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Testing for series convergence.

  1. Sep 30, 2012 #1
    1. The problem statement, all variables and given/known data

    [itex]\sum[/itex]([itex]\frac{2n}{2n+1}[/itex])n2

    (The sum being from n=1 to ∞).


    2. Relevant equations



    3. The attempt at a solution

    Used exponent properties to get ([itex]\frac{2n}{2n+1}[/itex])2n. Using the root test, the nth root of an = lim n->∞([itex]\frac{2n}{2n+1}[/itex])2 = 1. However, the root test is indeterminate if the limit = 1.

    The ratio test didn't work out, and it doesn't seem to be similar to any series, such as geometric, tunnel, etc.

    The integral test seems like it would be very complicated, but if someone thinks it's the right way to go, I'll give it a shot.

    Would it work to say that ([itex]\frac{2n}{2n+1}[/itex])n2 < ([itex]\frac{2n}{2n+1}[/itex]), so if ([itex]\frac{2n}{2n+1}[/itex]) converges, then ([itex]\frac{2n}{2n+1}[/itex])n2 also does by squeeze convergence? Would that also work if it diverges?

    Any help would be great, thanks!!
     
  2. jcsd
  3. Sep 30, 2012 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Try a really simple test, the nth term test. If the limit of the nth term of your series as n->infinity is not zero, then the sum of the series can't exist.
     
    Last edited: Sep 30, 2012
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