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

[mmilton]

Homework Statement

Is the sequence {((-1)^n)/2n} convergent? If so, what is the limit?

Homework Equations

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.

 

[Ray Vickson]

Homework Statement

Is the sequence {((-1)^n)/2n} convergent? If so, what is the limit?

Homework Equations

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.

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

 

[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.

 

[mmilton]

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

 

[DryRun]

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.

 

[LCKurtz]

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

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.

 

[Ray Vickson]

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

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

 

[Whovian]

(The series Ʃ(-1)^n/(2n) also converges, by the "alternating series test"---but that is a totally different question!)

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.

 

[LCKurtz]

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.

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.

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.

 

[Mark44]

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.

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:
$$\sum_{n = 1}^{\infty}\frac{(-1)^n}{2n} $$

 

[mmilton]

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.

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.

 

[mmilton]

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

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

 

[Whovian]

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:
$$\sum_{n = 1}^{\infty}\frac{(-1)^n}{2n} $$

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

 

[HallsofIvy]

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.