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Math Help - Series

  1. #1
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    Series

    Is it true that  \sum_{n=1}^{\infty} {a_{n}}^2 is convergent if  \sum_{n=1}^{\infty} {a_{n}} is convergent? Justify your answer.

    Any hints?
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  2. #2
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    Remember that if \vert a \vert \leq 1 then a^2 \leq \vert a \vert. Now also remember if the series is convergent what is \lim_{n \to \infty} \ a_n ?

    Although now that I think of it, I'm not sure it works in general if the series does not converge absolutely...if it does, following the above idea suffices, but if the series converges only conditionally I'm not so sure.
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  3. #3
    Senior Member Sampras's Avatar
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    Quote Originally Posted by Jose27 View Post
    Remember that if a \leq 1 then a^2 \leq a. Now also remember if the series is convergent what is \lim_{n \to \infty} \ a_n ?

    Although now that I think of it, I'm not sure it works in general if the series does not converge absolutely...if it does, following the above idea suffices, but if the series converges only conditionally I'm not so sure.

    no if  a = -8 , then  a^2 = 64 \not \leq -8 ....
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  4. #4
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    Yes, by the comparison test.
    Since \lim_{n\to\infty}a_n=0 then there exists N such that |a_n|<1\Longrightarrow a_n^2<|a_n| for all n>N. Then apply the comparison test, but you can assume that a_n\ge 0.


    EDIT:** Sorry but the answer is NO. Take a_n=\frac{(-1)^n}{\sqrt{n}} **
    Last edited by putnam120; June 1st 2009 at 08:27 AM.
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  5. #5
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    Well, if the series is convergent then  \lim_{n \to \infty} \ a_n = 0 . But how do I use this to answer the question?
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  6. #6
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    oh ok comparison test. thanks!
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  7. #7
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    Quote Originally Posted by putnam120 View Post
    Yes, by the comparison test.
    Since \lim_{n\to\infty}a_n=0 then there exists N such that |a_n|<1\Longrightarrow a_n^2<|a_n| for all n>N.

    So \sum_{n=N}^\infty a_n^2<\sum_{n=N}^\infty a_n<\infty.
    But there's a problem, if for instance \{a_n \} alternates sign, comparison test no longer applies since \{a_n^2 \} is always positive, but \{a_n \} isn't. Here's an example if \{a_n \}= \{\frac{(-1)^n}{n} \} then 1-\frac{1}{2} \ngeq 1+\frac{1}{4}
    Last edited by Jose27; May 31st 2009 at 08:35 PM.
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  8. #8
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    Quote Originally Posted by nmatthies1 View Post
    Is it true that  \sum_{n=1}^{\infty} {a_{n}}^2 is convergent if  \sum_{n=1}^{\infty} {a_{n}} is convergent? Justify your answer.

    Any hints?
    it's false. counter-example: a_n=\frac{(-1)^n}{n^c}, where c is any real number in the interval (0, 1/2].
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  9. #9
    Super Member Deadstar's Avatar
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    Just thought i'd add in this one point...

    If \sum_n a_n is a convergent series of POSITIVE TERMS then \sum_n a_n^2 converges...

    Proof,

    a_n must be bounded, i.e. for some M > 0 we have an a_n \leq M for all n. Then a_n^2 \leq Ma_n so by comparison test a_n^2 converges.
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