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Let $\{a_n\}$ be a sequence defined by $0<a_1<{\pi \over 2}$, $a_{n+1}={a_n\over \sin a_n}$.

$Attempt:$ $a_1>0$ and $\sin a_1>0$ and therefore the sequence begins positive and remains positive since sine preserves sign. $a_{n+1}-a_n={a_n\over \sin a_n}-a_n=a_n({1-\sin a_n\over \sin a_n})$. By showing $0<a_n<{\pi \over 2}$ I can prove that the sequence is bounded, monotonically increasing and its supremum, ${\pi\over 2}$ is also the limit. Except I don't know how to show those steps. I would appreciate your reply.

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  • $\begingroup$ It of course is, but it is complicated to show it only with $a_1$... $\endgroup$ – Meitar Abarbanel Jan 28 '15 at 2:04
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Hint: Show the sequence is bounded using induction.

If $0 < a_n < \pi/2$, then applying a well-known inequality,

$$ \frac{2a_n}{\pi } \leqslant \sin a_n \leqslant a_n,$$

and

$$ 1 \leqslant a_{n+1} =\frac{a_n}{\sin a_n} \leqslant \frac{\pi}{2}.$$

Also, $0 < a_n < \pi/2 \implies 1 - \sin a_n > 0 \implies a_{n+1} > a_n.$

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