infty }{\Big1-{\frac {1}{s_{2n+1}}}+{\frac {1}{s_{2n+1}^{2}}}{\Big )}$
(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
    ${\frac {1}{q_{n}(q_{n}+q_{n+1})}}\leq {\Big |}C-{\frac {p_{n}}{q_{n}}}{\Big |}\leq {\frac {1}{q_{n}q_{n+1}}}$
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.