Is the sequence {((-1)^n)/2n} convergent? (I think that it does)

1. Apr 26, 2012

mmilton

1. The problem statement, all variables and given/known data
Is the sequence {((-1)^n)/2n} convergent? If so, what is the limit?

2. Relevant equations

3. The attempt at a solution

I'm thinking that it is convergent by the alternating series test, but I am not certain. The limit part I'm not sure how to go about it. Is it simply the sum, or is that something completely different. Thanks so much for any hints or suggestions.

2. Apr 26, 2012

Ray Vickson

Do mean "sequence" or "series"---your post used both words! In some cases it makes a great difference; for example, the sequence {1/n} converges, but the series Ʃ1/n diverges.

RGV

3. Apr 26, 2012

Whovian

I think it would be a bad idea to use the Alternating Series test, the limit of the series of the absolute value of this diverges.

4. Apr 26, 2012

mmilton

It is a sequence. So, does that preclude using the alternating series test (i.e. is that only used on series)?

5. Apr 26, 2012

sharks

Try to write all your equations in latex form, as it's much clearer and easier to understand: $$\sum \frac{(-1)^n}{2n}$$
First thing that you should notice, is the presence of the alternating sign. Apply the theorem and use modulus.

6. Apr 26, 2012

LCKurtz

Yes. Plot a few points of your sequence on the x axis and see if you can figure out if it converges and if so, to what. Once you see graphically what it is doing you should be able to prove it.

7. Apr 26, 2012

Ray Vickson

The alternating series test is used for SERIES---that is, for a series where the sequence of terms is alternating in sign. If you really mean the sequence {(-1)^n/(2n)}, then that converges because the terms --> 0 as n --> ∞. (The series Ʃ(-1)^n/(2n) also converges, by the "alternating series test"---but that is a totally different question!)

RGV

8. Apr 26, 2012

Whovian

Huh? I thought by the AST, we'd get that if Ʃ1/2n converges, then so does the original series. Which it doesn't. So we have to use some other test.

9. Apr 26, 2012

LCKurtz

First, the OP keeps claiming that he has a sequence problem, not a series. I admit, since he keeps using the term "sum" here and there, I'm not sure I believe him. Secondly, if it really is an alternating series, you need to re-look at what the AST says. The alternating series converging does not mean the series of absolute values does. Read RGV's explanation again.

10. Apr 26, 2012

Staff: Mentor

I think that you are confused about what the alternating series test says.

The AST can be used to determine the convergence of this series:
$$\sum_{n = 1}^{\infty}\frac{(-1)^n}{2n}$$

11. Apr 26, 2012

mmilton

Yes, it is a sequence. I only used "sum" because I was initially confused and was asking whether I could use AST on the sequence.

12. Apr 26, 2012

mmilton

Thank you Ray! Great explanation. I understand it now.

13. Apr 27, 2012

Whovian

Yep. And, for some reason, I thought that it said that if $$\sum\left|a_n\right|$$ converged, so did $$\sum\left(a_n\right)$$.

14. Apr 27, 2012

HallsofIvy

Staff Emeritus
It certainly is true that if $\sum |a_n|$ converges, then $\sum a_n$ converges. But the "alternating series theorem" says more than that. It says that if ${a_n}$ is a sequence of positive numbers that converges to 0, then $\sum (-1)^n a_n$ converges.