Convergence of Alternating Series?

LuisRodg · 2008-04-07 03:55:01

So I was given an alternating series:

According to my book, given an alternating series, theres 2 types of convergence tests to apply.

So if:

Then the series converges.

So I saw that Ak decreases, but its limit is not equal to 0, its equal to 1. So this test is inconclusive.

Second test for alternating series to to do the abs ratio test. I went on to take the abs of the series:

So:

So the limit is equal to 1 as well and this test is inconclusive. If the limit was <1 then I could say that the series converges absolutely which implies that it converges. But it doesnt.

So we have:

The harmonic series diverges so Ak diverges as well so in my quiz I just said that the series diverges absolutely. Is this right?

Last edited by LuisRodg (2008-04-07 03:56:47)

TheDude · 2008-04-07 06:00:17

First of all, you'd better not start that sum at k=0 or you'll be in trouble.

Secondly, I'm not that familiar with these tests you used, but I'm pretty sure that your line of reasoning is wrong.

may diverge, but that doesn't guarantee that your sum will diverge too, unless I'm mistaken.

Assuming I'm right and that you can't evaluate the limit that way, my suggestion would be to combine every other term in the series and then evaluate that series. Let k be an odd number, then we'd get this:

This gives us the difference between 2 consecutive terms in the series, so we won't have to deal with the -1^k any more. Rewriting this into a form we can use with summation notation we get

You have to change the terms a little bit since we're counting by 2 instead of by 1. Hopefully this will be a little easier to work with.

Edit: If all of my tinkering around was correct then the sum will, in fact converge. This can be see by comparing it to the sum of the reciprocals of the squares: 1 + 1/4 + 1/9 + 1/16 ... = pi^2/6.

Last edited by TheDude (2008-04-07 06:09:34)

mathsyperson · 2008-04-07 06:28:06

Here's how I see it:

By the Leibniz Alternating Test which Luis mentioned, the second term of the RHS converges.
Hence, the entire series converges iff ∑(-1)^k does.

However, the sequence of partial sums of this is 1, 0, 1, 0, 1, 0, ...
It is clearly oscillating and not converging to anything. Hence the series in the original question is divergent.
(It is bounded though)

TheDude · 2008-04-07 08:31:17

Well, something's screwy here. mathsy's and my alterations seem to be equivalent, yet we're drawing different conclusions from them. For example, in mathsy's series at n = 6, the sum is -37/60. Similarly, in my series at k = 2 (which is the same point as n = 6 in mathsy's series) the sum is again -37/60, so I'm confident that our series are the same.

However, our evaluation of them is completely different. Mine is bounded by

, which is convergent. However, I can't find a single weakness in mathsy's method either. I'm confused.

LuisRodg · 2008-04-07 09:13:34

[when I started the series from k=0 it was just a mistake when writing the LaTeX code.]

If your confused. Im even more confused

Just kidding, I understand it all but this series topic just seems too broad. Theres like a zillion convergence tests to apply, and just deciding which one to use becomes a question of its own. I guess practice will solve that.

Anyways, this was a question in a quiz. Its 3 questions like this in 10 mins so my guess is that its not supposed to be that complex. Maybe we are missing something?

Anyways, from my conclusions I answered that the series diverges *absolutely* which is true. I know this doesnt imply it diverges.

If a series converges absolutely then it converges, but if it diverges absolutely you cannot imply it diverges.

An example will be the alternating harmonic series:

so Bk is the harmonikc series which diverges which implies that Ak diverges absolutely but it doesnt imply it diverges since we know that the alternating harmonic series converges to ln(2) if im not mistaken.

Last edited by LuisRodg (2008-04-07 09:16:41)

Ricky · 2008-04-07 09:15:26

Couple of theorems you guys may want to know about:

Theorem:

Definition:

Theorem:

Personally I find it an abuse of notation to say that the sum converges to infinity, but that just means the sum diverges in the positive direction. Similarly with negative infinity. Rudin's book is the standard though, so who am I to argue?

TheDude · 2008-04-07 09:50:33

After thinking about it, I'm definitely wrong. My method basically ignores every other term in the series, which isn't going to give an accurate picture of an alternating series. Thinking about it, every other term is more than 1 away from the term before it, so the series can't possibly converge.

mathsyperson · 2008-04-07 10:57:03

Ricky wrote:

Theorem:

This is probably what was meant to be used for the question. In that case, a_n clearly doesn't converge to 0 and so doesn't converge. That's all that's needed for a proof.

I almost said that earlier, but got confused and thought that that only worked if all the a_n's were positive (which isn't true).

Ricky · 2008-04-07 11:23:21

I'm fairly certain your post #3 is wrong as well. You can't just separate terms "math-nilly". In order to do what you did, you need to look at the sequence of partial sums, split them apart there, and show that sequence doesn't converge.

\/\/illy is apparently blocked by the filter...

mathsyperson · 2008-04-07 11:39:29

I don't see the mistake in #3.

There's nothing wrong with saying that ∑(a_n+b_n) = ∑(a_n) + ∑(b_n), is there?

Ricky · 2008-04-07 14:31:11

There isn't... so long as the sum is finite. When it's infinite, that's not a trivial statement. So far I have yet to see any proof of it. The proof would probably just involve partial sums as I said before, and perhaps I'll try to sketch one up later tonight. Maybe I'm just forgetting that theorem, I dunno.

Ricky · 2008-04-07 15:30:39

The proof was much more trivial than I'd thought it would be. It's almost by definition. As it turns out, all cases are trivial to the point where proof isn't even needed except for one:

And the proof of this is almost by definition. We have:

What we really mean when we write this is:

However, by our limit theorems, we know that this is really:

By assumption, the first term is alpha. So this is:

Certainly this sequence does not tend toward anything, otherwise b_n would, which would result in a contradiction. Therefore, we get that the sequence diverges.

So easier than I thought, but I stand by what I said earlier about it needing justification. I just don't believe I've ever seen it proved.

mathsyperson · 2008-04-08 00:41:35

Mmm, I had a similar thought process when I made my original post. So I knew it was right, I just didn't explain it rigorously enough. Sorry.

I've just realised that #10 isn't always true though, because two divergent sequences can add to a convergent one. (Trivial example: a_n and -a_n)

LuisRodg · 2008-04-09 02:54:08

I got my quiz back and my professor gave me full points. Remember that my final answer consisted of saying "this series diverges absolutely". I did know that if a series diverges absolutely you can't imply that it diverges conditionally.

Anyways, the solution he gave on the board was:

Since the limit is not equal to 0 then it diverges. So basically he used the Divergence Test which says that if the limit of the series is not 0 then the series diverges. I did use this test and I saw that it gave me 1 (which is probably why he gave me the full points) but I didnt draw any conclusions because I took the limit of just k/(k+1) and thats not the actual series. The actual series is alternating.

So from his conclusions and his answer on the board you can deduce this:

But isnt this wrong? b_k is the absolute of a_k so by finding that the limit of b_k is not equal to 0, your actually finding that b_k diverges. But b_k is the absolute of a_k and if a series diverges absolutely it doesnt imply it diverges conditionally.

So my professor is wrong?

Please tell me if you can't follow my reasoning as im very intrigued by this.

Last edited by LuisRodg (2008-04-09 03:05:51)

TheDude · 2008-04-09 03:13:56

You're confusing series and sequences.

and are sequences, is a series. Specifically, is the sequence of individual terms in the series, and is the absolute value of the individual terms in the series.

Using this distinction you can see that if

then the series cannot converge.

To see this, consider a sequence

. This is called a sequence of partial sums (I think). The limit as this new sequence approaches infinity will be the same as the infinite series we're working with. Note that we can rewrite this new sequence recursively like this: . From a high level we can see that if diverges, then will diverge as well. This is by no means rigorous, but it should give you something to start with.

Last edited by TheDude (2008-04-09 03:26:59)

LuisRodg · 2008-04-09 03:23:10

Im not sure I understand. if b_{k} diverges, how does it imply that a_{k} cannot converge? Isnt this implying that if a series diverges absolutely then it diverges conditionally? (Which is wrong).

For example. The alternating harmonic series converges:

So the alternating harmonic series diverges absolutely but it converges conditionally and my professor's solution implied that if it diverged absolutely then it must diverge conditionally (which is a contradiction):

Could you please clarify this for me? Im very confused.

Last edited by LuisRodg (2008-04-09 03:26:55)

TheDude · 2008-04-09 03:44:11

Sorry, I made some notational errors in my last post that I've edited, I don't know if they'll help or not. In case they don't I'll try to explain it again.

Im not sure I understand. if b_{k} diverges, how does it imply that a_{k} cannot converge? Isnt this implying that if a series diverges absolutely then it diverges conditionally? (Which is wrong).

Again, you're confusing sequences and series. I don't have a proof of this, but my belief is that if the sequence b_{k} diverges (that is to say, a_{k} diverges absolutely) then a_{k} will diverge.

This is, however, NOT true for series, which you showed.

However, in order for a series to converge it's terms must converge to 0. This can easily be seen by considering the definitions of series and limits. A series has a limit L if and only if for every real number e > 0 there exists a natural number N such that

for every n > N.

Now, remember that

.

Now, let's assume that a_{k} does not converge to 0. If it does not then we can always find an e > 0 such that

, which means that the series cannot have a limit.

To reiterate, don't confuse series with sequences. A series can diverge absolutely and converge conditionally, but a sequence can't (as far as I know). In order for a series to converge the sequence of it's individual terms must converge to 0.

Last edited by TheDude (2008-04-09 03:45:55)

LuisRodg · 2008-04-09 03:53:00

Thanks a lot for the explanation. Now im clear. You have mentioned that I was confusing sequences for series and that is exactly the root of my confusion.

Thanks. It really helps to understand the topic when you have this kinds of discussions.

Ricky · 2008-04-09 11:57:26

In a rush, so I didn't get to read all of Dude's post. Just wanted to post this general theorem:

If:

Then:

Good exercise is to prove it, though it should be rather easy.

Math Is Fun Forum

#1 2008-04-07 03:55:01

Convergence of Alternating Series?

#2 2008-04-07 06:00:17

Re: Convergence of Alternating Series?

#3 2008-04-07 06:28:06

Re: Convergence of Alternating Series?

#4 2008-04-07 08:31:17

Re: Convergence of Alternating Series?

#5 2008-04-07 09:13:34

Re: Convergence of Alternating Series?

#6 2008-04-07 09:15:26

Re: Convergence of Alternating Series?

#7 2008-04-07 09:50:33

Re: Convergence of Alternating Series?

#8 2008-04-07 10:57:03

Re: Convergence of Alternating Series?

#9 2008-04-07 11:23:21

Re: Convergence of Alternating Series?

#10 2008-04-07 11:39:29

Re: Convergence of Alternating Series?

#11 2008-04-07 14:31:11

Re: Convergence of Alternating Series?

#12 2008-04-07 15:30:39

Re: Convergence of Alternating Series?

#13 2008-04-08 00:41:35

Re: Convergence of Alternating Series?

#14 2008-04-09 02:54:08

Re: Convergence of Alternating Series?

#15 2008-04-09 03:13:56

Re: Convergence of Alternating Series?

#16 2008-04-09 03:23:10

Re: Convergence of Alternating Series?

#17 2008-04-09 03:44:11

Re: Convergence of Alternating Series?

#18 2008-04-09 03:53:00

Re: Convergence of Alternating Series?

#19 2008-04-09 11:57:26

Re: Convergence of Alternating Series?

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