Quick Question on Limit Comparison Test

In summary, the conversation revolves around whether the limit comparison test can be used on alternating series. Some believe it can be used as long as the series is compared to a positive function and absolute values are used, while others believe it can only be applied to positive series. The question also arises as to why the LCT does not work for the alternating harmonic series, which is known to converge. Some suggest using the alternating series test instead.
  • #1
Anonymous217
355
2
Can the limit comparison test work on alternating series? I looked at the conditions online and it said that the series a_n and b_n must be > 0 in order to use it. To be specific it said, "Suppose that a_n > 0 and b_n>0 for all n> or = to N (N a positive integer)."
I asked my teacher about this and he said that it does work, and yet... I don't know. Maybe I misinterpreted the conditions.
 
Physics news on Phys.org
  • #2
When dealing with alternating series, the comparison can be carried out with absolute values of the series in question. The comparison series must have positive terms.
 
  • #3
Isn't there an alternating series test for just such situations?
 
  • #4
To be particular, I'm curious about being able to use the Limit Comparison test for
[tex]\sum_{n=1}^{\infty} \frac{(-1)^n}{n}[/tex]
I understand that it's easier to do with the AST, but I tried this with the LCT and it didn't turn out nice.

I compared it to 1/n.
[tex]\lim_{n\rightarrow \infty} |\frac{1}{n} * \frac{n}{(-1)^n}|[/tex]
[tex]\lim_{n\rightarrow \infty} |\frac{1}{(-1)^n}|[/tex]
[tex] = 1[/tex]
1/n diverges, so (-1)^n / n must diverge..? That's what I got using this test that is, and it's wrong. What did I do incorrectly with my math?
 
  • #5
The limit comparison test doesn't use absolute values. Since limn→∞ 1/(-1)n doesn't exist/diverges, this comparison test doesn't help.
 
  • #6
Also, the alternating harmonic series does converge... even if I forgot exactly why, or to what value... is it log 2?
 
  • #7
It is log(2) and comes from the http://en.wikipedia.org/wiki/Mercator_series" .
 
Last edited by a moderator:
  • #8
Bohrok said:
The limit comparison test doesn't use absolute values. Since limn→∞ 1/(-1)n doesn't exist/diverges, this comparison test doesn't help.
My first question asked if you could use the LCT on an alternating or negative series. Mathman replied that it's possible as long as you compare it to a positive function and while using absolute values. That's what I tried doing, and that's what my second post is about. Something seems off about the LCT or I made a mistake.

Char. Limit said:
Also, the alternating harmonic series does converge... even if I forgot exactly why, or to what value... is it log 2?
I'm aware that it does converge, because it should converge conditionally by using the AST. However, I'm wondering why the LCT doesn't work for this if apparently you are able to use it on a negative or alternating series. And it's -log(2), not log(2) by the way.
 
Last edited:
  • #9
I also believe it will work with alternating series as long as the two series you're comparing alternate with the same sign so their limit converges to a number (since it doesn't use absolute values), unlike 1/(-1)n which oscillates between -1 and 1.
 
  • #10
The comparison tests can be applied only on the postive series.
 
  • #11
I'm getting mixed answers here.. :(

@Bohrok: But if I use an alternating series as a comparison for another alternating series, how will I know if the series I compared converges or diverges? And if you say AST, that makes the LCT useless because you could've done AST on the original series.

@Sweet GirL: I'm starting to believe in this. However, my teacher also said you can use it on alternating series by putting absolute values, but it doesn't seem like it works.
 
  • #12
My math teacher agrees with Sweet Girl. LCT can only be done on positive series.

Sorry, log(.5) then.
 
  • #13
Okay. I'll go challenge my math teacher. And by challenge, I mean ask how to do the above series using LCT and subtly hint that he's wrong.
 

What is the Limit Comparison Test?

The Limit Comparison Test is a method used in calculus to determine the convergence or divergence of a series by comparing it to a known series with a known convergence or divergence.

When should I use the Limit Comparison Test?

The Limit Comparison Test should be used when the series you are trying to evaluate does not have a clear pattern or when other tests, such as the Ratio Test or the Root Test, do not provide a conclusive result.

How do I perform the Limit Comparison Test?

To perform the Limit Comparison Test, you must first find a known series with a known convergence or divergence. Then, you must take the limit of the ratio of the two series. If the limit is equal to a positive number, then the two series have the same convergence or divergence. If the limit is equal to 0 or infinity, then the two series have opposite convergence or divergence. If the limit is inconclusive, then the test cannot be used.

What is the difference between the Limit Comparison Test and the Comparison Test?

The Limit Comparison Test compares the ratio of two series, while the Comparison Test compares the individual terms of two series. The Limit Comparison Test is more versatile and can be used in situations where the Comparison Test cannot be applied.

Can the Limit Comparison Test be used to determine the exact value of a series?

No, the Limit Comparison Test only determines the convergence or divergence of a series. It does not provide an exact value for the series. To find the exact value, you must use other methods such as the Geometric Series or the Integral Test.

Similar threads

Replies
6
Views
667
Replies
3
Views
2K
  • Calculus
Replies
3
Views
1K
Replies
1
Views
142
Replies
9
Views
882
Replies
3
Views
956
  • Calculus
Replies
2
Views
1K
Replies
1
Views
1K
  • Calculus
Replies
4
Views
2K
  • Calculus and Beyond Homework Help
Replies
1
Views
251
Back
Top