Limit Comparison Test for series

AI Thread Summary
The discussion centers on applying the limit comparison test to the series sum 1/sqrt(3n-2) from n=1 to infinity. The suggested comparison series is sum 1/sqrt n, known to be divergent. Initially, the user calculated the limit and found it to be 0, indicating a misunderstanding. Upon further reflection, they realized that the correct limit comparison yields a result of 1/sqrt3, confirming that the original series diverges. The limit comparison test effectively demonstrates the divergence of the series.
bcjochim07
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I have been given the series
sum 1/sqrt(3n-2) from n=1 to infinity

I am supposed to use the limit comparison test, and the comparison series my book suggests is sum 1/sqrt n from n=1 to infinity, which I know is a divergent p series

However, when I take the limit of one divided by the other I come up with 0.

lim n -> infinity for (sqrt n)/(sqrt 3n-2) = 0

What am I doing wrong?
 
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Nevermind, I figured it out.

This Limit Comparison test shows that the series diverges because the limit is 1/sqrt3.
 
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