Results 1 to 12 of 12

Math Help - Prove a sequence is nonincreasing by induction

  1. #1
    Junior Member
    Joined
    Sep 2009
    Posts
    62

    Prove a sequence is nonincreasing by induction

    I'm stuck on this....

    Problem:
    Show that (tn) is a nonincreasing sequence. That is, use induction to prove that t_{n} \geq t_{n+1} where t_{n+1}=\left(1-\frac{1}{(n+1)^2}\right)t_{n}.

    I have the base case so, 1 \geq \frac{3}{4}. Then assume t_{k} \geq t_{k+1}.
    But I'm stuck at this part. I don't know how to arrive at t_{k+1} \geq t_{k+2}. I tried messing with the inequalities and stuff, but I'm still not sure.

    Edit: The only thing I could really come up with was .....

    (1-\frac{1}{(k+1)^2}) \geq (1-\frac{1}{(k+1)^2})(1-\frac{1}{(k+2)^2} then use the fact t_{k+1}=1-\frac{1}{(k+1)^2} and so \frac{t_{k+1}}{t_{k}}=1-\frac{1}{(n+1)^2}.
    Last edited by JSB1917; November 13th 2012 at 11:56 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,969
    Thanks
    1788
    Awards
    1

    Re: Prove a sequence is nonincreasing by induction

    Quote Originally Posted by JSB1917 View Post
    Problem:
    Show that (tn) is a nonincreasing sequence. That is, use induction to prove that t_{n} \geq t_{n+1} where t_{n+1}=1-\frac{1}{(n+1)^2}.
    You better double check the problem!
    t_{n+1}=1-\frac{1}{(n+1)^2} is an increasing sequence.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Sep 2009
    Posts
    62

    Re: Prove a sequence is nonincreasing by induction

    Yeah, sorry, it's actually t_{n+1}=\left(1-\frac{1}{(n+1)^2}\right)t_{n} I forgot the t_{n} at the end. I put that in the original post now.
    Last edited by JSB1917; November 13th 2012 at 11:56 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,969
    Thanks
    1788
    Awards
    1

    Re: Prove a sequence is nonincreasing by induction

    Quote Originally Posted by JSB1917 View Post
    Yeah, sorry, it's actually t_{n+1}=\left(1-\frac{1}{(n+1)^2}\right)t_{n} I forgot the t_{n} at the end. I put that in the original post now.

    Suppose you know that t_k\ge t_{k+1}.
    Then multiply: t_k\left( {1 - \frac{1}{{(k + 1)^2 }}} \right)\ge t_{k+1}\left( {1 - \frac{1}{{(k + 1)^2 }}} \right)
    But  \left( {1 - \frac{1}{{(k + 1)^2 }}} \right) \ge \left( {1 - \frac{1}{{(k + 2)^2 }}} \right)
    So t_{k+1}\ge t_{k+2}
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Sep 2009
    Posts
    62

    Re: Prove a sequence is nonincreasing by induction

    Quote Originally Posted by Plato View Post
    Suppose you know that t_k\ge t_{k+1}.
    Then multiply: t_k\left( {1 - \frac{1}{{(k + 1)^2 }}} \right)\ge t_{k+1}\left( {1 - \frac{1}{{(k + 1)^2 }}} \right)
    But  \left( {1 - \frac{1}{{(k + 1)^2 }}} \right) \ge \left( {1 - \frac{1}{{(k + 2)^2 }}} \right)
    So t_{k+1}\ge t_{k+2}
    Yeah, but I don't think it's correct that  \left( {1 - \frac{1}{{(k + 1)^2 }}} \right) \ge \left( {1 - \frac{1}{{(k + 2)^2 }}} \right)

    Because....
     (k+1)^2 \leq (k+2)^2, then
     -\frac{1}{(k+2)^2} \geq -\frac{1}{(k+1)^2} and so
     1-\frac{1}{(k+2)^2} \geq 1-\frac{1}{(k+1)^2}
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,969
    Thanks
    1788
    Awards
    1

    Re: Prove a sequence is nonincreasing by induction

    Quote Originally Posted by JSB1917 View Post
    Yeah, but I don't think it's correct that  \left( {1 - \frac{1}{{(k + 1)^2 }}} \right) \ge \left( {1 - \frac{1}{{(k + 2)^2 }}} \right)
    Because....
     (k+1)^2 \leq (k+2)^2, then
     -\frac{1}{(k+2)^2} \geq -\frac{1}{(k+1)^2} and so
     1-\frac{1}{(k+2)^2} \geq 1-\frac{1}{(k+1)^2}
    You are correct about that. Sorry, I got lost in my own notation.
    I still have your OP in my head. It is odd how the brains holds on to things.
    Last edited by Plato; November 13th 2012 at 01:22 PM.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Sep 2009
    Posts
    62

    Re: Prove a sequence is nonincreasing by induction

    Quote Originally Posted by Plato View Post
    You are correct about that. Sorry, I got lost in my own notation.
    I still have your OP in my head.
    I know that we need to work backwards from t_{k+2}\le t_{k+1}.
    I think using  \left(1-\frac{1}{(k+2)^2}\right) \geq \left(1-\frac{1}{(k+1)^2}\right) that then

    \left(1-\frac{1}{(k+2)^2}\right)t_{k} \geq \left(1-\frac{1}{(k+2)^2}\right)t_{k+1} and so,

    \left(1-\frac{1}{(k+2)^2}\right)t_{k} \geq  \left(1-\frac{1}{(k+1)^2}\right)t_{k} \geq \left(1-\frac{1}{(k+2)^2}\right)t_{k+1} and so

    t_{k+1} \geq t_{k+2}

    That seems right to me. Thoughts?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,969
    Thanks
    1788
    Awards
    1

    Re: Prove a sequence is nonincreasing by induction

    Quote Originally Posted by JSB1917 View Post
    \left(1-\frac{1}{(k+2)^2}\right)t_{k} \geq  \left(1-\frac{1}{(k+1)^2}\right)t_{k} \geq \left(1-\frac{1}{(k+2)^2}\right)t_{k+1} and so Thoughts?
    I do not follow that. As I said above, the brain just get stuck on an idea and will not let it go.

    Think about this. We know t_{k + 1}  = \left( {1 - \frac{1}{{(k + 1)^2 }}} \right)t_k .

    Consider t_{k + 1}  - t_k  = \left( {1 - \frac{1}{{(k + 1)^2 }}} \right)t_k  - t_k  = \left( { - \frac{{t_k }}{{(k + 1)^2 }}} \right) < 0.

    What does that tell us?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member
    Joined
    Sep 2009
    Posts
    62

    Re: Prove a sequence is nonincreasing by induction

    Quote Originally Posted by Plato View Post
    I do not follow that. As I said above, the brain just get stuck on an idea and will not let it go.

    Think about this. We know t_{k + 1}  = \left( {1 - \frac{1}{{(k + 1)^2 }}} \right)t_k .

    Consider t_{k + 1}  - t_k  = \left( {1 - \frac{1}{{(k + 1)^2 }}} \right)t_k  - t_k  = \left( { - \frac{{t_k }}{{(k + 1)^2 }}} \right) < 0.

    What does that tell us?
    Aside from it's negative, not sure.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,969
    Thanks
    1788
    Awards
    1

    Re: Prove a sequence is nonincreasing by induction

    Quote Originally Posted by JSB1917 View Post
    Aside from it's negative, not sure.
    Surely t_{k+1}-t_k<0 says t_{k+1}<t_k
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Junior Member
    Joined
    Sep 2009
    Posts
    62

    Re: Prove a sequence is nonincreasing by induction

    Quote Originally Posted by Plato View Post
    Surely t_{k+1}-t_k<0 says t_{k+1}<t_k
    yeah...I know that, but where is this leading? I'm just not seeing it.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Super Member
    Joined
    Mar 2010
    Posts
    993
    Thanks
    244

    Re: Prove a sequence is nonincreasing by induction

    If we have:

    t_{k + 1} = \left( {1 - \frac{1}{{(k + 1)^2 }}} \right)t_k

    then since 1 - \frac{1}{{(k + 1)^2 }}<1, t_{k + 1} < t_k and therefore the sequence is decreasing.

    - Hollywood
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: November 7th 2011, 10:35 PM
  2. Replies: 4
    Last Post: September 21st 2010, 12:35 PM
  3. Replies: 10
    Last Post: June 29th 2010, 01:10 PM
  4. Replies: 2
    Last Post: August 28th 2009, 03:59 AM
  5. Fibonacci sequence - prove by induction
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: December 23rd 2008, 06:19 PM

Search Tags


/mathhelpforum @mathhelpforum