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Math Help - Proof with absolute and conditional convergence

  1. #1
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    Proof with absolute and conditional convergence

    hello, i've stumbled upon another proof. the problem i s:

    Prove that if sigma ace of n converges absolutely, then sigma (ace of n)^2 also converges. Then show by giving a counterexample that sigma (ace of n)^2 need not converge if sigma ace of n is only conditionally convergent.

    For the first part, can I simply state that if sigma ace of n converges absolutely, it is implied that it converges. If sigma ace of n converges, then the square also converges because p>2 (p-series)?
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  2. #2
    MHF Contributor matheagle's Avatar
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    what is sigma ace?
    sigma=summation I guess
    and is ace a_n?
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  3. #3
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    yes, sorry about that. i dont know how to code the math signs.
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  4. #4
    MHF Contributor matheagle's Avatar
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    I never heard of ace b4. The counterexample is

    \sum_{n=1}^{\infty}{(-1)^n\over \sqrt n}.
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  5. #5
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    thank you! so does that mean that my proof works?
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  6. #6
    MHF Contributor matheagle's Avatar
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    The proof of part 1 is...

    Since \sum_{n=1}^{\infty}|a_n| converges, we have |a_n|\to 0, so just find N where |a_n|<1, whenever n \ge N.

    Then \sum_{n=N}^{\infty}a_n^2=\sum_{n=N}^{\infty}|a_n|\  cdot |a_n|< \sum_{n=N}^{\infty}|a_n|<\infty<br />
.

    And of course \sum_{n=1}^{\infty}a_n^2=\sum_{n=1}^{N-1}a_n^2+\sum_{n=N}^{\infty}a_n^2<\infty.
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  7. #7
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by wyhwang7 View Post
    thank you! so does that mean that my proof works?

    nope
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