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Is it ok to do this with the ratio test for series?/

  1. Nov 5, 2007 #1
    Is it ok to do this with the ratio test for series??/

    I had the series from 1 to infinity of:

    n(-3)^(n)/(2^(n-1))


    by applying the root test, i got:

    lim as n-->infinity [ 3(n+1)/2n] , so put the 3/2 outside and let the (n+1)/n be n/n --> which means the limit would yield 3/2...

    does anyone know if this is how I am supposed to do this?
     
  2. jcsd
  3. Nov 5, 2007 #2

    dynamicsolo

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    I'm reading the general term to be

    [tex] \frac{n (-3)^{n}}{2^{n-1}}[/tex].

    This is not suitable for the Root Test because of that dang factor of n in the numerator. So I'm presuming you meant the Ratio Test. And yes, I get the same ratio you did. The limit is 3/2, so this series doesn't converge.
     
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