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Thread: convergent or divergent?

  1. #1
    Apr 2009
    dallas, tx

    convergent or divergent?

    trying to find divergence or convergence of:
    the series from n=2 to infinity of (1)/((n^p)-(n^q)) (0<q<p) using limit comparison test
    Using limit comparison I have

    i used 1/n^p

    used reciprical and multiplied by original
    (1)/((n^p)-(n^q)) x (n^p/1)

    simplified and ended up with this and cannot figure out what to do or if it is even right
    (1)/(1-(1/n^p)) found the limit to be 1 then is it divergent by nth term test?
    Last edited by calculushelptx; Apr 7th 2009 at 10:24 PM. Reason: clarify
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  2. #2
    MHF Contributor chisigma's Avatar
    Mar 2009
    near Piacenza (Italy)
    I suppose that p and q are both integers. In such case the general term a_{n} of the series is...

    a_{n}= \frac {1}{n^{p}-n^{q}}

    If you apply the 'comparison test' in order to establish if the series \sum_{n=2}^{\infty} a_{n} convergs or not you obtain...

    \lim_{n \rightarrow \infty}\frac{a_{n+1}}{a_{n}}= 1

    ... so that the series can converge or not... you don't know!...

    An easy analysis however shows that for p=1, q=0 the series diverges and in all other cases converges...

    Kind regards

    \chi \sigma
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  3. #3
    MHF Contributor matheagle's Avatar
    Feb 2009
    Where are we assuming that p and q are integers?
    I would do the limit comparison to \sum {1\over n^p}.
    The limit of the ratio does go to one since 0<q<p.
    So your series converges whenever p exceeds one, integer or not.
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