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Series/Convergence help

  1. Dec 7, 2005 #1
    help determining where the series here is abs. convergent.

    its Sin(2n)/n^2, i thought about the ratio test but it gets nasty, is there a easier way?
    nm, i think it is convergence if I use the comparison test? Sound right?


    what about (-3)^n/n!, i used the ratio test, and got 0, which means it abs. convergent since 0<1, but i dont think i did it right .
    I did:
    Lim n-> infinity : (-3^(n+1)/(n+1)!)(n!/-3^n)= -3 lim n!/n+1!....
    does that look right?
     
    Last edited: Dec 7, 2005
  2. jcsd
  3. Dec 7, 2005 #2

    Pyrrhus

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    Homework Helper

    On the 1st one: what about the comparison test to its absolute value series?

    On the 2nd one: Consider the Ratio Test or d'Alambert's ratio.
     
    Last edited: Dec 7, 2005
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