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Math Help - convergence

  1. #1
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    convergence

    so confused!
    if sum (a_n) with a_n > 0 is convergent then is sum (a_n)^2 always convergent? how do we prove it or what is a counter example for it
    Last edited by sandy; April 25th 2010 at 05:21 AM.
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  2. #2
    MHF Contributor chisigma's Avatar
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    Simple counterexample: the sum \sum_{n=1}^{\infty}\frac{1}{n^{2}} converges, the sum \sum_{n=1}^{\infty}\frac{1}{n} diverges...

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    \chi \sigma
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  3. #3
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    thankyou

    hi thanks for your help but then what is the answer to
    if sum (a_n) is convergent then sum (a_n)^2 that is sum of a_n squared is always convergent? counter example or proving it
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  4. #4
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    Quote Originally Posted by sandy View Post
    hi thanks for your help but then what is the answer to
    if sum (a_n) is convergent then sum (a_n)^2 that is sum of a_n squared is always convergent? counter example or proving it
    I think chisigma got the question wrong way around.

    Think about it this way, if \sum a_n < \infty, then \left (\sum a_n \right)^2 < \infty.

    What does that tell you about  \sum a_n^2?

    Hint: x^2+y^2 \leq x^2+y^2+2xy=(x+y)^2 for x,y positive.
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  5. #5
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by sandy View Post
    hi thanks for your help but then what is the answer to
    if sum (a_n) is convergent then sum (a_n)^2 that is sum of a_n squared is always convergent? counter example or proving it
    That is true only if \forall n  a_{n}>0 or \forall n  a_{n}<0. An example: \sum_{n=1}^{\infty} \frac{(-1)^{n}}{\sqrt{n}} converges, \sum_{n=1}^{\infty} \frac{1}{n} diverges...

    Kind regards

    \chi \sigma
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  6. #6
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    \sum\limits_n {\frac{{( - 1)^n }}{{\sqrt n }}} converges.
    Does \sum\limits_n {\frac{{1 }}{{ n }}} ?
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