Limit as n->infinity of: [(-1)^n]*[(3n-1)/(2n+1)]

[MarkSA]

Hello,

I'm doing alternating series problems. I have a particular problem that the alternating series test cannot be used for since the limit of b(sub n), in this case (3n-1)/(2n+1), does not equal 0.
To 'solve' this, I need to do the divergence test. (if limit is nonexistant or not = 0, it's divergent)

The problem is i'm not sure how to take the limit of [(-1)^n]*[(3n-1)/(2n+1)], and I think I may need to show the work for that. How would I take the limit of such a beast? The product is what is throwing me off.

 

[sgtpepper]

I think there are two tests for alternating series. The first is, as I think you're aware, the limit of the function (we'll call it a(n) ) has to be zero. So for this test, we would always evaluate the absolute value of the function, getting this:

lim as n approaches infinity of (3n - 1)/(2n + 1)

(see, the absolute value renders the (-1)^n an unneccessary part of a(n) )

okay, so since the leading terms are to the same power (n^1), the limit of the function is the ratio of their coefficients:

3/2

Since the limit of a(n) is not 0, we can already skip the next test and determine this function to be divergent. However, if the limit were zero, we would move to the next test, which is that if the absolute value of a(n + 1) is less than or equal to the absolute value of a(n), then the function is convergent, which in this case is not true (see my work below), so this function is definitely divergent.

http://tinyurl.com/2whxqa

Hope this helps and that I didn't make any mistakes!

 

[MarkSA]

Yes that helped, thank you. I wasn't aware they were doing absolute value, but that makes sense. That's great to hear, it will save me a lot of work to just be able to state that it's divergent by the divergence test.