Is it possible to construct a number (by way of an infinite series or a continued fraction say) having any, possibly non-integer, irrationality measure $>2$ ? It is known that this can be done for integer irrationality measures: the Champernowne constant in base b has irrationality measure equal to b, for example. There are also some seemingly good candidates like the family of series $ \sum_{k=0}^{\infty} 10^{-\lfloor b^k \rfloor} $ for $ b >1$, but they have irrationality measure at least b, not exactly b. Any help would be greatly appreciated.
Do there exist numbers with non-integer irrationality measure
diophantine-approximationirrationality-measurenumber theory
Related Solutions
It is well-known that if $\lambda > 2$, then for all sufficiently large $q$, the inequality $\left| x-\frac{p}{q} \right| \le \frac1{q^\lambda}$ implies that $\frac{p}{q} = \frac{p_n}{q_n}$ is a convergent of the continued fraction expansion of $x$, in which case $\frac{1}{2 q_n q_{n+1}} \le \left| x-\frac{p_n}{q_n} \right| \le \frac{1}{q_n q_{n+1}}$. This means that $q_{n+1} \ge \frac{q_n^{\lambda-1}}{2}$. From the recursive formula $q_{n+1} = a_{n+1} q_n + q_{n-1} \le (a_{n+1}+1)q_n$ for the denominators of the continued fraction convergents (where $a_n$ are the coefficients of the continued fraction) this means the inequality implies that $a_{n+1}+1 \ge \frac{q_n^{\lambda-2}}{2}$. On the other hand, if $a_{n+1} \ge q_n^{\lambda-2}$, then $\left| x-\frac{p_n}{q_n} \right| \le \frac{1}{q_n^\lambda}$.
Now with these inequalities one can easily construct a continued fraction expansion such that $\mu = \sup G_x = 3$, but $3 \notin G_x$. Roughly, the idea is to recursively define something like $a_{n+1} = \frac{q_n^{1-1/n}}{3}$ (for $n \ge n_0$, and rounded down to the next integer), which then implies that for every $\epsilon>0$ one has $a_{n+1} \ge q_n^{1-\epsilon}$ for $n$ sufficiently large, but always $a_{n+1} \le \frac{q_n}{3} < \frac{q_n}2$.
In this posting it is proved that the rationality measure of $x\in\mathbb{R}\setminus\mathbb{Q}$ is indeed the same as $\lambda=\limsup_n\lambda_n$, where $\lambda_n$ is the sequence of exponents defined in the OP.
Throughout this posting, $\frac{p_n}{q_n}$ denotes the $n$-th convergent of $x=[a_0,a_1,\ldots]$.
Observe that the set $R_x$ defined in the OP is either empty, or an interval such that $$(\mu(x),\infty)\subset R_x\subset [\mu(x),\infty)$$ As in the OP, let $\lambda_n$ be such that $$\Big|x-\frac{p_n}{q_n}\Big|=\frac{1}{q^{\lambda_n}_n}$$
Recall that $$\frac{1}{2q_nq_{n+1}}<\frac{1}{(q_{n+1}+q_n)q_n}<\Big|x-\frac{p_n}{q_n}\Big|<\frac{1}{q_nq_{n+1}}<\frac{1}{q^2_n} $$ From this, it follows that $\lambda_n>2$ and so, $$\lambda:=\limsup_n\lambda_n\geq2$$
Claim I: If $\lambda<\infty$, then $\mu(x)\leq \lambda$. For any $\varepsilon>0$ there exists $N_\varepsilon\in\mathbb{N}$ such that \begin{align} \lambda_n<\lambda+\varepsilon \qquad\text{whenever}\quad n\geq N_\varepsilon\tag{0}\label{zero} \end{align} If $\lambda+\varepsilon<\mu(x)$, then there are infinitely many solutions $(p,q)\in\mathbb{Z}\times\mathbb{N}$, $q\geq1$, to the inequality \begin{align} \Big|x-\frac{p}{q}\Big|\leq \frac{1}{q^{\mu+\varepsilon}}\tag{1}\label{one} \end{align} Then, for such solution $(p,q)$ with $q\geq \max(2^{1/\varepsilon}, q_{N_\varepsilon})$ $$\Big|x-\frac{p}{q}\Big|\leq \frac{1}{q^{\mu}q^{\varepsilon}}<\frac{1}{2q^\lambda}\leq \frac{1}{2q^2}$$ By Lagrange's criteria for continued fractions (see for example Khintchine, A. Ya., Continued Fractions, Dover Publications, NY, 1997, pp. 30) it follows that $\frac{p}{q}$ is a convergent of $x$, that is, $\frac{p}{q}=\frac{q_n}{q_n}$ for some $n\geq N_\varepsilon$. But then, $$\Big|x-\frac{p_n}{q_n}\Big|=\frac{1}{q^{\lambda_n}_n}<\frac{1}{q^{\lambda+\varepsilon}_n}$$ contradicting \eqref{zero}. Therefore $\mu(x)\leq\lambda+\varepsilon$ for all $\varepsilon>0$, i.e., $$\mu(x)\leq \lambda$$ and in particular, $\mu(x)<\infty$.
Claim II: If $\mu(x)<\infty$, then $\lambda\leq\mu(x)$. By definition of $\mu(x)$, given $\varepsilon>0$, $\mu(x)+\varepsilon\in R_x$. Thus, there is $Q_\varepsilon\in\mathbb{N}$ such that \begin{align} \Big|x-\frac{p}{q}\Big|>\frac{1}{q^{\mu(x)+\varepsilon}}\qquad\text{for all}\quad p\in\mathbb{Z}, \,q\geq Q_\varepsilon\tag{2}\label{two} \end{align} Choose $N$ large enough so that $q_N>Q_\varepsilon$. Then \begin{align} \frac{1}{q^{\lambda_n}}=\Big|x-\frac{p_n}{q_n}\Big|>\frac{1}{q^{\mu(x)+\varepsilon}}\qquad\text{whenever}\quad n\geq N\tag{3}\label{three} \end{align} Hence, $\lambda_n<\mu(x)+\varepsilon$ for all $n\geq N$ and so, $\lambda=\limsup_n\lambda_n\leq\mu(x)+\varepsilon$ for all $\varepsilon>0$, i.e., $$\lambda\leq\mu(x)$$ and in particular, $\lambda<\infty$.
Putting things together, we have that $$\mu(x)=\lambda=\limsup_n\lambda_n$$
Best Answer
According to and using the notation of http://mathworld.wolfram.com/IrrationalityMeasure.html , all you need is $\limsup\frac{\ln a_{n+1}}{\ln q_n}=\mu-2$. As $q_n\to \infty$ is guaranteed, you can recursively let $a_{n+1}=\lceil q_n^{\mu-2}\rceil$.