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Series test for convergence/divergence
05-11-2018, 06:28 PM
Post: #9
RE: Series test for convergence/divergence
(05-11-2018 04:25 PM)Benjer Wrote:  Hi Parisse,

I see that the value of the main term is n^(-3/2) but I don't know the connection between that and whether the series converges or diverges. I guess what I'm asking is, if I apply the series command to some function, what am I looking for in the output to see if the series converges or diverges? I tried some different series that I know are convergent and divergent, but I couldn't see a pattern in the output that helped me connect the two.

Thank you for your time.
n^(-3/2)==1/n^(3/2) therefore as n gets larger
OR
limit(n^(-3/2),n,∞)
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RE: Series test for convergence/divergence - toshk - 05-11-2018 06:28 PM



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