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Math Help - two sequences converging to 0

  1. #1
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    two sequences converging to 0

    Let (an) and (bn) be two sequences converging to 0. Prove that the sequence (anbn) converges to 0. (Must prove the statement directly from the definition of convergence.)


    I don't know how to prove by definition of convergence.I'm a bit confused with "definition of limit" and "definition of convergence". Are they same which is the "epsilon-delta definition".
    Please help me,thanks a lot.
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  2. #2
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    I believe that it is the same as the epsilon delta proof. Basically, given epsilon > 0, you need to find an N such that n> N implies the magnitude of an times bn is < epsilon. To do this you need to use the epsilon-delta definitions of an and bn's convergences.
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  3. #3
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    \lim_{n\to\infty} a_n = 0
    \lim_{n\to\infty} b_n = 0

    \forall   \epsilon > 0  \exists n_\epsilon \in N: \forall n'\ge n_\epsilon   | a_n' - 0 | <  \epsilon same holds for \b_n

    \ |a_n\cdot b_n -  0 | = |a_n||b_n|
    since \forall   \epsilon > 0  \exists n_\epsilon \in N: \forall n'\ge n_\epsilon   | a_n' - 0 | <  \epsilon set \epsilon = \sqrt\epsilon '
    so \exist n_\epsilon : \forall n' \ge \n_\epsilon |a_n| < \epsilon = \sqrt\epsilon ' same for b_n, lets call the resulting \ n_\epsilon for \ b_n m_\epsilon

    Let \ k = \max  ( n_\epsilon , m_\epsilon )
    so \forall n'\ge k |a_n'\cdot b_n | = |a_n|\cdot |b_n| < \ (\sqrt\epsilon ' )^2 = \epsilon'

    or you can use the fact that all 1) sequences converge if and only if they are cauchy in R (a complete space) and 2) all cauchy sequences are bounded to prove that. use the identity
    \ |a_n\cdot b_n - a_m\cdot\b_m| = |a_n\cdot b_n - a_m\cdot b_n + b_n\cdot a_m - a_m\cdot b_m| = |(a_n\cdot b_n - a_m\cdot b_n) + (b_n\cdot a_m + b_m\cdot a_m) |
    by triangle inequality this simplifies too \ | a_n\cdot b_n - a_m\cdot b_m| \le |b_n|\cdot |a_n - a_m| + |a_m|\cdot |b_n - b_m|
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