Lower bound on the irrationality measure of $\pi$

Question

There seems to be a lot of work on the upper bound for the irrationality measure of $\pi$, but I could not find anything on a lower bound except the general $\mu(\pi)\geq2$. Looks like it is the best one known, is it? What is the conventional wisdom on what $\mu(\pi)$ should be? Does a bound follow from some grand conjecture (Schanuel's, Riemann hypothesis, etc.)?

I am interested because Alekseyev proved (Theorem 2) that $\frac{1}{n^u\sin{n}}$ converges to $0$ if $\mu(\pi)<u+1$ and diverges if $\mu(\pi)>u+1$. If we replace $\sin{n}$ by a sequence of independent random variables $\xi_n$ uniformly distributed on $[-1,1]$ then $\frac{1}{n^u\xi_n}$ converges to $0$ almost surely iff $\sum\frac{1}{n^u}<\infty$. Thus, if $\mu(\pi)>2$ then $\sin{n}$ is not 'uniformly distributed' on $[-1,1]$. On the other hand, if $\mu(\pi)=2$ then $\pi$ is just as irrational as $e$ and irrational algebraic numbers, while some people seem to believe that it should be 'more irrational' than that.

Answer 1

The expected value is $2$, simply because almost all real numbers have irrationality measure $2$. Note that this is "almost all" in the sense of measure theory. In the sense of topological category, it's the other way around: the set of reals of measure $\infty$ is residual. As I pointed out recently, it's a bit mysterious why measure theory makes better predictions than topological category for "naturally defined" numbers like $e$ or $\pi$, but experimentally, this appears to be the case, as the upper bound on the irrationality measure of $\pi$ is finite.