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So I have to determine if $\sum_2^{\infty} \frac{(-1)^n}{ln(n)}$ absolutely converges, conditionally converges, or diverges.

So first I tried the Alternating Series Test, because that is what you do first (right???). (Also I just started understanding mathjax more, so excuse the formatting).

Alternating Series Test

lim n-> infinity ($\frac{1}{ln(n)}$) = 0

and it's decreasing as well, so that means its convergent.

One question I have here is if one of these attribute of the alternating series test fails, does that mean it's divergent or I just can't use the test?

Ratio Test

Now to find if it's absolute convergence or conditional convergence, I did the ratio test, but I got 1. That means I can't use this test.

enter image description here

Root Test

I don't see how I can use the root test here because I just raised everything to the 1/n power and I'm stuck.

What should I do?

I hope someone can clear up my confusions. Thanks.

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It converges conditionally and not absolutely because $\displaystyle \sum_{n=2}^\infty \left|(-1)^n\cdot \dfrac{1}{\ln n}\right| = \displaystyle \sum_{n=2}^\infty \dfrac{1}{\ln n} \geq \displaystyle \sum_{n=2}^\infty \dfrac{1}{n} = \infty$, and you've done the first part that series is convergent by the alternating series test.

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  • $\begingroup$ ohhh! so you used the comparison test! $\endgroup$ – Elsa Apr 21 '15 at 7:24
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    $\begingroup$ yes, i do and it works doesn't it? $\endgroup$ – DeepSea Apr 21 '15 at 7:25
  • $\begingroup$ yes it does! Thanks for clarifying that part. but what if the alternating series test failed in the first place? $\endgroup$ – Elsa Apr 21 '15 at 7:27
  • $\begingroup$ Then you would have to use other methods. $\endgroup$ – DeepSea Apr 21 '15 at 7:32
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    $\begingroup$ You can use absolute vaue on the $a_n$ and o another test on it. $\endgroup$ – DeepSea Apr 21 '15 at 7:43
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You are correct that the alternating series gives you the answer you want about convergence (although not about absolute convergence).

In general, these tests are phrased as: "If such-and-such a condition is true, then the series converges." If you like, think of this in terms of predicates: you have a statement of the form $P \implies Q$. However, given that, $\sim\! P$ (the negation of $P$) tells you nothing at all.

Note that if the condition fails, then this simply gives us no information. It could converge or diverge; in fact your example here is a perfect example---you know the series converges (by the alternating series test), but the ratio test tells you nothing at all.

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  • $\begingroup$ thanks for your answer :). so what if the alternating series test failed and did not tell me that it converged? I can't use any other test on it other than the n-th term test for divergence. $\endgroup$ – Elsa Apr 21 '15 at 7:24

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