# A continued fraction for $\sqrt{2}\Bigg({e^\sqrt{2}-1 \above 1.5pt e^\sqrt{2}+1 }\Bigg)$

This question is about the integer sequence A123168. The title of the sequence reads: Continued fraction for - $$\sqrt{2}\Bigg({e^\sqrt{2}-1 \above 1.5pt e^\sqrt{2}+1 }\Bigg)$$ A comment claims that 'this continued fraction shows that $e^\sqrt{2}$ is irrational". I thought Lindemann & Weierstrass theorem established the irrationality of $e^\sqrt{2}$ so what is the motivation behind this particular continued fraction and the irrationality of $e^\sqrt{2}$? How do you write that continued fraction out?

note: I am not asking for a proof of the irrationality of $e^\sqrt{2}$.

• "How do you write that continued fraction out?" : $[0;0, 1, 6, 5, 14, 9, 22, 13, 30, 17, 38, 21, 46,...$ – Dietrich Burde Sep 28 '16 at 12:45
• ^ sure but I meant explicitly using fraction notation. – Antonio Hernandez Maquivar Sep 28 '16 at 12:47
• Well, you know, $[a; b, c, d, ...]$ means $a+1/(b+1/(c+1/(d+...)))$. – Dietrich Burde Sep 28 '16 at 12:49
• Not quite sure but $$0 + \cfrac{1}{0 + \cfrac{1}{6 + \cfrac{1}{5+ \cfrac{1}{14 + \cdots}}}}$$ – Antonio Hernandez Maquivar Sep 28 '16 at 13:07

By Gauss' continued fraction, $$\tanh(x)=\cfrac{x}{1+\cfrac{x^2}{3+\cfrac{x^2}{5+\cfrac{x^2}{7+\ldots}}}}\tag{1}$$ hence by evaluating at $x=\frac{1}{\sqrt{2}}$, $$\sqrt{2}\tanh\left(\frac{1}{\sqrt{2}}\right)=\color{green}{\sqrt{2}\,\frac{e^{\sqrt{2}}-1}{e^{\sqrt{2}}+1}} = \cfrac{1}{1+\cfrac{1/2}{3+\cfrac{1/2}{5+\cfrac{1/2}{7+\ldots}}}}\tag{2}$$ and by converting the RHS of $(2)$ into an ordinary continued fraction, $$\color{green}{\sqrt{2}\,\frac{e^{\sqrt{2}}-1}{e^{\sqrt{2}}+1}} = \cfrac{1}{1+\cfrac{1}{6+\cfrac{1}{5+\cfrac{1}{14+\ldots}}}}=\color{green}{[0;1,6,5,14,9,22,13,30,\ldots]}\tag{3}$$ proving that the LHS of $(3)$ is an irrational number (since the continued fraction has an infinite number of terms) and not an algebraic number over $\mathbb{Q}$ with degree $2$ (otherwise, by Lagrange's theorem, the continued fraction would be periodic from some point on). Assuming that $e^{\sqrt{2}}=\frac{p}{q}\in\mathbb{Q}$, we have that the LHS of $(3)$ is a rational multiple of $\sqrt{2}$, hence an algebraic number over $\mathbb{Q}$ with degre $2$. We have already proved that is not our case, hence $$(3)\implies \color{red}{e^{\sqrt{2}}\not\in\mathbb{Q}}.\tag{4}$$

• Although a closed formula is given in the description in the sequence - I believe I have one also. That was my motivation for the question. – Antonio Hernandez Maquivar Sep 28 '16 at 14:35
• @AnthonyHernandez: in a explicit way, the $n$-th term of the continued fraction of $\sqrt{2},\frac{e^{\sqrt{2}}-1}{e^{\sqrt{2}}+1}$ is $2n-1$ if $n$ is odd and $4n-2$ if $n$ is even. I am assuming the zeroth term is $0$. – Jack D'Aurizio Sep 28 '16 at 14:41
• I write $s_n ={n(n-1)(n+3) \above 1.5 pt 2}$ and $t_n = {n+1 \choose 2}$. We can interpret $s_n$ as the sum of sets of the form $\{2,3\}$, $\{5,6,7\}$, $\{9,10,11,12\}$, and so on. Surely $t_n$ are the so called "triangular numbers". I write $a_n =(s_n,t_n)$ the greatest common divisor of $s_n$ and $t_n$. Then for odd $n$ we have that $a_n =$A123168 – Antonio Hernandez Maquivar Sep 28 '16 at 15:17
• Perhaps this continued fraction is algebraic of degree ${} > 2$, but more likely it is transcendental. – GEdgar Oct 4 '16 at 12:31
• I just mean that, in your explanation, do not say it is an algebraic number with degree ${} > 2$, instead say it is not an algebraic number with degree $2$. – GEdgar Oct 4 '16 at 12:42