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Is it ok to do this with the ratio test for series??/ |
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| Nov5-07, 02:18 PM | #1 |
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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? |
| Nov5-07, 02:29 PM | #2 |
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Recognitions:
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[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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