In mathematics, Cahen's constant is defined as the value of an infinite series of unit fractions with alternating signs:
Here denotes Sylvester's sequence, which is defined recursively by
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:
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]
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 is known: it is where the sequence of coefficients
is defined by the recurrence relation All the partial quotients of this expansion are squares of integers. Davison and Shallit made use of the continued fraction expansion to prove that 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 that . Indeed, we have , and if holds for some , then
where we used the recursion for in the first step respectively the recursion for in the final step. As a consequence, holds for every , from which it is easy to conclude that
.
Cahen's constant has best approximation order . That means, there exist constants such that the inequality has infinitely many solutions , while the inequality has at most finitely many solutions . This implies (but is not equivalent to) the fact that has irrationality measure 3, which was first observed by Duverney & Shiokawa (2020).
To give a proof, denote by the sequence of convergents to Cahen's constant (that means, ). [5]
But now it follows from and the recursion for that
for every . As a consequence, the limits
(recall that ) both exist by basic properties of infinite products, which is due to the absolute convergence of . Numerically, one can check that . Thus the well-known inequality
yields
for all sufficiently large . Therefore has best approximation order 3 (with ), where we use that any solution to
is necessarily a convergent to Cahen's constant.
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