Determine whether the sequence converges or diverges. an =(-1)^n [(n+2)/(3n-1)]

- anonymous

Determine whether the sequence converges or diverges. an =(-1)^n [(n+2)/(3n-1)]

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- anonymous

Have you learned alternating series test?

- anonymous

not really.

- anonymous

but yeah

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- anonymous

o.k. Its been a while, let me dust off the old brain....

- anonymous

lol, it's alright, take all the time you need ^_^

- anonymous

It 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

I'm not sure if I did, does it have to do with finding the absolute value of an? as an approaches infinity?

- anonymous

not the absolute value. The (-1)^n gives a hint that its alternating series.

- anonymous

yes I know it is and it alternates between -1 and 1

- anonymous

Usually when you see that then you know that its alternating series.

- anonymous

o.k

- anonymous

yep :)

- anonymous

well 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

I've done that before, but thanks anyways :)

- anonymous

I'll figure something out ^_^

- anonymous

oh, o.k. Sorry for not being able to help you.

- anonymous

no, thank you for your time dear, atleast you've tried :)

- anonymous

I very much appreciate it ^_^

- anonymous

yeah. Ill keep looking at it. I hate it when I use to know something and cannot figure it out.

- anonymous

lol, 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

Have you done the test for divergence?

- anonymous

you mean discussing the convergence/divergence of a sequence ,yes

- anonymous

Well 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

I did find the limit

- anonymous

I first multiplied (-1)^n to the fraction, then simplified + found the limit.

- anonymous

The way I remember doing it is that you do not look at the (-1)^n . Just (n+2)/(3n-1). 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

If 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

you 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

No, just the second limit.

- anonymous

find that limt.

- anonymous

why? what about the first limit?

- anonymous

The (-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

That 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

Well good luck on your midterm. Sorry if I caused any confusion.

- anonymous

thank 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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