1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Convergence of a sequence

  1. Feb 4, 2010 #1
    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.

    [tex] 0 < a_0 < \frac {\pi}{2} [/tex]


    [tex]sin(a_n)= \frac {a_n}{a_{n+1}} [/tex]


    and so [tex] 1 > sin(a_0)= \frac {a_0}{a_{1}} [/tex] therefore [tex]a_{1}> a_0 [/tex]

    At first I thought this was simple and the sequence converges to [tex] \frac {\pi}{2} [/tex]
    But I realised that the inequality can hold for any n, i.e [tex] a_{n+1}> a_n [/tex] 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. jcsd
  3. Feb 4, 2010 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Convergence of a sequence
  1. Sequence Convergence (Replies: 6)

  2. Sequence Convergence (Replies: 2)

Loading...