Cahen's constant

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

${\displaystyle C=\sum _{i=0}^{\infty }{\frac {(-1)^{i}}{s_{i}-1}}={\frac {1}{1}}-{\frac {1}{2}}+{\frac {1}{6}}-{\frac {1}{42}}+{\frac {1}{1806}}-\cdots \approx 0.64341054629.}$

Here ${\displaystyle (s_{i})_{i\geq 0}}$ denotes Sylvester's sequence, which is defined recursively by

${\displaystyle {\begin{array}{l}s_{0}~~~=2;\\s_{i+1}=1+\prod _{j=0}^{i}s_{j}{\text{ for }}i\geq 0.\end{array}}}$

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:

${\displaystyle C=\sum {\frac {1}{s_{2i}}}={\frac {1}{2}}+{\frac {1}{7}}+{\frac {1}{1807}}+{\frac {1}{10650056950807}}+\cdots }$

This constant is named after Eugène Cahen (also known for the Cahen–Mellin integral), who was the first to introduce it and prove its irrationality.^[1]

Continued fraction expansion[]

The majority of naturally occurring^[2] mathematical constants have no known simple patterns in their continued fraction expansions.^[3] Nevertheless, the complete continued fraction expansion of Cahen's constant ${\displaystyle C}$ is known: If we form the sequence

0, 1, 1, 1, 2, 3, 14, 129, 25298, 420984147, ... (sequence A006279 in the OEIS)

defined by the recurrence relation

${\displaystyle a_{0}=0,~a_{1}=1,~a_{n+2}=a_{n}\left(1+a_{n}a_{n+1}\right)~\forall ~n\in \mathbb {Z} _{\geqslant 0}}$

then Cahen's constant has the (simple) continued fraction expansion

${\displaystyle C=\left[a_{0}^{2};a_{1}^{2},a_{2}^{2},a_{3}^{2},a_{4}^{2},\ldots \right]=[0;1,1,1,4,9,196,16641,\ldots ]}$

It is remarkable that all the partial quotients of this expansion are squares of integers. Davison and Shallit made use of the continued fraction expansion to prove that ${\displaystyle C}$ is transcendental.^[4]

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 ${\displaystyle n\geq 1}$ that ${\displaystyle 1+a_{n}a_{n+1}=s_{n-1}}$. Indeed, we have ${\displaystyle 1+a_{1}a_{2}=2=s_{0}}$, and if ${\displaystyle 1+a_{n}a_{n+1}=s_{n-1}}$ holds for some ${\displaystyle n\geq 1}$, then

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

${\displaystyle C=[0;1,1,1,s_{0}^{2},s_{1}^{2},(s_{0}s_{2})^{2},(s_{1}s_{3})^{2},(s_{0}s_{2}s_{4})^{2},\ldots ]}$.

Best approximation order[]

Given the continued fraction expansion of Cahen's constant ${\displaystyle C}$, it can be easily deduced that ${\displaystyle C}$ has best approximation order 3 (even though this fact does not seem to appear in the literature): That means, there exist constants ${\displaystyle K_{1},K_{2}>0}$ such that the inequality

${\displaystyle 0<{\Big |}C-{\frac {p}{q}}{\Big |}<{\frac {K_{1}}{q^{3}}}}$ has infinitely many solutions ${\displaystyle {\frac {p}{q}}\in \mathbb {Q} }$, while the inequality ${\displaystyle 0<{\Big |}C-{\frac {p}{q}}{\Big |}<{\frac {K_{2}}{q^{3}}}}$ has at most finitely many solutions ${\displaystyle {\frac {p}{q}}\in \mathbb {Q} }$.

Note that this implies (but is not equivalent to) the fact that ${\displaystyle C}$ has irrationality measure 3, which was first observed by Duverney & Shiokawa (2020).

To give a proof, denote by ${\displaystyle (p_{n}/q_{n})_{n\geq 0}}$ the sequence of convergents to Cahen's constant (that means, ${\displaystyle q_{n-1}=a_{n}{\text{ for every }}n\geq 1}$).^[5]

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

${\displaystyle {\frac {a_{n+2}}{a_{n+1}^{2}}}={\frac {a_{n}\cdot s_{n-1}}{a_{n-1}^{2}\cdot s_{n-2}^{2}}}={\frac {a_{n}}{a_{n-1}^{2}}}\cdot {\frac {s_{n-2}^{2}-s_{n-2}+1}{s_{n-1}^{2}}}={\frac {a_{n}}{a_{n-1}^{2}}}\cdot {\Big (}1-{\frac {1}{s_{n-1}}}+{\frac {1}{s_{n-1}^{2}}}{\Big )}}$ for every ${\displaystyle n\geq 1}$. As a consequence, the limits

${\displaystyle \alpha :=\lim _{n\to \infty }{\frac {q_{2n+1}}{q_{2n}^{2}}}=\prod _{n=0}^{\infty }{\Big (}1-{\frac {1}{s_{2n}}}+{\frac {1}{s_{2n}^{2}}}{\Big )}}$ and ${\displaystyle \beta :=\lim _{n\to \infty }{\frac {q_{2n+2}}{q_{2n+1}^{2}}}=2\cdot \prod _{n=0}^{\infty }{\Big (}1-{\frac {1}{s_{2n+1}}}+{\frac {1}{s_{2n+1}^{2}}}{\Big )}}$

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

${\displaystyle {\frac {1}{q_{n}(q_{n}+q_{n+1})}}\leq {\Big |}C-{\frac {p_{n}}{q_{n}}}{\Big |}\leq {\frac {1}{q_{n}q_{n+1}}}}$

yields

${\displaystyle {\Big |}C-{\frac {p_{2n+1}}{q_{2n+1}}}{\Big |}\leq {\frac {1}{q_{2n+1}q_{2n+2}}}={\frac {1}{q_{2n+1}^{3}\cdot {\frac {q_{2n+2}}{q_{2n+1}^{2}}}}}<{\frac {1}{q_{2n+1}^{3}}}}$ and ${\displaystyle {\Big |}C-{\frac {p_{n}}{q_{n}}}{\Big |}\geq {\frac {1}{q_{n}(q_{n}+q_{n+1})}}>{\frac {1}{q_{n}(q_{n}+2q_{n}^{2})}}\geq {\frac {1}{3q_{n}^{3}}}}$

for all sufficiently large ${\displaystyle n}$. Therefore ${\displaystyle C}$ has best approximation order 3 (with ${\displaystyle K_{1}=1{\text{ and }}K_{2}=1/3}$), where we use that any solution ${\displaystyle {\frac {p}{q}}\in \mathbb {Q} }$ to

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

is necessarily a convergent to Cahen's constant.

Notes[]

References[]

• Cahen, Eugène (1891), "Note sur un développement des quantités numériques, qui présente quelque analogie avec celui en fractions continues", Nouvelles Annales de Mathématiques, 10: 508–514
• Davison, J. Les; Shallit, Jeffrey O. (1991), "Continued fractions for some alternating series", Monatshefte für Mathematik, 111 (2): 119–126, doi:10.1007/BF01332350, S2CID 120003890
• Borwein, Jonathan; van der Poorten, Alf; Shallit, Jeffrey; Zudilin, Wadim (2014), Neverending Fractions: An Introduction to Continued Fractions, Australian Mathematical Society Lecture Series, vol. 23, Cambridge University Press, doi:10.1017/CBO9780511902659, ISBN 978-0-521-18649-0, MR 3468515
• Duverney, Daniel; Shiokawa, Iekata (2020), "Irrationality exponents of numbers related with Cahen's constant", Monatshefte für Mathematik, 191 (1): 53–76, doi:10.1007/s00605-019-01335-0, MR 4050109, S2CID 209968916

External links[]

• Weisstein, Eric W., "Cahen's Constant", MathWorld
• "The Cahen constant to 4000 digits", Plouffe's Inverter, Université du Québec à Montréal, archived from the original on March 17, 2011, retrieved 2011-03-19
• "Cahen's constant (1,000,000 digits)", Darkside communication group (in Japan), retrieved 2017-12-25