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Math Help - convergence help

  1. #1
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    convergence help

    Suppose that the summation from k=1 to infinity of a_k converges absolutely. Prove that the series summation from k=1 to infinity of (a_k)^2 converges.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by friday616 View Post
    Suppose that the summation from k=1 to infinity of a_k converges absolutely. Prove that the series summation from k=1 to infinity of (a_k)^2 converges.
    Problem: Suppose that \sum_{n=1}^{\infty}a_n converges absolutely. Prove that \sum_{n=1}^{\infty}a^2_n converges as well.

    Proof: Since, a_n\to0 by definition there exists some N\in\mathbb{N} such that N\le n\implies |a_n|<1. Therefore, for N\le n\implies a^2_n=\left|a_n\right|^2<|a_n|. Therefore \sum_{n=N}^{\infty}a^2_n<\sum_{n=N}^{\infty}|a_n|, and since \sum_{n=1}^{\infty}a_n converges absolutely, \sum_{n=N}^{\infty}|a_n| does as well. Therefore, since \sum_{n=1}^{\infty}a^2_n=\sum_{n=1}^{N}a^2_n+\sum_  {n=N}^{\infty}a^2_n both of which converge it is clear that \sum_{n=1}^{\infty}a^2_n converges.
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  3. #3
    Moo
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    You seem to be sulking quantifiers

    Also, just a remark, maybe we can discuss about it... Isn't it more logical to put n\geq N instead of N\leq n ? Because we're "conditioning" n, while N is already fixed.
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Moo View Post
    You seem to be sulking quantifiers

    Also, just a remark, maybe we can discuss about it... Isn't it more logical to put n\geq N instead of N\leq n ? Because we're "conditioning" n, while N is already fixed.
    Honestly, I think it's semantics. I like the less than or equal sign to be facting left. If you feel that one has a different connotation thatn the other, well then I am sorry.
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