# Thread: Problem with limit/direct comparison for series

1. ## Problem with limit/direct comparison for series

Hey guys, for some reason I am really struggling with the whole limit and direct comparison tests for series. Many times, I can't decide what to compare my series with.

For example: (I am currently stuck on) -- the series from 2 to infinity of 2 / 7n ln(n)+4n
I am assuming that I would use the limit comparison test (right??), which I am thinking would be ln(n)/n (a divergent series), and then divide An/Bn as limit goes to infinity? So then I am thinking the answer would be divergent?

Appreciate any feedback, thanks

2. ## Re: Problem with limit/direct comparison for series

Originally Posted by Beevo
Hey guys, for some reason I am really struggling with the whole limit and direct comparison tests for series. Many times, I can't decide what to compare my series with.

For example: (I am currently stuck on) -- the series from 2 to infinity of 2 / 7n ln(n)+4n
I am assuming that I would use the limit comparison test (right??), which I am thinking would be ln(n)/n (a divergent series), and then divide An/Bn as limit goes to infinity? So then I am thinking the answer would be divergent?

Appreciate any feedback, thanks
So that we know what you are talking about, please put some brackets where they're needed.

3. ## Re: Problem with limit/direct comparison for series

Originally Posted by Prove It
So that we know what you are talking about, please put some brackets where they're needed.
I have no idea how to type in the proper mathematical symbols on forums. Basically, the problem states: sigma (series) from n=2 to infinity, of 2 / 7n*ln(n) + 4n
Which means 2 over 7n times ln(n) plus 4.

Would I use the limit comparison test of say ln(n)/n? Then I am assuming it would be divergent.

4. ## Re: Problem with limit/direct comparison for series

Is this what you mean

$\displaystyle \sum_{n=2}^{\infty} \frac{2}{7n\log(n)+4n}$?

You type it like this \sum_{n=2}^{\infty} \frac{2}{7n\log(n)+4n} with tex /tex around it (with square brackets).

5. ## Re: Problem with limit/direct comparison for series

Originally Posted by Vlasev
Is this what you mean

$\displaystyle \sum_{n=2}^{\infty} \frac{2}{7n\log(n)+4n}$?

You type it like this \sum_{n=2}^{\infty} \frac{2}{7n\log(n)+4n} with tex /tex around it (with square brackets).
Yeah that's it, thanks dude. So basically:

$\displaystyle \sum_{n=2}^{\infty} \frac{2}{7n\ln(n)+4n}$

Would I take the limit comparison test with bn = $\displaystyle \frac{ln(n)}{n}$ ?? Would that then be divergent??

Any feedback appreciated.