Thread: Power series; 2
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vela
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Mar4-12, 10:27 PM
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Quote Quote by arl146 View Post
Wait wait, why do I have to show that my series is less than the one we are comparing against?
You have to show it because that's one of the conditions required for the comparison test to apply.

Can't you just say that since the degree of the n on the bottom is bigger that the whole fraction is smaller?? I don't get how I would show that .. Do I just plug in different values of n for that?
No, you can't just compare the degree of the denominators. Take the two series ##\sum \frac{1}{n^2}## and ##\sum \frac{n+1}{n^2}##. They both n2 in the denominator, but the first one converges while the second doesn't.

You can't just plug in a few values for n. You have to show that the series you're working with is less than 1/n2 after some point, that is when n>N for some N. I'm sure your book has examples showing how to apply the comparison test.

Ok um I don't see anything in the book that is similar to the x=3 one I don't where else in the book I'd find those conditions you talk about. I don't get it. I mean I get that it won't work since its +,-,+- but how do you show for this one by comparing? And do you still compare with the 1/n^2 ?
The conditions I'm talking about have to do with the test itself, and it's the one you mentioned. The comparison test only works for a non-negative series, and the x=3 series doesn't satisfy that requirement. That means, you can't use the comparison test on that series.

Absolute convergence when the value of the limit of the series with absolute value signs is < 1
No, this is wrong. Look up what it means and what absolute convergence implies. This is the key to figuring out if the x=3 series converges.