# Alternating Series Test Conditions

1. Nov 10, 2013

### Lebombo

1. The problem statement, all variables and given/known data

This is what I understand about Alternating Series right now:

If I have an alternate series, I can apply the alternative series test.

$\sum(-1)^{n}a_{n}$

Condition 1: Nth term test on $a_{n}$

Condition 2: 0 < $a_{n+1} ≤ a_{n}$

If condition 1 is positive or ∞, convergence is inconclusive, try another test.

If condition 1 is 0, then go to condition 2.

If condition 2 is false, then convergence is inconclusive, try another test.

If condition 2 is true, then series is convergent, go to absolute convergence test.

Absolute convergence test

$\sum |a_{n}|$

Choose a test to find convergence or divergence.

If convergent, Alternating series is absolutely convergent.

If divergent, Alternating series is conditionally convergent.

So my questions come from applying Alternating Series Conditions 1 and 2.

Question 1:
If condition 1 is positive or ∞ (instead of 0), then the test is said to be inconclusive, but then what test can then be applied? It looks like Condition 1 is sort of the nth term test, so I would be inclined to think that a positive or ∞ would imply divergence. So if it doesn't imply divergence (only inconclusiveness), what then?

Question 2:
If condition 1 turns out to be 0 and condition 2 is tested and turns out false, the test is said to be inconclusive, but then, again, what test can be applied thereafter?

2. Nov 10, 2013

### Lebombo

P.S.

If condition 1 or 2 of the Alternating Series test fails and shows inconclusive, can I then apply the Ratio test to the Alternating Series to test for convergence/divergence?

EDIT: UPDATE:

From what I've just learned, I don't necessarily have to do the Alternating Series Test on an Alternating Series. I can just skip it and jump directly to the Absolute Convergence Test and use any other test to determine convergence/divergence, such as ratio, comparison, integral test. Alternating Series Test is not a mandatory test for Alternating Series. Is this in fact true?

Last edited: Nov 10, 2013
3. Nov 10, 2013

### LCKurtz

I suppose your condition 1 is really whether or not $a_n\to 0$. If not the series diverges and if so go to condition 2.

You can test for absolute convergence with any of the standard tests. Most text problems will be solved when you have done all that. If you actually run into a series in your homework where nothing works, it's time to ask your teacher and see if he can solve it.

4. Nov 10, 2013

### LCKurtz

Yes. You would usually test for absolute convergence first, since if it works, you are done. But if it isn't absolutely convergent and it is alternating, try the alternating series stuff.

5. Nov 10, 2013

### vela

Staff Emeritus
This isn't quite correct. If the limit isn't equal to 0, you can conclude that the series diverges. There's no need to do additional tests.

6. Nov 11, 2013

### Lebombo

Are you sure vela. I read that you can never conclude divergence from the alternating series test alone. The only thing you can do is conclude inconclusive and try a new test.

7. Nov 11, 2013

### jbunniii

If the terms don't converge to zero, then the series diverges, whether or not it is alternating.

If $a_n \rightarrow 0$ but not monotonically, then you can't conclude anything from the alternating series test. The series may converge or diverge.

8. Nov 11, 2013

### Lebombo

okay, would I be understanding you correctly if I said:

Given the alternating series Ʃ((-1)^n)(a_{n}) , if I apply the Nth term test on the entire thing, ((-1)^n)(a_{n}), it will diverge if the limit is anything other than zero.

However, given the same series Ʃ((-1)^n)(a_{n}), if I apply condition 1 of the alternating series test, which is the Nth term test on just the (a_{n}) portion, the test is inconclusive if the limit is anything other than zero.

9. Nov 11, 2013

### vela

Staff Emeritus
No. If $a_n$ doesn't go to 0, then $(-1)^n a_n$ isn't going to go to 0 either. If the terms of the sequence do not converge to 0, the series diverges, period.