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Math Help - Finding Value of P in Infinite Series

  1. #1
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    Finding Value of P in Infinite Series

    \sum_{n=1}^{\infty} \frac{4n^6}{3n^8+8}^{p}
    Find all values of p for which the infinite series converges.

    Can any give me any advice or help on how to solve this problem? I have no idea where to begin.
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  2. #2
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    Re: Finding Value of P in Infinite Series

    Quote Originally Posted by Beevo View Post
    \sum_{n=1}^{\infty} \frac{4n^6}{3n^8+8}^{p}
    Find all values of p for which the infinite series converges.
    Can any give me any advice or help on how to solve this problem? I have no idea where to begin.
    I think that you mean \sum_{n=1}^{\infty}\left( \frac{4n^6}{3n^8+8}\right)^{p}

    If so, then note that \left( \frac{4n^6}{3n^8+8}\right)^{p}\le \left( \frac{4}{3n^2}\right)^{p}

    For that p-series what is the answer?
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  3. #3
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    Re: Finding Value of P in Infinite Series

    Let \sum_{n=1}^{\infty} (\frac{4n^6}{3n^8+8})^{p} = \sum_{n=1}^{\infty} s_n where  s_n = (\frac{1}{n^2} \frac{4}{3+\frac{8}{n^8}})^{p} by Pringsheim Theorem we have the necessary condition that  \lim_{n \rightarrow \infty} n.s_n \rightarrow 0 \implies  \lim_{n \rightarrow \infty} \frac{1}{n^{2p-1}} (\frac{4}{3+\frac{8}{n^8}})^{p} = \lim_{n \rightarrow \infty} \frac{1}{n^{2p-1}} \lim_{n \rightarrow \infty} (\frac{4}{3+\frac{8}{n^8}})^{p} = (\frac{4}{3})^p \lim_{n \rightarrow \infty} \frac{1}{n^{2p-1}} this limit \rightarrow 0 iff 2p-1 > 0 \implies p > \frac{1}{2} .
    Last edited by kalyanram; September 9th 2012 at 10:48 AM.
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    Re: Finding Value of P in Infinite Series

    Quote Originally Posted by kalyanram View Post
    Let \sum_{n=1}^{\infty} (\frac{4n^6}{3n^8+8})^{p} = \sum_{n=1}^{\infty} s_n where  s_n = (\frac{1}{n^2} \frac{4}{3+\frac{8}{n^8}})^{p} by Pringsheim Theorem we have the necessary condition that  \lim_{n \rightarrow \infty} n.s_n \rightarrow 0 \implies  \lim_{n \rightarrow \infty} \frac{1}{n^{2p-1}} (\frac{4}{3+\frac{8}{n^8}})^{p} = \lim_{n \rightarrow \infty} \frac{1}{n^{2p-1}} \lim_{n \rightarrow \infty} (\frac{4}{3+\frac{8}{n^8}})^{p} = (\frac{4}{3})^p \lim_{n \rightarrow \infty} \frac{1}{n^{2p-1}} this limit \rightarrow 0 iff 2p-1 > 1 \implies p>1 \. We need p>1 .
    @kalyanram.
    Why did you spoil the question for the student?
    Do you want the student to learn? The student will never learn if you simply hand the complete solution.
    Last edited by Plato; September 9th 2012 at 10:51 AM.
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  5. #5
    Member kalyanram's Avatar
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    Re: Finding Value of P in Infinite Series

    Quote Originally Posted by Plato View Post
    @kalyanram.
    Why did you spoil the question for the student?
    Do you want the student to learn? The student will never learn if you simply hand the complete solution.
    I was typesetting in latex when you replied to the post. I was not aware of your post; wouldn't have simplified it if in that case.
    ~Kalyan.
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  6. #6
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    Re: Finding Value of P in Infinite Series

    Thank you for both the responses, but yeah, I was trying to figure out the problem on my own and the solution kind of spoiled it.

    I don't why, but I am struggling to grasp the concept of series (especially the comparison tests). I felt Cal I was a breeze, but but Cal II is a different story so far. I am putting in at least 2-3 hours every other day of studying, but still having trouble keeping up. This is frustrating.
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