Thread: power series; 2
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vela
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#47
Mar4-12, 11:21 PM
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Quote Quote by arl146 View Post
Ok well I dont know how to show that it is less than 1/n^2? Literally the book says: "5/(2n^2+4n+2) < 5/(2n^2) because the left side has a bigger denominator. We know that summation 5/(2n^2) = (5/2)* summation (1/n^2) is convergent (p-series with p=2>1). Therefore *the series mentioned for this example* is convergent by the comparison test." it really doesn't show anything else..
Do you understand the logic behind the book's argument here?

Ohhhh I gotcha I can't use te comparison test onthe x=3 one, you should have just said that, that hurts my brain a little less haha (just kidding). So I use the alternating series test right ? (this is all coming together, slowly, but getting there). soooo to be convergent according to the alternating series test, it has to satisfy two things: (i.) b(n+1) <= b(n) [which is really b sub n not b of n]. Which our series does. Because (n+1)/((n+1)^3+1) is less than n/(n^3+1). And has to satisfy (ii.) lim of b sub n must equal 0, which is does!
You have to show (n+1)/((n+1)^3+1) < n/(n^3+1) if you want to use the alternating-series test.

Also, I did look up absolute convergence in my book. Oh well it just says the series is absolutely convergent if the series of absolute values is convergent. So to me that means nothing, like I get nothing out of that? Can you explain how to apply that. When I start getting values I don't know how to tell if it's convergent.
That's the definition of absolute convergence. You need to know that so when the term comes up, you know what's being talked about.

Now look in the book for theorems that apply to absolutely convergent series to see why it might apply to this problem.