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Math Help - convergence of sequence

  1. #1
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    convergence of sequence

    (Xn) = (1 + 1/n^2)^n^2

    i) does it converge?
    ii) find the limit.

    attempt at a solution for i):
    - prove that it is monotonic and bounded
    => convergence
    - bounded:
    |(Xn)| <= M , all n in Naturals
    (1 + 1/n^2)^n^2 <= M
    n^2ln((n^2 + 1)/n^2) <= lnM
    n^2ln(n^2 + 1)-n^2ln(n^2) <= lnM
    n^2ln(n^2 + 1)-n^2ln(n^2) <= n^2ln(n^2 + 1) so,
    n^2ln(n^2 + 1) <= lnM
    not sure how to continue...

    ii) have yet to try

    thanks
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  2. #2
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    Quote Originally Posted by sprinks13 View Post
    (Xn) = (1 + 1/n^2)^n^2

    i) does it converge?
    ii) find the limit.

    attempt at a solution for i):
    - prove that it is monotonic and bounded
    => convergence
    - bounded:
    |(Xn)| <= M , all n in Naturals
    (1 + 1/n^2)^n^2 <= M
    n^2ln((n^2 + 1)/n^2) <= lnM
    n^2ln(n^2 + 1)-n^2ln(n^2) <= lnM
    n^2ln(n^2 + 1)-n^2ln(n^2) <= n^2ln(n^2 + 1) so,
    n^2ln(n^2 + 1) <= lnM
    not sure how to continue...

    ii) have yet to try

    thanks
    By choosing n^2 terms of the sequence \left( 1 + \frac{1}{n} \right)^n we form the subsequence \left( 1 + \frac{1}{n^2} \right)^{n^2}. Since \left( 1 + \frac{1}{n} \right)^n \to e it means each and every subsequence converges to the same limit.
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