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Math Help - So, how exactly does a p-series converge

  1. #1
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    So, how exactly does a p-series converge

    I'm studying the p-series convergence test right now , I understand what the theorem says but I don't understand why it works. I understand that a harmonic series [sum(1/k)] diverges and I understand why it diverges. But to me the p-series [sum(a/k^p)] is basically the same kind of series but the terms will just approach zero a bit faster.
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  2. #2
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    Say the p-series is

    \sum_{k=1}^{\infty}\frac{a}{k^p}.

    Imagine the terms of the p-series drawn as rectangles on the positive x-axis. These rectangles can be shifted to fall under the curve defined by

    y=\frac{a}{x^p}.

    Therefore, the total sum of the area of these rectangles -- that is, the value of the sum \sum_{k=2}^{\infty}\frac{a}{k^p} -- is less than the value of the integral

    \int_1^{\infty}\frac{a}{x^p}\,dx.

    Because this integral is finite (for p>1), the series converges.

    Edit: Replaced k=1 with k=2 in the above expression.
    Last edited by Scott H; November 7th 2009 at 06:21 PM.
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  3. #3
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    Quote Originally Posted by dmoreau View Post
    I'm studying the p-series convergence test right now , I understand what the theorem says but I don't understand why it works. I understand that a harmonic series [sum(1/k)] diverges and I understand why it diverges. But to me the p-series [sum(a/k^p)] is basically the same kind of series but the terms will just approach zero a bit faster.

    Besides the integral test you can use Cauchy's Condensation Test: if a sequence \{a_n\}_{n=1}^\infty is positive and monotone converging to zero, then the series \sum\limits_{n=1}^\infty a_n\,\,and\,\,\sum\limits_{n=1}^\infty 2^na_{2^n} converge and diverge together.

    So take the p-series \sum\limits_{n=1}^\infty\frac{a}{n^p} . By CCT, this series converges iff converges the series

    \sum\limits_{n=1}^\infty\frac{2^na}{2^{np}}=a\sum\  limits_{n=1}^\infty\left(\frac{1}{2^{p-1}}\right)^n and this last is a geometric series which we know converges iff its quotient is between -1 and 1, i.e. iff

    \frac{1}{2^{p-1}}<1\Longleftrightarrow2^{p-1}>1\Longleftrightarrow p-1>0 and we're done

    Tonio
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