Results 1 to 4 of 4

Math Help - Help! Sequence question

  1. #1
    Newbie
    Joined
    Jan 2008
    Posts
    3

    Help! Sequence question

    For the sequence a1= 1, a(n+1)= 1 + 2a(n)/ 1 + a(n) n=1,2,3,..

    we need to show that the sequence is bounded from above by 100.

    I talked to my prof and he said to use induction with a(k) < 100 to show a(k+1) < 100.

    if I set it up, i get
    1 + 2a(k)/ 1+ a(k) < 100
    1 + 2a(k) < 100 (1+ a(k))
    1 + 2a(k) < 100 + 100 a(k)
    -99 < 98 a(k)

    which is where I am stuck. How would I use this to show that a(n) is bounded by 100?

    Please help!!!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Sep 2007
    Posts
    127
    Hello

    Ok, I'll just sketch the proof for you.

    n=1
    a_1 = 1 < 100

    Therefore true for n=1, assume n=k is true, that is,

     a_{k+1} = \frac{1+2a_k}{1+a_k} < 100

    Consider n=k+1

     a_{k+2} = \frac{1+2a_{k+1}}{1+a_{k+1}}

     = \frac{1 + \frac{2+4a_k}{1+a_k}}{1 + \frac{1+2a_k}{1+a_k}}

    after a bit of simple algebra, you'll arrive at

     a_{k+2} = \frac{3+5a_k}{2+3a_k} < \frac{3+6a_k}{3+3a_k} since  a_k > 1 , \forall k \in N

    But  \frac{3 + 6a_k}{3+3a_k} = \frac{1 + 2a_k}{1+ a_k} < 100 by the induction hypothesis.

    so  a_{k+2} < 100

    Therefore, if n=k is true, then n=k+1 is true. But since n=1 is true, this implies than n=2 is true, which implies n=3 is true..etc. Hence  a_{n+1} < 100 for all natural numbers n, which, by definition, implies that 100 is an upper bound.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jan 2008
    Posts
    3
    The only problem with that proof is that it is stating that ak+2<ak+1 when in fact ak+2 > ak+1 because {an} is increasing
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Dec 2007
    From
    Melbourne
    Posts
    428
    a_{k+2} = \frac{3+5a_k}{2+3a_k} < \frac{3+6a_k}{3+3a_k} since a_k > 1, \forall k \in N
    The inequality here doesn't work eg a_k = 1.

    I'm not sure what approach your professor was expecting but here is a method of proof:

    a_{k+1} = \frac {1+2a_k}{1+a_k}
    =\frac {2a_k+2-1}{a_k+1}
    = 2 - \frac {1}{a_k+1}
    <2 since a_k>-1.
    so a_k<100
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Sequence question
    Posted in the Number Theory Forum
    Replies: 6
    Last Post: May 6th 2009, 02:47 PM
  2. sequence question
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: April 29th 2009, 02:27 PM
  3. tell me the sequence in the below question;
    Posted in the Discrete Math Forum
    Replies: 5
    Last Post: April 14th 2009, 09:14 PM
  4. sequence question
    Posted in the Calculus Forum
    Replies: 5
    Last Post: April 12th 2009, 12:27 AM
  5. Sequence question
    Posted in the Algebra Forum
    Replies: 1
    Last Post: May 1st 2008, 09:21 PM

Search Tags


/mathhelpforum @mathhelpforum