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P-series math problem

  1. Jun 24, 2005 #1
    Just a quick question on this series.

    [tex]
    \sum\limits_{n = 1}^\infty {\frac{1}{{n^p }}}
    [/tex]

    It coverges of p > 1 and diverges for all other values of p.

    In one of the examples in my book some things are said and there is a line which says observe that the series ( [tex]\sum\limits_{n = 1}^\infty {\frac{1}{{\sqrt {n + 1} }}} [/tex]) ) is a p-series with p = (1/2) < 1 so it diverges. But there is an extra "1" inside the radical so can it still be regarded as a p-series? Also, [tex]\frac{1}{{\sqrt {n + 1} }} < \frac{1}{{\sqrt n }}[/tex] since n is a natural number. So the comparison test wouldn't really tell us anything about the 1/sqrt(n+1) series would it, because the series with 1/sqrt(n) diverges.

    Even so, can it still be concluded that 1/sqrt(n+1) is a p-series with p = (1/2) < 1 and so it diverges? The comparison test wouldn't seem to work and I can't really think of any other series to compare to 1/sqrt(n+1). Well apart from something like 1/((n)^(9/10)) but my book seems to have drawn that the conclusion that the series involving 1/sqrt(n+1) is divergent by comparison with 1/sqrt(n). Is that a valid approach?
     
  2. jcsd
  3. Jun 24, 2005 #2

    OlderDan

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    Try rewriting

    [tex]\sum\limits_{n = 1}^\infty {\frac{1}{{\sqrt {n + 1} }}} [/tex]

    in terms of m = n + 1. You may want to have a stand alone term in addition to the resulting infinite series to make a comparison.
     
  4. Jun 24, 2005 #3
    hello there

    the comparision test should tell you enough think of the rieman zeta function

    [tex]\sum\limits_{n = 1}^\infty {\frac{1}{n^s} [/tex]

    it converges for all s>1 and diverges for all s<=1, by the way try the "limit comparision test" ,you should try and compare it to something that behaves like it such as 1/sqrt(n) which is divergent.

    steven
     
  5. Jun 24, 2005 #4
    Yeah, that might have been the method that my book used. It makes sense to use the limit comparison test for this particular series. Thanks for the help.
     
  6. Jun 24, 2005 #5

    matt grime

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    why comparison? nothing to do with comparison, it's far simpler than that

    one sum is

    1 + (1/2)^{1/2} + (1/3)^{1/2}+(1/4)^{1/2}+...

    the other is

    (1/2)^{1/2} + (1/3)^{1/2}+(1/4)^{1/2}+...

    so obvlously on diverges iff the other does.
     
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