Results 1 to 3 of 3

Thread: 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:

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

    satisfies $\displaystyle 0 < a_n <= 2$ 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...


    $\displaystyle a_1 \geq 2$

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

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

  3. #3
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    5
    Quote Originally Posted by FalconPUNCH! View Post
    Show that the sequences defined by:

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

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

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

    $\displaystyle
    a_{n+1}=\phi+\varepsilon=\frac{1}{3-a_n}>\frac{1}{1-\phi}
    $

    So:

    $\displaystyle
    [\phi(3-\phi)-1] + \varepsilon (3-\phi) >0
    $

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

    $\displaystyle \varepsilon (3-\phi)>0$

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

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

    Now consider:

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

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

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

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

    $\displaystyle
    a=\frac{1}{3-a}
    $

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

    RonL
    Last edited by CaptainBlack; Apr 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: Oct 6th 2009, 01:20 PM
  2. Sequence proof
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: Oct 5th 2009, 04:59 PM
  3. Sequence proof help
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: Sep 23rd 2009, 05:08 AM
  4. Sequence proof
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: Sep 23rd 2009, 12:37 AM
  5. Proof of sequence
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Nov 11th 2008, 07:04 AM

Search Tags


/mathhelpforum @mathhelpforum