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Math Help - Limit of sequence

  1. #1
    Len
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    Limit of sequence



    Using n>=N ==> | A_n - L | < e

    Most the proofs I read say to start with | a_n - 1 | < e and work backwards to find N but I have done so and I cannot eliminate then n term. I am stuck and any tips / help would be hugely appreciated.

    I end up getting

    1+2n - (-1)^n
    -------------- < e
    n^2 +2n+1
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Len View Post


    Most the proofs I read say to start with | a_n - 1 | < e and work backwards to find N but I have done so and I cannot eliminate then n term. I am stuck and any tips / help would be hugely appreciated.

    I end up getting

    1+2n - (-1)^n
    -------------- < e
    n^2 +2n+1
    Your sequence definition makes no sense. The left hand side is independent of n, but you are using that to define your a_n.

    -Dan
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  3. #3
    Len
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    Quote Originally Posted by topsquark View Post
    Your sequence definition makes no sense. The left hand side is independent of n, but you are using that to define your a_n.

    -Dan
    Sorry I wrote that wrong. I am trying to show that A_n converges to 1.

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  4. #4
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    You want to prove a_n \to 1.

    Thus, we have to show \left| \frac{n^2+(-1)^n}{(n+1)^2} - 1 \right| < \epsilon.

    Thus, \left| \frac{n^2 + (-1)^n - (n+1)^2}{(n+1)^2}\right| < \epsilon.

    Thus, \left| \frac{n^2 + (-1)^n - n^2 - 2n - 1}{(n+1)^2} \right| < \epsilon \implies \left| \frac{(-1)^n - 2n}{(n+1)^2} \right| < \epsilon.

    But note, \left| \frac{(-1)^n - 2n }{(n+1)^2} \right| \leq \frac{2n+1}{n^2} \leq \frac{2n + n}{n^2} \leq \frac{3}{n}.

    So if we can make 3/n < \epsilon then |a_n - 1| < \epsilon. And this of course can be made by choosing N > 3/\epsilon.
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  5. #5
    Len
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    Quote Originally Posted by ThePerfectHacker View Post
    You want to prove a_n \to 1.

    Thus, we have to show \left| \frac{n^2+(-1)^n}{(n+1)^2} - 1 \right| < \epsilon.

    Thus, \left| \frac{n^2 + (-1)^n - (n+1)^2}{(n+1)^2}\right| < \epsilon.

    Thus, \left| \frac{n^2 + (-1)^n - n^2 - 2n - 1}{(n+1)^2} \right| < \epsilon \implies \left| \frac{(-1)^n - 2n}{(n+1)^2} \right| < \epsilon.

    But note, \left| \frac{(-1)^n - 2n }{(n+1)^2} \right| \leq \frac{2n+1}{n^2} \leq \frac{2n + n}{n^2} \leq \frac{3}{n}.

    So if we can make 3/n < \epsilon then |a_n - 1| < \epsilon. And this of course can be made by choosing N > 3/\epsilon.


    Thanks for help but my last question is do i need the = between

    |a_n - 1| <= 3/n . I do not think they are ever equal.
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