I was riding my bike yesterday and thinking about the following expression:

Suppose $$S_3 = \sin^2(\frac\pi2) + \sin^2(\sin^2(\frac\pi2)) + \sin^2(\sin^2(\sin^2(\frac\pi2)))$$ which serves to define an $$S_n$$ that has $n$ of these terms, the last of which being $n$ successive applications of $\sin^2$.

Now, you will appreciate that $$\lim\limits_{n \to \infty}{S_n}$$ goes "somewhere". I have no idea how to calculate it, and what type of maths would describe its calculation. All I know is that it seems to drift close to the religious rational number $\frac73$ - although as someone pointed out, not quite that number.

During the bike ride I was actually interested in building an expression $\lim\limits_{n \to \infty}(S_n-C)^{F(n)}$ and see if it combines $e$ and $\pi$ in a new expression. Note that for instance $$\lim\limits_{n \to \infty}{(1+\frac{1}{n^2})^{n^2}} = e$$ so even the quick "convergence" of $S_n$ can bloom open to something interesting when other operators are involved.

Now, sitting at my desk I realise that I don't even know where to start with that original idea, as I don't even know how to compute the limit.

Can anyone point me in a direction, does this hyperoperator for sin^2 converge? How does one prove it? Are there more things I could read about sine hyperoperators like these?

I'm probably overlooking something basic - I'm not that knowledgable in maths. Thanks for any hints!

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    $\begingroup$ A good estimate of the sum is $2.3284669142910280888416748949659$. You don't seem to have computed a lot. The series converges very fast, as the successive terms are obtained by squaring. $\endgroup$ – Yves Daoust Jul 14 '19 at 20:11
  • $\begingroup$ Calculated 5 terms, like you :) $\endgroup$ – buddhabrot Jul 14 '19 at 20:12
  • $\begingroup$ I edited the question a little based on this discussion, thanks for pointing out there's no religion involving 7/3. $\endgroup$ – buddhabrot Jul 14 '19 at 20:25
  • $\begingroup$ $\lim_\limits{n \to \infty}(1+\frac{1}{n^2})^{n^2}=e.$ $\endgroup$ – Yves Daoust Jul 14 '19 at 20:28
  • $\begingroup$ Ah oops, of course. Sorry, I'll edit that, my point stays the same ("mastering the convergance"). thx again $\endgroup$ – buddhabrot Jul 14 '19 at 20:29

If $x\le1$, $\sin^2(x)\le x\sin(1)\le1$ and the series is bounded by a converging geometric series of common ratio $\sin(1)$.

Below a plot of the first partial sums.

enter image description here

  • $\begingroup$ Is there then a nice expression for the limit? $\endgroup$ – buddhabrot Jul 14 '19 at 20:52
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    $\begingroup$ @buddhabrot: no. (Unless Ramanujan worked on that.) $\endgroup$ – Yves Daoust Jul 14 '19 at 20:54
  • $\begingroup$ That's incredibly disappointing. I will keep hunting based on your common ratio hint. $\endgroup$ – buddhabrot Jul 14 '19 at 20:56
  • $\begingroup$ @buddhabrot: this is a gross upper bound, it will lead you nowhere. This problem is intractable. $\endgroup$ – Yves Daoust Jul 14 '19 at 20:59
  • $\begingroup$ Yes I realise but even this simple ingredient on breaking the recursion has got me thinking. Still, I don't feel like accepting the answer, it avoids the meat, and want to enable other viewpoints that get closer. I have since changed it to "How does it converge". $\endgroup$ – buddhabrot Jul 14 '19 at 21:00

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