Given $0 < a_1 < \frac{\pi}{2}$, for $n\geq 1$ we define $$a_{n+1} = \dfrac{{a_n}}{\sin({a_n})}.$$ I need to show that $\lim_{n\rightarrow\infty}{a_n} = L$, and then I need to find $L$.
I can't find a way to bound $\{a_n\}_n$ so that $$0 <a_n < \frac{\pi}{2}$$ I've tried to use every convergence test I know, but I always have the same problem which I want to bound $\{a_n\}_n$ and can't to that without bounding $\{\sin(a_n)\}_n$.
Note that for $x\in (0,\pi/2)$, $$x<f(x):=\frac{x}{\sin(x)}<\frac{\pi}{2}.$$ The left inequality is trivial while the one on the right is equivalent to $2x/\pi<\sin(x)$ which holds because $\sin(x)$ is concave in $(0,\pi/2)$.
Therefore, since $a_{n+1}=f(a_n)$ and $a_1\in (0,\pi/2)$, it follows that the sequence $(a_n)$ is strictly increasing and bounded above by $\pi/2$.
This shows that the limit $\lim_{n\rightarrow\infty}{a_n}$ exists. Are you able to find it?