How in the world do you determine the following series' convergence/divergence specifically using the limit comparison test?

from n=1 to n=infinity

ln (2n+1)/ (n^2 + 2n)

I'm bamboozled and befuddled!

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- Mar 10th 2009, 08:39 AMKaitosanFreaky series using limit comparison test!
How in the world do you determine the following series' convergence/divergence specifically using the limit comparison test?

from n=1 to n=infinity

ln (2n+1)/ (n^2 + 2n)

I'm bamboozled and befuddled! - Mar 10th 2009, 09:08 AMOpalg
I would write the n'th term as $\displaystyle \frac{\ln(2n+1)}{n^2+2n} = \frac{\ln(2n+1)}{\sqrt n}\frac{\sqrt n}{n^2+2n}$. Then because "ln(n) goes to infinity slower than any positive power of n", it follows that the first fraction becomes small. The second fraction, $\displaystyle \frac{\sqrt n}{n^2+2n}$, is approximately $\displaystyle n^{-3/2}$. So do the limit comparison test, comparing the given series with the convergent series $\displaystyle \sum n^{-3/2}$.

- Mar 10th 2009, 10:03 AMKaitosan
Hm. I understand things better now. Thanks a lot.

- Mar 11th 2009, 10:17 AMKaitosan
Also, what's to prevent me from using, say, "n" instead of sqrt(n)? In that way, it will appear that the series diverges....

I know it's against the rule to bump topics but I don't care, I really need help. I might as well post another topic.

Anyways, I followed Opalag's directions but, strangely, I got zero. According to the limit comparison test, if the number equals zero or infinity, then the result is inconclusive. Please clarify?

Also, why would I not use like n^2 instead of sqrt(n)? If I had done that then the series appears to diverge!

Please help. Thanks a lot! - Mar 11th 2009, 11:35 AMOpalg
Okay, what the limit comparison test says is that if $\displaystyle \textstyle\sum a_n$ and $\displaystyle \textstyle\sum b_n$ are series of positive terms, $\displaystyle \lim_{n\to\infty}\frac{a_n}{b_n}$ exists and $\displaystyle \textstyle\sum b_n$ converges, then$\displaystyle \textstyle\sum a_n$ converges. (The test is sometimes stated in the form that if the limit exists

__and is nonzero__then $\displaystyle \textstyle\sum a_n$ converges__if and only if__$\displaystyle \textstyle\sum b_n$ converges, but that is not what is needed for this example.)

If $\displaystyle a_n = \frac{\ln(2n+1)}{n^2+2n}$, and you choose $\displaystyle b_n = 1/n^p$ for some p between and 2, then $\displaystyle \textstyle\sum b_n$ will converge (because p>1), and $\displaystyle \lim_{n\to\infty}\frac{a_n}{b_n}$ will be 0 (because p<2).

I hope that makes it clearer. (If you get an infraction for bumping, tell the Moderator that I think you were justified on this occasion.) - Mar 11th 2009, 12:00 PMKrizalid
Or we can bound the general term: we have $\displaystyle \ln x< 2\sqrt{x}-2,\,\forall\,x>1,$ thus for $\displaystyle n\ge1$ it's $\displaystyle \frac{\ln (2n+1)}{n^{2}+2n}< \frac{2\sqrt{2n+1}-2}{n^{2}+2n}< \frac{4\sqrt{n}}{n^{2}}=\frac{4}{n^{3/2}},$ then your series converges since $\displaystyle \sum\limits_{n=1}^{\infty }{\frac{1}{n^{3/2}}}<\infty .$

- Mar 11th 2009, 12:02 PMKaitosan

Why must p be between 1 and 2? You simplified the series so as to exclude the ln expression in order to dig up the right-handed series for the b series. In that process of simplification, you can input any number in the bottom left and and the top right, and each number brings a different convergence/divergence result!

Also, you seems to contradict yourself. If an/bn is nonzero and finite and if bn converges, then an converges. But you says an/bn converges to zero!

Let me repeat my understanding for the limit comparison test-

To evaluate the series an, pick bn based on an's highest ordered term in each numerator and denominator. Then structure an/bn and look for the limit. If the limit is between zero and infinity and if bn converges, then an/bn converges. But an/bn converges to zero! I'm so confused.

Please don't make me bump this topic again. ;) - Mar 11th 2009, 12:15 PMOpalg
Please read what I said. I stated very carefully the form of comparison test that I was using. I said that if the ratio $\displaystyle a_n/b_n$ converges (to any limit, including the possibility that the limit might be zero) and if $\displaystyle \textstyle\sum b_n$ converges, then $\displaystyle \textstyle\sum a_n$ converges.

If you take $\displaystyle a_n = \frac{\ln(2n+1)}{n^2+2n}$ and $\displaystyle b_n = 1/n^{3/2}$, then $\displaystyle \frac{a_n}{b_n} = \frac{n^{3/2}\ln(2n+1)}{n^2+2n} = \frac{\ln(2n+1)}{n^{1/2} + 2n^{-1/2}} \to 0$ as $\displaystyle n\to\infty$. therefore, by the limit comparison test in the form that I stated, $\displaystyle \textstyle\sum a_n$ converges. - Mar 11th 2009, 12:31 PMKaitosan
Looks like we found our "communication hitch." Based on what it looks like, there are two possibilities. First, my book is wrong. Or, secondly, you may be slightly off the definition of the limit comparison (I say this in humility, there's always a good chance that I'm wrong).

My book gives the definition of the limit comparison test in the following -

"Let an and bn be positive-termed series. Let L = lim (n to infinity) an/bn. If 0 < L < +infinity, then an and bn either both converge or both diverge. If L = 0 or L = +infinity, then the test is inconclusive."

The issue here is the domain of the limit comparison test and whether it includes zero. It's there, I copied exactly as my book puts it. What's going on?

I really appreciate the fact that you replied speedily! - Mar 11th 2009, 12:46 PMOpalg
I can only quote what I said before:

Your book uses the version in red, but the slightly different version that I was using is used by some other authors, and is more useful for dealing with the example in this thread.

In fact, I think that your book is somewhat misleading in claiming that the test is inconclusive when the limit is zero or infinity. As my version of the test shows, it is possible to draw conclusions in these cases. However, in those situations the test can only be used to get an implication in one direction; in these cases it is not an "if and only if" test. - Mar 11th 2009, 12:59 PMKaitosan
Aaaah! I see! Thank you so much, I understand. I'll keep in mind that there are two "versions" of the limit comparison test then. Thanks for not giving me up. (Cool)