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Thread: Series

  1. #1
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    Series

    "Suppose $\displaystyle \sum a_n, \sum b_n$ are convergent. Let $\displaystyle c_n=\sum_{k=0}^n a_kb_{n-k}$ and let $\displaystyle \sum c_n$ converge. Prove that$\displaystyle (\sum_0^\infty a_n) (\sum_0^\infty b_n) = (\sum_0^\infty c_n).$"

    I can't make any headways. Any hints would be appreciated.
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  2. #2
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    Quote Originally Posted by Treadstone 71
    "Suppose $\displaystyle \sum a_n, \sum b_n$ are convergent. Let $\displaystyle c_n=\sum_{k=0}^n a_kb_{n-k}$ and let $\displaystyle \sum c_n$ converge. Prove that$\displaystyle (\sum_0^\infty a_n) (\sum_0^\infty b_n) = (\sum_0^\infty c_n).$"

    I can't make any headways. Any hints would be appreciated.
    Perhaps you should look at its partial sums.
    Let $\displaystyle _aS_n$ represent the nth partial sum of $\displaystyle a_k$ let $\displaystyle _bS_n$ represent the nth partial sum of $\displaystyle b_k$ and $\displaystyle _cS_n$ represent the nth partial sum of $\displaystyle c_k$.
    Thus, if we can show that the partial sums of these two series' are equal then they converge to the same number.
    Notice that:
    $\displaystyle _aS_n=a_0+a_1+...+a_n$
    $\displaystyle _bS_b=b_0+b_1+...+b_n$
    Then,
    $\displaystyle \left(_aS_n\right)\left(_bS_n\right)=(a_0+...+a_n) (b_0+...+b_n)=c_0+c_1+...+c_n$=$\displaystyle _cS_n$
    (I might have made a mistake on the last step where I claim it is $\displaystyle _cS_n$ but I do not see anything wrong)
    Q.E.D.
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  3. #3
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    The last step is definately not equal. On one side you have n^2 terms, on the other, you have 1+2+3+...+n terms.
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