
#19
Feb2312, 01:58 PM

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#20
Feb2312, 03:46 PM

P: 343

x4<1 converge
x4>1 diverge 



#21
Feb2312, 03:51 PM

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What about if x  4 = 1? The Ratio test doesn't cover this situation, so you have to investigate it as a special case. 



#22
Feb2312, 08:00 PM

P: 343

interval? .... i dont know O_O




#23
Feb2312, 08:06 PM

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#24
Feb2312, 08:07 PM

P: 343

oh, x < 5 ... so just (infinity, 5) ?




#25
Feb2312, 08:48 PM

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You should review working with absolute values in equations and inequalities. 



#26
Feb2312, 08:51 PM

P: 343

ohhhhhhhh, i dont know why I KEEP just looking over the absolute values. sooo .... (4, 5) ?




#27
Feb2312, 08:54 PM

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What about 3.5? Doesn't it satisfy x  4 < 1?




#28
Feb2312, 09:08 PM

P: 343

ohh right .... so (3, 5) because since it's not a square bracket it doesn't include the 3 right
but anything after 3 and before 4 would really never actually equal 1 which satisfies the less than sign 



#29
Feb2412, 12:13 AM

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OK, this establishes that the radius of convergence is 1. In the interval (3, 5), the series converges absolutely. If you aren't clear on what this means, look up the definition of absolute convergence in your book. To finish the problem you need to check the two endpoints: x = 3 and x = 5. Substitute these numbers in your formula for the series and see what you get. Obviously (??) you won't be able to use the Ratio Test, but you should be able to use what other facts you know about series to say whether the series converges (either absolutely or conditionally) or diverges, at each of these values. If you aren't clear on conditional convergence, look up that term as well. 



#30
Feb2612, 01:17 AM

P: 343

I know that, I was explaining how I understood why it is (3,5) instead of (4,5).
I think at x=5 it absolutely converges. It seems to get closer and closer to one and theyre all positive values so Looking for absolute convergence with the absolute values doesn't change that. For x=3 I think it coverages just conditionally. Not absolute because the values don't seem to get closer to a Specific value when you add the absolute value signs... And I don't think it diverges because The values are ,+,,+ and so on an they seem to end up going towards a specific number Although, I've never really understood when something converges or diverges 



#31
Feb2612, 01:52 AM

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You need to be able to prove your conclusions. You can't just look at a few of the partial sums and guess that it's going to converge or not.
What series do you get when you set x=5? 



#32
Feb2612, 07:04 PM

P: 343

i got summation [n(1)^n] / [n^3 + 1] for x=5. and the same for x=3 just with the negative, [n(1)^n] / [n^3 + 1]
and thats when n=1 to infinity 



#33
Feb2612, 07:12 PM

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[tex]\sum_{n = 1}^{\infty}\frac{n\cdot 1^n}{n^3 + 1}[/tex] Does that series converge or diverge? (You should simplify it first.) Why? What about when x = 3? Same questions. 



#34
Feb2612, 07:20 PM

P: 343

no, not n^3 + 3, its n^3 + 1 ... right
i said that it absolutely converges. but the other person said i need to prove it. i dont know how i do that 



#35
Feb2612, 08:53 PM

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#36
Feb2812, 05:31 PM

P: 343

you mean like, root test, alternating series test, integral test .. those kinds? wont the root test work for this?



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