# How does this hyperoperator for sin^2 converge and what's the limit?

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!

• 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. – Yves Daoust Jul 14 '19 at 20:11
• Calculated 5 terms, like you :) – buddhabrot Jul 14 '19 at 20:12
• I edited the question a little based on this discussion, thanks for pointing out there's no religion involving 7/3. – buddhabrot Jul 14 '19 at 20:25
• $\lim_\limits{n \to \infty}(1+\frac{1}{n^2})^{n^2}=e.$ – Yves Daoust Jul 14 '19 at 20:28
• Ah oops, of course. Sorry, I'll edit that, my point stays the same ("mastering the convergance"). thx again – 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 $$\sin(1)$$.