The irrationality measure $\mu(x)$ of a number $x\in\mathbb{R}\setminus\mathbb{Q}$ is defined as the inifimum of the set
$$R_x=\{\mu>0: \text{for some}\, Q_\mu>0, \, \Big|x-\frac{p}{q}\Big|>\frac{1}{q^\mu}\, \text{for all integers}\, p\in\mathbb{Z}, \text{and}\, q\geq Q_\mu\}$$
that is, $\mu$ is the smallest number for which the inequality $\Big|x-\frac{p}{q}\Big|\leq\frac{1}{q^\mu}$ has at most a finite number of solutions $(p,q)\in\mathbb{Z}\times\mathbb{N}$, $q\geq1$.
In the article by Sondow, J. "Irrationality Measures, Irrationality Bases, and a Theorem of Jarnik." Proceedings of Journées Arithmétiques, Graz 2003 in the Journal du Theorie des Nombres Bordeaux, the author proves that
$$\begin{align}
\mu(x)\stackrel{(1)}{=}1+\limsup_n\frac{\log(q_{n+1})}{\log(q_n)}\stackrel{(2)}{=}2+\limsup_n\frac{\log(a_{n+1})}{\log(q_n)}\tag{*}\label{irr-mea}
\end{align}$$
where $x=[a_0,a_1,\ldots]$ is the simple continued fraction of $x$ and the rations $p_n/q_n$, called convergents, are define as
$$
\begin{align}
p_n&=p_{n-1}a_n+p_{n-2}\\
q_n&=q_{n-1}a_n+q_{n-2}
\end{align}
$$
$p_{-2}=0=q_{-1}$, $q_{-2}=1=p_{-1}$.
Identity (2) in \eqref{irr-mea} is straight forward.
In the course of the proof, the author defines an exponent $\lambda_n>0$ such that
$$\Big|x-\frac{p_n}{q_n}\Big|=\frac{1}{q^{\lambda_n}}$$
Using the properties of the convergents $p_n/q_n$ and Liouville's theorem he establishes that
$$\begin{align}
\lambda=\limsup_n \lambda_n=1+\frac{\log(q_{n+1})}{\log(q_n)}\tag{**}\label{author}\end{align}$$
The problem is to show that (1) holds, that is that
$\mu(x)=\lambda$. Any reference or solution to this will be appreciated.
Best Answer
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 $$\mu(x)=\lambda=\limsup_n\lambda_n$$