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anonymous
 5 years ago
Determine whether the sequence converges or diverges. an =(1)^n [(n+2)/(3n1)]
anonymous
 5 years ago
Determine whether the sequence converges or diverges. an =(1)^n [(n+2)/(3n1)]

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Have you learned alternating series test?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0o.k. Its been a while, let me dust off the old brain....

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol, it's alright, take all the time you need ^_^

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It looks like you have to do the alternating series test. I'm looking over my old notes....but you said you have not lean this yet?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm not sure if I did, does it have to do with finding the absolute value of an? as an approaches infinity?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0not the absolute value. The (1)^n gives a hint that its alternating series.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes I know it is and it alternates between 1 and 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Usually when you see that then you know that its alternating series.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well I am really rusty on my cal 2. You can google paul's online math notes to get better clarification. I used his notes to get through cal 1 cal 3.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I've done that before, but thanks anyways :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'll figure something out ^_^

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh, o.k. Sorry for not being able to help you.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no, thank you for your time dear, atleast you've tried :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I very much appreciate it ^_^

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah. Ill keep looking at it. I hate it when I use to know something and cannot figure it out.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol, it's alright, I'll visit my prof tomorrow before the midterm, he has submitted a special office our just in case :) it's alright.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Have you done the test for divergence?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you mean discussing the convergence/divergence of a sequence ,yes

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well i was looking up videos from patrickjmt.com and he gives examples about alternating series and he points out how its a two step process in figuring out whether it diverges or converges. He recommends first taking the limit of an and see whether it does not exists or it does not equal zero, if you find that its either of these then in is divergent.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I first multiplied (1)^n to the fraction, then simplified + found the limit.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The way I remember doing it is that you do not look at the (1)^n . Just (n+2)/(3n1). This will now be your bn. Take this limit. It it does not exist or does not equal to zero then it is considered divergent.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If it does equal to zero, then you have to check whether bn+1 is less than bn. If you pove this also, then you know that it is convergent.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you mean I have to divide the question into 2 parts? like this:\[ \lim_{n \rightarrow \infty}(1)^n . \lim_{n \rightarrow \infty}(n+2)/(n+3)\] ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No, just the second limit.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0why? what about the first limit?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The (1)^n tells us that its alternating series, so we have to show whether our bn which is what we define as (n+2)/(n+3) has a limit which equals to 0. It it does that we also have to prove that bn+1 is less than bn. By satisfying these two conditions than we can say that its convergent. If while we take the limit as n goes to infinity of bn, we see that the limit does not exist or that it does not equal to zero than we can say that it is divergent.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0That is how I would approach this problem. Just check w/ your professor like you said you were going to do. I am not an expert at this, just recovering info from what i learned.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well good luck on your midterm. Sorry if I caused any confusion.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thank you, I think I got what you've meant, all I have to do is review what you've said again, thanks dear ^_^
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