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Math Help - Proving Convergence

  1. #1
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    Proving Convergence

    Let (
    an) be a sequence which converges to 0. Use the definition of a convergent sequence to prove that the sequence (an^2) to 0.

    Definition: There exists an L that is Real s.t. for every E>0 there exists N that is a positive integer s.t. for every n that is a positive integer n>N implies that |an - L| < E

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    Quote Originally Posted by ah-bee View Post
    Let (
    an) be a sequence which converges to 0. Use the definition of a convergent sequence to prove that the sequence (an^2) to 0.

    Definition: There exists an L that is Real s.t. for every E>0 there exists N that is a positive integer s.t. for every n that is a positive integer n>N implies that |an - L| < E

    Since a_n\to 0 it is bounded, i.e. there is A>0 so that |a_n|\leq A.

    This means, |a_n^2| = |a_n||a_n|\leq A|a_n| < A\epsilon for n>N.
    Last edited by ThePerfectHacker; May 5th 2008 at 07:46 PM.
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    i dont understand how works. can u explain to me how it works?
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    Quote Originally Posted by ah-bee View Post
    i dont understand how works. can u explain to me how it works?
    There exists A>0 so that |a_n|\leq A. Because convergent sequences are bounded.

    To prove that a_n^2 \to 0 we need to show |a_n^2 - 0| < \epsilon for any \epsilon > 0 where N is sufficiently large.

    Give me any \epsilon > 0, then since a_n \to 0 it means |a_n - 0| = |a_n|< \frac{\epsilon}{A} for n>N.

    But then if n>N we have |a_n^2 - 0| = |a_n^2| = |a_n||a_n|\leq A|a_n| < A\cdot \frac{\epsilon}{A} = A.
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    thanks, i understand it now, so if i were to get a question saying if an converges to k, prove an^2 converges to k^2, i would approach it in the same way?
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  6. #6
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    Quote Originally Posted by ah-bee View Post
    thanks, i understand it now, so if i were to get a question saying if an converges to k, prove an^2 converges to k^2, i would approach it in the same way?
    Not exactly in the same way. But it should be similar.
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