Quick Question on Limit Comparison Test

  • #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.
 

Answers and Replies

  • #2
mathman
Science Advisor
8,087
550
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
Char. Limit
Gold Member
1,216
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Isn't there an alternating series test for just such situations?
 
  • #4
Anonymous217
355
2
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
Bohrok
867
0
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
Char. Limit
Gold Member
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Also, the alternating harmonic series does converge... even if I forgot exactly why, or to what value... is it log 2?
 
  • #7
Bohrok
867
0
It is log(2) and comes from the http://en.wikipedia.org/wiki/Mercator_series" [Broken].
 
Last edited by a moderator:
  • #8
Anonymous217
355
2
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.

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
Bohrok
867
0
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
Sweet_GirL
24
0
The comparison tests can be applied only on the postive series.
 
  • #11
Anonymous217
355
2
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
Char. Limit
Gold Member
1,216
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My math teacher agrees with Sweet Girl. LCT can only be done on positive series.

Sorry, log(.5) then.
 
  • #13
Anonymous217
355
2
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.
 

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