We define the irrationality measure of a number $x\in\mathbb{R}$ as the supremum of the $r\in\mathbb{R}$ such that exists $c\in \mathbb{R}$ verifying
$$|x-\frac{p}{q}|\geq \frac{1}{q^r}$$
I was trying to prove that the irrationality measure of the Euler's constant $e$ is 2 using simple results of continued fractions (such as the expression of the denominators), and the simple continued fraction of $e$,
$$[2;1,2,1,1,4,1,1,6,1,1,8…]$$
Thank you very much for your help.
Best Answer
Let $\alpha\in\Bbb R\setminus\Bbb Q$. It can be shown that (iirc this paper contains an error but the conclusions are all correct): $$\begin{align}\mu(\alpha)=1+\limsup_{n\to\infty}&\frac{\ln(q_{n+1})}{\ln(q_n)}=2+\limsup_{n\to\infty}\frac{\ln(a_n)}{\ln(q_{n-1})}\\0\le\limsup_{n\to\infty}\frac{\ln a_{n+1}}{\ln(F_n\prod_{j=1}^na_j)}&\le\mu(\alpha)-2\le\limsup_{n\to\infty}\frac{\ln a_{n+1}}{\ln(F_n+\prod_{j=1}^na_j)}\end{align}$$Where $\mu$ is the irrationality measure, $F_n$ is the $n$th Fibonnaci number and $\alpha=[a_0;a_1,a_2,\cdots]$ as an infinite simple continued fraction (SCF) (this representation is unique) with convergents $p_n/q_n$, indexing these so that the recurrence $q_{n+1}=a_{n+1}q_n+q_{n-1}$ holds.
As a corollary, any number $\alpha$ whose SCF satisfies $a_n=\exp(o(n))$, in particular any $\alpha$ whose SCF has coefficients which grow in polynomial time - as is the case for $e$ - must have irrationality measure $2$.
I haven't looked at this for a while, so I'm sorry to say I can't quite remember if - for the specific case of numbers whose SCF coefficients grow slowly - there is a more elementary way to deduce $\mu(\alpha)=2$. However this paper is not super technical, it only requires a few lemmas of the theory of SCFs (which I believe could be found in Hardy's book on number theory). I can review my notes if you wish.