If $a_1=1$ and $a_n=\sec (a_{n-1})$ then what does the proportion of positive terms approach, as $n\to\infty$?

Question

Consider the sequence $a_1=1$ and $a_n=\sec (a_{n-1})$ for $n>1$.

What does the proportion of positive terms approach, as $n\to\infty$?

At first I thought the limiting proportion might be $\frac{1}{\sqrt {2\pi}}\approx 0.3989$, but perhaps not.

Answer 1

$\DeclareMathOperator{\arcsec}{arcsec}$ A quick heuristic argument: we can assume that values of secant mod $\pi$ are equidistributed. Then using symmetry, consider the graph of secant over the $[0, \pi/2]$ interval; we just need to look at the subintervals that go to $[0, \pi/2], [3\pi/2, 5\pi/2],\ldots$ and see what portion of that interval they make up. This suggests that the value should be $\frac{2}{\pi}\left(\arcsec(\frac\pi2)+\sum_{i=1}^\infty\left(\arcsec(\frac{(4i+1)\pi}2)-\arcsec(\frac{(4i-1)\pi}2)\right)\right)$ (where all values of $\arcsec$ here are taken in that $[0,\pi/2]$ interval) but I'd be surprised if there were a closed form for that.