- #1

Anonymous217

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I asked my teacher about this and he said that it does work, and yet... I don't know. Maybe I misinterpreted the conditions.

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- Thread starter Anonymous217
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- #1

Anonymous217

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I asked my teacher about this and he said that it does work, and yet... I don't know. Maybe I misinterpreted the conditions.

- #2

mathman

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- #3

Char. Limit

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Isn't there an alternating series test for just such situations?

- #4

Anonymous217

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[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

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- #6

Char. Limit

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- #7

Bohrok

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It is log(2) and comes from the http://en.wikipedia.org/wiki/Mercator_series" [Broken].

Last edited by a moderator:

- #8

Anonymous217

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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._{n→∞}1/(-1)^{n}doesn't exist/diverges, this comparison test doesn't help.

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

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- #10

Sweet_GirL

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The comparison tests can be applied only on the postive series.

- #11

Anonymous217

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@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

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Sorry, log(.5) then.

- #13

Anonymous217

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