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Math Help - Convergence problem

  1. #1
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    Convergence problem

    Show that if  \sum_{n=0}^\infty a_n converges conditionally, then  \sum_{n=0}^\infty n^2a_n diverges.

    Just by looking, I think  n^2a_n is unbounded so the second summation diverges, but I don't know how to show this. I know  a_n \leq c as a result of convergence so my best attempt so far is to say  a_n/n \leq c/n \leq cn^2 but I don't know if I'm going in the right direction.
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  2. #2
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    Suppose that \sum n^2a_n converges. Then you have that \lim_{n\to\infty}|n^2a_n|\to{0}. Now choose \epsilon=1, then there exists N such that n>N\Longrightarrow |n^2a_n|<1\Leftrightarrow |a_n|<\frac{1}{n^2}. But this contradicts the fact that \sum a_n converges conditionally. Just in case you don't see it, use the comparison test.

    Using this approach you can conclude that \sum_{n=0}^{\infty}n^{1+\alpha}a_n diverges for \alpha>0.
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  3. #3
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    Equivalently, if \sum a_n converges, then \sum \frac {|a_n|}{n^2} converges, which is clear by comparison. All I have done is taken contrapositives and renamed the terms.
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  4. #4
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    Quote Originally Posted by cubrikal View Post

    Show that if  \sum_{n=0}^\infty a_n converges conditionally, then  \sum_{n=0}^\infty n^2a_n diverges.

    Just by looking, I think  n^2a_n is unbounded so the second summation diverges, but I don't know how to show this. I know  a_n \leq c as a result of convergence so my best attempt so far is to say  a_n/n \leq c/n \leq cn^2 but I don't know if I'm going in the right direction.
    suppose \sum n^2a_n is convergent. then \lim_{n\to\infty} n^2a_n=0 and hence \lim_{n\to\infty} \frac{|a_n|}{\frac{1}{n^2}}=0. but then, by the limit comparison test, \sum |a_n| will be convergent. contradiction!
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