Results 1 to 3 of 3

Math Help - Sequence Proof

  1. #1
    Member FalconPUNCH!'s Avatar
    Joined
    Sep 2007
    From
    Daly City, CA
    Posts
    167

    Sequence Proof

    Show that the sequences defined by:

    a_1 = 2 ;  a_{n+1} = \frac{1}{3-a_n}

    satisfies  0 < a_n <= 2 and is decreasing. Deduce that the sequence is convergent and find it's limit.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hello,

    Use induction to prove it...


    a_1 \geq 2

    If a_n \geq 2 would a_{n+1}=\frac{1}{3-a_n} be  \geq 2?

    It's the same for a_n > 0
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by FalconPUNCH! View Post
    Show that the sequences defined by:

    a_1 = 2 ;  a_{n+1} = \frac{1}{3-a_n}

    satisfies  0 < a_n <= 2 and is decreasing. Deduce that the sequence is convergent and find it's limit.
    Put \phi=\frac{3-\sqrt{5}}{2} then \phi is a root of x^2-3x+1=0.
    (we will also require the other root \phi^*=\frac{3+\sqrt{5}}{2} later so we may as well anounce it here)

    Now suppose a_n>\phi, then let a_{n+1}=\phi+\varepsilon, and we have:

     <br />
a_{n+1}=\phi+\varepsilon=\frac{1}{3-a_n}>\frac{1}{1-\phi}<br />

    So:

     <br />
[\phi(3-\phi)-1] + \varepsilon (3-\phi) >0<br />

    But [\phi(3-\phi)-1]=-\phi^2+3\phi-1=0, so:

    \varepsilon (3-\phi)>0

    and as 3-\phi>0, we have \varepsilon>0, and so a_{n+1}>\phi.

    Hence as a_1>\phi, a_n>\phi for all n \in \mathbb{Z}_+ .

    Now consider:

    \frac{a_{n-1}}{a_n}=a_{n-1}(3-a_{n-1})

    This is greater than 1 for a_{n-1} between the roots of -x^2+3x-1 that is between \frac{3 \pm \sqrt{5}}{2} .

    That is \{a_n \} is decreasing while a_n<\frac{3+\sqrt{5}}{2}=\phi^* and a_n>\frac{3-\sqrt{5}}{2}=\phi and as 2< \frac{3+\sqrt{5}}{2}, \{a_n\} is a decreasing sequence and also bounded below, so converges.

    Now if \{a_n\} converges the limit a satisfies:

     <br />
a=\frac{1}{3-a}<br />

    which has roots \phi and \phi^*. But it cannot converge on \phi^* as the sequence is decreasing so the limit is \phi.

    RonL
    Last edited by CaptainBlack; April 22nd 2008 at 11:09 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Sequence proof
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: October 6th 2009, 01:20 PM
  2. Sequence proof
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: October 5th 2009, 04:59 PM
  3. Sequence proof help
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: September 23rd 2009, 05:08 AM
  4. Sequence proof
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: September 23rd 2009, 12:37 AM
  5. Proof of sequence
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 11th 2008, 07:04 AM

Search Tags


/mathhelpforum @mathhelpforum