 Quote by arl146
Wait wait, why do I have to show that my series is less than the one we are comparing against?
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You have to show it because that's one of the conditions required for the comparison test to apply.
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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?
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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 n
2 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/n
2 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.
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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 ?
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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.
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Absolute convergence when the value of the limit of the series with absolute value signs is < 1
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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.