prove that $\{a_{n}\} $ converges as $, a_{n+1}=\frac{a_{n}}{sin(a_{n})},\ \forall n\geq2$ and $0<a_{1}<\frac{\pi}{2}$

Question

let $\{a_{n}\}_{n=1}^{\infty} $ be a sequence defined as $$0<a_{1}<\frac{\pi}{2}$$ and $$ a_{n+1}=\frac{a_{n}}{\sin(a_{n})},\ \forall n\geq2$$

prove that $\{a_{n}\} $ converges, and find the limit of $\{a_{n}\}$.

Answer 1

Finding the limit is easy: set $a_{n+1}=a_n=\alpha$, and we get: $$\alpha=\frac{\alpha}{\sin\alpha}$$ We get $\sin\alpha=1$, which leads to $\alpha=\frac\pi2$

So it converges to $\frac\pi2$.

To prove it's convergence, show that when you start with $0<a_1<\frac{\pi}{2}$, that $a_{n+1}>a_n\forall n$ (It is as shown here)