# Lower bound on the irrationality measure of $\pi$

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.

• As far as I am aware, the conventional wisdom expects $\mu(\pi)=2$, so the lack of stronger lower bounds is not surprising. Aug 18, 2014 at 19:56
• $\sin n$ is not uniformly distributed, it is $n \mod 2\pi$ that is uniformly distributed in $[0,2\pi]$ but when you apply $\sin x$ the density changes. Apr 7, 2016 at 16:33

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.
• Thank you. But doesn't $e$ having a nice pattern in its continued fraction disqualify it from being "empirically random"? Although $\mu(e)=2$ could then be another argument for $\mu(\pi)=2$ since it has patterned generalized continued fractions too. Aug 18, 2014 at 22:45