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 (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

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

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

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

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

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

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.

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, retrieved 2017-12-25