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Thread: product of 2 series

  1. #1
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    product of 2 series

    given that we have 2 series a_n and b_n that converges, does that mean that the product of this two series a_n b_n also converges?

    what i said was that since both converge, the seq of finite partial sums for both will be bounded and hence
    using cauchy seq,

    sum from k = n to m of a_n b_n is also bounded.
    since the product is cauchy, it is convergent.

    is my reasoning right?

    thanks
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  2. #2
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    What if $\displaystyle a_n=b_n=\dfrac{(-1)^n}{\sqrt{n}}~?$
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  4. #4
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    of course if the statement adds the assumptions $\displaystyle a_n,b_n>0$ then it would be true.
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  5. #5
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    Quote Originally Posted by Krizalid View Post
    of course if the statement adds the assumptions $\displaystyle a_n,b_n>0$ then it would be true.
    If $\displaystyle a_n\ge 0$ and $\displaystyle \sum {a_n } $ converges, then can you show that $\displaystyle \sum {(a_n)^2 } $ also converges?

    Is it true that $\displaystyle 2a_nb_n\le (a_n)^2 +(b_n)^2~? $
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    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by Krizalid View Post
    of course if the statement adds the assumptions $\displaystyle a_n,b_n>0$ then it would be true.
    ... it is sufficient that only $\displaystyle a_{n}$ or $\displaystyle b_{n}$ is $\displaystyle \ge 0$ and the other simply tends to 0...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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  7. #7
    Grand Panjandrum
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    Quote Originally Posted by Plato View Post
    If $\displaystyle a_n\ge 0$ and $\displaystyle \sum {a_n } $ converges, then can you show that $\displaystyle \sum {(a_n)^2 } $ also converges?
    The conditions imply that for $\displaystyle $$n$ large enough $\displaystyle a_n<1$, so $\displaystyle a_n^2<a_n$. So the sequence of partial sums of $\displaystyle S_n=\sum_{n=0}^N {a_n^2 } $ is increasing and bounded by $\displaystyle K+\sum_{n=0}^N {a_n } $ for some positive number $\displaystyle $$K$, and so $\displaystyle S_N$ is increasing and bounded above ...

    Is it true that $\displaystyle 2a_nb_n\le (a_n)^2 +(b_n)^2~? $
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