Cahen's constant

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

$C=\sum _{i=0}^{\mathrm{\infty }}\frac{(-1{)}^i}{s_i-1}=\frac{1}{1}-\frac{1}{2}+\frac{1}{6}-\frac{1}{42}+\frac{1}{1806}-\cdots \approx \mathrm{0.643410546288...}$ Template:OEIS

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

$\begin{array}{c}{s}_0=2;\\ s_{i+1}=1+\prod _{j=0}^i{s}_j\text{ for }i\ge 0.\end{array}$

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 Template:Ill (also known for the Cahen–Mellin integral), who was the first to introduce it and prove its irrationality.Template:Sfnp

Continued fraction expansion

The majority of naturally occurringTemplate:Efn mathematical constants have no known simple patterns in their continued fraction expansions.Template:Sfnp Nevertheless, the complete continued fraction expansion of Cahen's constant $C$ is known: it is $$C=[a_0^2;a_1^2,a_2^2,a_3^2,a_4^2,\dots ]=[0;1,1,1,4,9,196,16641,\dots ]$$ where the sequence of coefficients Template:Bi is defined by the recurrence relation $$a_0=0,a_1=1,a_{n+2}=a_n(1+a_n{a}_{n+1})\mathrm{\forall }n\in {\mathbb{Z}}_{⩾0}.$$ 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.Template:Sfnp

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

$C=[0;1,1,1,s_0^2,s_1^2,(s_0{s}_2{)}^2,(s_1{s}_3{)}^2,(s_0{s}_2{s}_4{)}^2,\dots ]$.

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 $0<|C-\frac{p}{q}|<\frac{K_1}{q^3}$ has infinitely many solutions $(p,q)\in \mathbb{Z}\times \mathbb{N}$, while the inequality $0<|C-\frac{p}{q}|<\frac{K_2}{q^3}$ 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 Template:Harvtxt.

To give a proof, denote by $(p_n/q_n{)}_{n\ge 0}$ the sequence of convergents to Cahen's constant (that means, $q_{n-1}=a_n\text{ for every }n\ge 1$).

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

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

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

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

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

$\frac{1}{q_n(q_n+q_{n+1})}\le |C-\frac{p_n}{q_n}|\le \frac{1}{q_n{q}_{n+1}}$

yields

$|C-\frac{p_{2n+1}}{q_{2n+1}}|\le \frac{1}{q_{2n+1}{q}_{2n+2}}=\frac{1}{q_{2n+1}^3\cdot \frac{q_{2n+2}}{q_{2n+1}^2}}<\frac{1}{q_{2n+1}^3}$ and $|C-\frac{p_n}{q_n}|\ge \frac{1}{q_n(q_n+q_{n+1})}>\frac{1}{q_n(q_n+2q_n^2)}\ge \frac{1}{3q_n^3}$

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

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

is necessarily a convergent to Cahen's constant.

References

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External links

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