For a real number $\alpha$, let the irrationality measure $\mu(\alpha) \in \mathbb{R}\cup \{\infty\}$ be defined as the supremum of all real numbers $\mu$ such that 
$$ \left| \alpha-\frac{p}{q}\right| \le \frac{1}{q^\mu} $$ has infinitely many solutions $\frac{p}{q} \in \mathbb{Q}$, where $q \ge 1$.
It's known that if $\alpha$ is rational, then $\mu(\alpha)=1$ and if $\alpha$ is algebraic and irrational, then $\mu(\alpha)=2$ by Roth's Theorem. The set of all $\alpha$ such that $\mu(\alpha) >2$ has Lebesgue measure 0 by Khinchin's Theorem.


One can explicitly write down real numbers $\alpha$ such that $\mu(\alpha) = \infty$ (e.g. Liouville's constant) and transcendental real numbers $\alpha$ such that $\mu(\alpha)=2$ (e.g. $\alpha =e$, see also [this thread](https://mathoverflow.net/questions/89600/numbers-with-known-irrationality-measures)). It's also possible to find upper bounds on $\mu(\alpha)$ for some real numbers such as $\pi$. But I don't know of a single example of a real number $\alpha$ whose irrationality measure is known, finite and greater than 2.


Question: Is there an example of a real number whose irrationality measure $\mu(\alpha)$ is known exactly and satisfies $2<\mu(\alpha)<\infty$?



Numbers with known finite irrationality measure greater than 2
The answer is yes - see for example



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> > > Yann Bugeaud *Diophantine approximation and Cantor sets* Math. Ann. (2008) **341**:677–684
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