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Thread: Prove a sequence is nonincreasing by induction

  1. #1
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    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 $\displaystyle t_{n} \geq t_{n+1}$ where $\displaystyle t_{n+1}=\left(1-\frac{1}{(n+1)^2}\right)t_{n}.$

    I have the base case so, $\displaystyle 1 \geq \frac{3}{4}$. Then assume $\displaystyle t_{k} \geq t_{k+1}$.
    But I'm stuck at this part. I don't know how to arrive at $\displaystyle t_{k+1} \geq t_{k+2}$. 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 .....

    $\displaystyle (1-\frac{1}{(k+1)^2}) \geq (1-\frac{1}{(k+1)^2})(1-\frac{1}{(k+2)^2}$ then use the fact $\displaystyle t_{k+1}=1-\frac{1}{(k+1)^2}$ and so $\displaystyle \frac{t_{k+1}}{t_{k}}=1-\frac{1}{(n+1)^2}.$
    Last edited by JSB1917; Nov 13th 2012 at 10:56 AM.
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  2. #2
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    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 $\displaystyle t_{n} \geq t_{n+1}$ where $\displaystyle t_{n+1}=1-\frac{1}{(n+1)^2}.$
    You better double check the problem!
    $\displaystyle t_{n+1}=1-\frac{1}{(n+1)^2}$ is an increasing sequence.
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  3. #3
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    Re: Prove a sequence is nonincreasing by induction

    Yeah, sorry, it's actually $\displaystyle t_{n+1}=\left(1-\frac{1}{(n+1)^2}\right)t_{n}$ I forgot the $\displaystyle t_{n}$ at the end. I put that in the original post now.
    Last edited by JSB1917; Nov 13th 2012 at 10:56 AM.
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    Re: Prove a sequence is nonincreasing by induction

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

    Suppose you know that $\displaystyle t_k\ge t_{k+1}$.
    Then multiply: $\displaystyle t_k\left( {1 - \frac{1}{{(k + 1)^2 }}} \right)\ge t_{k+1}\left( {1 - \frac{1}{{(k + 1)^2 }}} \right)$
    But $\displaystyle \left( {1 - \frac{1}{{(k + 1)^2 }}} \right) \ge \left( {1 - \frac{1}{{(k + 2)^2 }}} \right) $
    So $\displaystyle t_{k+1}\ge t_{k+2}$
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    Re: Prove a sequence is nonincreasing by induction

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

    Because....
    $\displaystyle (k+1)^2 \leq (k+2)^2$, then
    $\displaystyle -\frac{1}{(k+2)^2} \geq -\frac{1}{(k+1)^2} $ and so
    $\displaystyle 1-\frac{1}{(k+2)^2} \geq 1-\frac{1}{(k+1)^2} $
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    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 $\displaystyle \left( {1 - \frac{1}{{(k + 1)^2 }}} \right) \ge \left( {1 - \frac{1}{{(k + 2)^2 }}} \right) $
    Because....
    $\displaystyle (k+1)^2 \leq (k+2)^2$, then
    $\displaystyle -\frac{1}{(k+2)^2} \geq -\frac{1}{(k+1)^2} $ and so
    $\displaystyle 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; Nov 13th 2012 at 12:22 PM.
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  7. #7
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    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 $\displaystyle t_{k+2}\le t_{k+1}$.
    I think using $\displaystyle \left(1-\frac{1}{(k+2)^2}\right) \geq \left(1-\frac{1}{(k+1)^2}\right) $ that then

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

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

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

    That seems right to me. Thoughts?
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  8. #8
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    Re: Prove a sequence is nonincreasing by induction

    Quote Originally Posted by JSB1917 View Post
    $\displaystyle \left(1-\frac{1}{(k+2)^2}\right)t_{k} \geq \left(1-\frac{1}{(k+1)^2}\right)t_{k} \geq \left(1-\frac{1}{(k+2)^2}\right)t_{k+1}$ 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 $\displaystyle t_{k + 1} = \left( {1 - \frac{1}{{(k + 1)^2 }}} \right)t_k $.

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

    What does that tell us?
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    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 $\displaystyle t_{k + 1} = \left( {1 - \frac{1}{{(k + 1)^2 }}} \right)t_k $.

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

    What does that tell us?
    Aside from it's negative, not sure.
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    Re: Prove a sequence is nonincreasing by induction

    Quote Originally Posted by JSB1917 View Post
    Aside from it's negative, not sure.
    Surely $\displaystyle t_{k+1}-t_k<0$ says $\displaystyle t_{k+1}<t_k$
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  11. #11
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    Re: Prove a sequence is nonincreasing by induction

    Quote Originally Posted by Plato View Post
    Surely $\displaystyle t_{k+1}-t_k<0$ says $\displaystyle t_{k+1}<t_k$
    yeah...I know that, but where is this leading? I'm just not seeing it.
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  12. #12
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    Re: Prove a sequence is nonincreasing by induction

    If we have:

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

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

    - Hollywood
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