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Math Help - sum of the product of convergent series

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    sum of the product of convergent series



    I'm not really sure what further to do with arbitrary series.

    Theorem one states that if { a_n } and { b_n} have limits A and B respectively, then { a_nb_n } has the limit AB.
    Last edited by davismj; January 26th 2010 at 05:05 PM. Reason: Clarification
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by davismj View Post


    I'm not really sure what further to do with arbitrary series.

    Theorem one states that if { a_n } and { b_n} have limits A and B respectively, then { a_nb_n } has the limit AB.
    Consider that since \sum_{n=1}^{\infty}a_n converges we must have that \lim\text{ }\sqrt[n]{|a_n|}\leqslant 1. And so \lim\text{ }\sqrt[n]{|a_n\cdot z^n|}=|z|\lim\text{ }\sqrt[n]{a_n}\leqslant |z|<1.
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    Quote Originally Posted by Drexel28 View Post
    Consider that since \sum_{n=1}^{\infty}a_n converges we must have that \lim\text{ }\sqrt[n]{|a_n|}\leqslant 1. And so \lim\text{ }\sqrt[n]{|a_n\cdot z^n|}=|z|\lim\text{ }\sqrt[n]{a_n}\leqslant |z|<1.
    Wouldn't it be easier to say that since | a_n| and | z^n| become arbitrarily close ( \varepsilon) to 0, their product also becomes arbitrarily close to zero, and thus the series is Cauchy?
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by davismj View Post
    Wouldn't it be easier to say that since | a_n| and | z^n| become arbitrarily close ( \varepsilon) to 0, their product also becomes arbitrarily close to zero, and thus the series is Cauchy?
    I am not sure what you're saying, but I feel as though you are not considering that just because the summand converges to zero does not mean the series converges.
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    Quote Originally Posted by Drexel28 View Post
    I am not sure what you're saying, but I feel as though you are not considering that just because the summand converges to zero does not mean the series converges.


    Is this better. the basic assumption is that since \sum a_n converges, for some N the sum from n to infinty is less than epsilon. Since \sum|z^n| is a geometric sum, its sum also converges (although less than 1 is sufficient), and so the product of these two will be less than epsilon as well. Is that kinda sorta right for an undergraduate introduction to complex analysis course?
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by davismj View Post


    Is this better. the basic assumption is that since \sum a_n converges, for some N the sum from n to infinty is less than epsilon. Since \sum|z^n| is a geometric sum, its sum also converges (although less than 1 is sufficient), and so the product of these two will be less than epsilon as well. Is that kinda sorta right for an undergraduate introduction to complex analysis course?
    The product of two series isn't the series of the products . Look at polynomials for example
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    Quote Originally Posted by Drexel28 View Post
    The product of two series isn't the series of the products . Look at polynomials for example
    Ugh. So in other words, it would get me dirty looks from a college algebra teacher, let alone complex analysis. I fail.

    How about this: Since |z| < 1, |z^n|a_n < a_n for all n.

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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by davismj View Post
    Ugh. So in other words, it would get me dirty looks from a college algebra teacher, let alone complex analysis. I fail.

    How about this: Since |z| < 1, |z^n|a_n < a_n for all n.

    You're getting warmer, but you need absolute convergence.
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