# Convergence of a sequence

1. Feb 4, 2010

### 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.
1. The problem statement, all variables and given/known data
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 realised 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

Last edited: Feb 4, 2010
2. Feb 4, 2010

### 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?