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Math Help - Does the converse hold?

  1. #1
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    Does the converse hold?

    Suppose that both series \Sigma a_k and \Sigma b_k are absolutely convergent. Show that then so too is the series \Sigma a_k b_k . Does the converse hold?
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  2. #2
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    Consider a_k= b_k= \frac{(-1)^k}{k}.
    Last edited by Jhevon; September 25th 2010 at 11:45 PM. Reason: fixed LaTeX
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  3. #3
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by rondo09 View Post
    Suppose that both series \Sigma a_k and \Sigma b_k are absolutely convergent. Show that then so too is the series \Sigma a_k b_k . Does the converse hold?
    Yust for clarity's sake: are You asking if, given a series \Sigma a_{k} b_{k} that absolutely converges, the same holds for \Sigma a_{k} and \Sigma b_{k} ?...

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    \chi \sigma
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  4. #4
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    Yes, the "converse" of "if A then B" is "if B then A".
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  5. #5
    MHF Contributor chisigma's Avatar
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    ... very well!... what about \displaystyle a_{k} = \frac{1}{k^{2}} and \displaystyle b_{k} = \sin k?... the series \displaystyle \sum_{k=1}^{\infty} \frac{\sin k}{k^{2}} converges absolutely... is the same also for \displaystyle \sum_{k=1}^{\infty} \sin k ?...

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  6. #6
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    Quote Originally Posted by chisigma View Post
    Yust for clarity's sake: are You asking if, given a series \Sigma a_{k} b_{k} that absolutely converges, the same holds for \Sigma a_{k} and \Sigma b_{k} ?...

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    \chi \sigma
    What am I asking if there are series \Sigma a_{k} and \Sigma b_{k} which are absolutely convergent, then show that the series \Sigma a_{k}b_{k} is also absolutely convergent. Then, P.S, does the converse hold?
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  7. #7
    MHF Contributor chisigma's Avatar
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    If \Sigma a_{k} and \Sigma b_{k} absolutely converge then is \displaystyle \lim_{k \rightarrow \infty} |a_{k}| = \lim_{k \rightarrow \infty} |b_{k}| =  0 , so that \forall k>K is \displaystyle |a_{k}\ b_{k}| < |a_{k}| or also \displaystyle |a_{k}\ b_{k}| < |b_{k}| and the series \Sigma a_{k}\ b_{k} converges...

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    \chi \sigma
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