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Math Help - Convergence proof

  1. #1
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    Convergence proof

    The proposition is as follows:

    If a_n\geq0  for all n \in \mathbb N and there is a p>1 such that $\lim_{n\to\infty} n^pa_{n} exists and is finite, then  \sum_{n=1}^\infty a_{n} converges.

    I'm honestly not even sure what angle to come at this from, since I can't see what good knowing that limit exists does in trying to prove the convergence of some series. If someone could just give me a nudge in the right direction here I should be OK.
    Last edited by Dinghy; March 1st 2012 at 03:19 PM.
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  2. #2
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    Re: Convergence proof

    Quote Originally Posted by Dinghy View Post
    The proposition is as follows:
    If a_n\geq0  for all n \in \mathbb N and there is a p>1 such that $\lim_{n\to\infty} n^pa_{n} exists and is finite, then  \sum_{n=1}^\infty a_{n} converges.
    This is a straightforward application of the limit comparison test.

    Note that if p>1 then \sum\limits_n {\frac{1}{{n^p }}} converges.
    Last edited by Plato; March 1st 2012 at 04:10 PM.
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  3. #3
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    Re: Convergence proof

    Quote Originally Posted by Plato View Post
    This is a straightforward application of the limit comparison test.

    Note that if p>0 then \sum\limits_n {\frac{1}{{n^p }}} converges.
    Legitimately embarrassed about not having seen that. This problem had been talked up so much that it didn't even occur to me that it could be such a straightforward application of one of the simpler convergence tests. One thing though, when you say p>0, you mean p>1, right?

    Thanks a lot.
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  4. #4
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    Re: Convergence proof

    Quote Originally Posted by Dinghy View Post
    you mean p>1, right?
    Yes, of course. Corrected!
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  5. #5
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    Re: Convergence proof

    Actually I just noticed, I don't think you can apply the limit comparison test, as technically  \sum_{n=1}^\infty a_{n} isn't necessarily a positive series. The proposition says that a_{n}\geq 0, whereas a positive series requires that a_{n}>0, so I think I'm back at square one.

    edit: but I guess the Limit Comparison test doesn't strictly require it to be a positive series, just that a_{n}\geq 0...

    Can't shake the feeling that my math professor is messing with me with the "  \sum_{n=1}^\infty a_{n} isn't necessarily a positive series!!" warning.
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    Re: Convergence proof

    Quote Originally Posted by Dinghy View Post
    Actually I just noticed, I don't think you can apply the limit comparison test, as technically  \sum_{n=1}^\infty a_{n} isn't necessarily a positive series. The proposition says that a_{n}\geq 0, whereas a positive series requires that a_{n}>0, so I think I'm back at square one.k

    edit: but I guess the Limit Comparison test doesn't strictly require it to be a positive series, just that a_{n}\geq 0...

    Can't shake the feeling that my math professor is messing with me with the "  \sum_{n=1}^\infty a_{n} isn't necessarily a positive series!!" warning.
    I don't understand what you are concerned about.
    Look at the link I provided.
    It requires that a_n\ge 0~\&~b_n\ge 0~.
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  7. #7
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    Re: Convergence proof

    Yeah, sorry, my Calc 2 professor defined the LCT really lazily (said that it only applies to positive series), and once I get something in my head it's really hard to get it out.

    Thanks for your help.
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