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Math Help - Convergent sum / divergent product

  1. #1
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    Oct 2010
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    Convergent sum / divergent product

    Hi guys,

    This is apparently a fairly well-known example, but I can't seem to figure it out. Anyone know this? Want

    $\sum_{n=1}^\infty a_n  < \infty$

    but

    \prod_{n=1}^\infty (1+a_n) \rightarrow \infty

    Now, we know that the sum will have to be conditionally convergent, as
    \sum_{n=1}^\infty |a_n| < \infty \Rightarrow \prod_{n=1}^\infty (1+a_n) < \infty
    and moreover, it is a theorem that if the sequence is square-summable, i.e.
    \sum_{n=1}^\infty |a_n|^2 < \infty
    then the product and the sum (without absolute values) converge and diverge together.

    The sequences
    a_n = (-3)^n
    a_n = \frac{(-1)^n}{n}
    a_n = \frac{(-1)^n}{\sqrt{n}}

    don't seem to work, although I might be wrong about this last one. what about

    a_n = \frac{i\cdot(-1)^n}{\sqrt{n}}
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  2. #2
    Newbie
    Joined
    Oct 2010
    Posts
    12

    Re: Convergent sum / divergent product

    Seems I've answered my own question as

    a_n = i^n/\sqrt{n} works!
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