Convergence of Sequence a_n = a_{n+1}/sin(a_n) and its Limit Calculation

[talolard]

Hello, this is a question we had on an exam and I can't figure it out. Our professors won't publish solutions so I'd be glad for your help.

Homework Statement

Prove the following series converges and calculate its limit.

0 < a_0 < \frac {\pi}{2}sin(a_n)= \frac {a_n}{a_{n+1}}and so 1 > sin(a_0)= \frac {a_0}{a_{1}} therefore a_{1}> a_0

At first I thought this was simple and the sequence converges to \frac {\pi}{2}
But I realized that the inequality can hold for any n, i.e a_{n+1}> a_n because we have no way of knowing by how much it is bigger. This one was on our exam and no one I talked to managed to overcome this little detail.
Some pointers would be greatly apreciated.
Thanks
Tal

 

[Dick]

It does converge to pi/2 and it is pretty simple. Consider the function f(x)=x/sin(x). First figure out what is the range of values of f(x) for x in (0,pi/2). I.e. look for maxs and mins of the function using critical points. Second, show f(x)>x for x in (0,pi/2). If you can show f(x) is bounded by pi/2 for x in (0,pi/2) then a_n is an increasing sequence bounded by pi/2. So a limit exists. If a limit exists then it must satisfy L=L/sin(L). What's the limit?