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Homework Help: What convergence/divergence test can I use on this series? - Calculus II

  1. Mar 2, 2006 #1
    What test can I use on the following series in order to determine
    if it converges or diverges? Looking at it graphically it appears to
    diverge but I cannot show it analytically.

    [tex]
    \sum\limits_{n = 1}^\infty {\frac{{n!}}{{3n! - 1}}}
    [/tex]

    Using the Ratio Test, here is I got thus far

    [tex]
    \left| {\frac{{a_{n + 1} }}{{a_n }}} \right| = \left| {\frac{{\left( {\frac{{(n + 1)!}}{{3(n + 1)! - 1}}} \right)}}{{\left( {\frac{{n!}}{{3n! - 1}}} \right)}}} \right|
    [/tex]

    [tex]
    = \frac{{(n + 1)!}}{{3(n + 1)! - 1}} \cdot \frac{{3n! - 1}}{{n!}}
    [/tex]

    [tex]
    = \frac{{(n + 1)n!}}{{3(n + 1)n! - 1}} \cdot \frac{{3n! - 1}}{{n!}}
    [/tex]

    [tex]
    = \frac{{(n + 1)(3n! - 1)}}{{3(n + 1)n! - 1}}
    [/tex]

    At this point, I don't see how the Ratio test will work. What test can I use on this series?

    Should I start off with long division to reduce the improper rational
    expression?

    http://img164.imageshack.us/img164/4656/longdiv7nr.jpg [Broken]

    So the equivalent sum would be

    [tex]
    \sum\limits_{n = 1}^\infty {\left( {\frac{1}{3} + \frac{1}{{3(3n! - 1)}}} \right)}
    [/tex]

    Which must diverge because of the 1/3 always being summed from
    1 to infinity.

    Is that the correct way to do it? If so, is it the only way? Am I
    missing a simple test that could be used?
     
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Mar 3, 2006 #2
    oops nevermind I looked again and it seems u already did that.
     
  4. Mar 3, 2006 #3

    benorin

    User Avatar
    Homework Helper

    Maybe comparison test. But the work have looks good to me.
     
  5. Mar 3, 2006 #4

    HallsofIvy

    User Avatar
    Science Advisor

    I would be inclined to use the fact that is an does not go to 0 then [itex]\Sum a_n[/itex] does not converge!
     
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