$$\sum_{n=1}^\infty \sin^{[n]}(1)$$ Where by $\sin^{[n]}(1)$ we mean $ \sin\left(\sin\left(\dots\sin(1)\right)\right)$ composed $n$ times.
Have tried the divergence test, which fails. Have tried Ratio test, also fails, as the limit is 1. Integral test, or root test do not seem promising. Help is appreciated
Once we prove the inequality $$\sin x > \frac{x}{1+x} \qquad (\star)$$ for $x \in (0,1]$, we can inductively show that $\sin^{[n]}(1) > \frac1n$ when $n\ge 2$. We have $\sin \sin 1 \approx 0.75 > \frac12$, and whenever $\sin^{[n]}(1) > \frac1n$, we have $$\sin^{[n+1]}(1) = \sin \sin^{[n]}(1) > \sin \frac1n > \frac{1/n}{1+1/n} = \frac1{n+1}.$$ Therefore, by the comparison test, $$\sum_{n=1}^\infty \sin^{[n]}(1) = \sin 1 + \sum_{n=2}^\infty \sin^{[n]}(1) > \sin 1 + \sum_{n=2}^\infty \frac1n,$$ which diverges.
To prove $(\star)$... well, to be honest, I just graphed both sides. But we can prove that $\sin x > x - \frac{x^3}{6}$ on the relevant interval by thinking about the error term in the Taylor series, and $x - \frac{x^3}{6} > \frac{x}{1+x}$ can be rearranged to $(x+1)(x -\frac{x^3}{6}) - x > 0$, which factors as $-\frac16 x^2(x-2)(x+3) > 0$.
Another possibility is to try to find an equivalent of the general term of this sequence : $\begin{cases} a_0=1\\ a_{n+1}=sin(a_n) \end{cases}$
Note that $f(x)=sin(x)$ has derivative $f'(x)=cos(x)$ which is positive on $[0,a_0]$ and also $<1$ on $]0,a_0[$ so $f$ is a contraction. From there it is easy to prove that $a_n\to 0$.
This means that $a_{n+1}\sim a_n$ when $n\to\infty$.
In this kind of problem we always search for an $\alpha$ such that $\mathbf{(a_{n+1})^\alpha-(a_n)^\alpha}$ does not depend of $\mathbf{a_n}$ (in the $\sim$ sense of the expression) so we are able to solve the recurrence.
From Taylor expansion $\displaystyle{(a_{n+1})^\alpha = \bigg(a_n - \frac{a_n^3}{6}+o(a_n^4)\bigg)^\alpha}=(a_n)^\alpha\bigg(1 - \frac{a_n^2}{6}+o(a_n^3)\bigg)^\alpha=(a_n)^\alpha\big(1-\frac{\alpha}{6}a_n^2+o(a_n^3)\big)$
So $(a_{n+1})^\alpha-(a_n)^\alpha=-\frac{\alpha}{6}(a_n)^{\alpha+2}+o((a_n)^{\alpha+3})\quad$ we see that we need $\alpha=-2$
Let's put $b_n=1/(a_n)^2,\qquad$ $b_n\to\infty$
We have $b_{n+1}-b_n=\frac 13+o(1/b_n)$ thus $b_n\sim\frac n3\qquad$
(more precisely $b_n=n/3+o(\ln(n))$ but it is not important at this point).
Finally $a_n\sim\sqrt\frac 3n,$ which is a term of a divergent serie so $\sum a_n$ diverges as well.
I wrote this answer for another question, but this question was pointed out in a comment, so I posted it here.
$|\,a_{n+1}|=|\sin(a_n)\,|\le|\,a_n|$. Thus, $|\,a_n|$ is decreasing and bounded below by $0$. Thus, $|\,a_n|$ converges to some $a_\infty$, and we must then have $\sin(a_\infty)=a_\infty$, which means that $a_\infty=0$.
Using $\sin(x)=x-\frac16x^3+O\!\left(x^5\right)$, we get $$ \begin{align} \frac1{a_{n+1}^2}-\frac1{a_n^2} &=\frac1{\sin^2(a_n)}-\frac1{a_n^2}\\ &=\frac{a_n^2-\sin^2(a_n)}{a_n^2\sin^2(a_n)}\\ &=\frac{\frac13a_n^4+O\!\left(a_n^6\right)}{a_n^4+O\!\left(a_n^6\right)}\\ &=\frac13+O\!\left(a_n^2\right)\tag1 \end{align} $$ Stolz-Cesàro says that $$ \lim_{n\to\infty}\frac{\frac1{a_n^2}}n=\frac13\tag2 $$ That is, $$ \bbox[5px,border:2px solid #C0A000]{a_n\sim\sqrt{\frac3n}}\tag3 $$ which means that the series diverges.
Motivation for $\boldsymbol{(1)}$
Note that $$ a_{n+1}-a_n=\sin(a_n)-a_n\sim-\frac16a_n^3\tag4 $$ which is a discrete version of $$ \frac{\mathrm{d}a_n}{a_n^3}=-\frac{\mathrm{d}n}6\tag5 $$ whose solution is $$ \frac1{a_n^2}=\frac{n-n_0}3\tag6 $$ so that $$ \frac1{a_{n+1}^2}-\frac1{a_n^2}=\frac13\tag7 $$ which suggests $(1)$.
