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Comparison tests for series

  1. Oct 16, 2011 #1
    1. The problem statement, all variables and given/known data
    [tex]\sum_{n=2}^{\infty}\frac{1}{n\sqrt{n^2-1}}[/tex]


    2. Relevant equations
    direct comparison test
    limit comparison test

    3. The attempt at a solution
    so i kind of cheated and looked at the back of my book and it says to compare with [itex]\frac{1}{n^{3/2}}[/itex]
    so i tried using the direct comparison test and tried to show that the original series converges if [tex]\frac{1}{n\sqrt{n^2-1}}<\frac{1}{n^{3/2}}[/tex] since [tex]\sum_{n=1}^{\infty}\frac{1}{n^{3/2}}[/tex] is a convergent p-series test
    i just dont know how to actually show [tex]\frac{1}{n\sqrt{n^2-1}}<\frac{1}{n^{3/2}}[/tex]
    or am i using the wrong test? limit comparison? by the way the only tests i've covered in my class are the divergence, p-series, integral, direct comparison, limit comparison tests and geometric and telescoping series
     
    Last edited by a moderator: Oct 16, 2011
  2. jcsd
  3. Oct 17, 2011 #2
    you know that [itex]n^{3/2} =n\sqrt{n}[/itex], right?
    now
    [tex]
    n \sqrt{n} < n \sqrt{n^2-1}[/tex]
    [tex]
    \sqrt{n} < \sqrt{n^2-1}[/tex]
    [tex]
    n < n^2-1[/tex]

    which is valid for all n >= 2
     
  4. Oct 17, 2011 #3

    LCKurtz

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    Since [itex]n\sqrt{n^2-1}[/itex] is of order [itex]n^2[/itex] this suggests the very easy limit comparison test with [itex]\sum\frac 1 {n^2}[/itex].
     
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