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

  • Thread starter frasifrasi
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  • #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?
 

Answers and Replies

  • #2
dynamicsolo
Homework Helper
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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?

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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