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Thread: 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; Apr 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 $\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^{2}}$ converges, the sum $\displaystyle \sum_{n=1}^{\infty}\frac{1}{n}$ diverges...

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    $\displaystyle \chi$ $\displaystyle \sigma$
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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 $\displaystyle \sum a_n < \infty$, then $\displaystyle \left (\sum a_n \right)^2 < \infty$.

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

    Hint: $\displaystyle x^2+y^2 \leq x^2+y^2+2xy=(x+y)^2$ 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 $\displaystyle \forall n$ $\displaystyle a_{n}>0$ or $\displaystyle \forall n$ $\displaystyle a_{n}<0$. An example: $\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n}}{\sqrt{n}}$ converges, $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n}$ diverges...

    Kind regards

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