# Homework Help: Convergence of Alternating Series

1. Jan 26, 2010

### Cairo

I need to test the series

∑_(n=1)^∞〖(-1)^(n+1) 1/(n+2x^2 )〗 x in R

for absolute, pointwise and uniform convergence.

I am unsure about absolute convergence. I thought about using the comparison test and the idea that 1/n+2x^2 > 1/n and so diverges. Is this correct?

I'm not sure how to go about pointwise convergence, but if shown, then I am sure I would be able to prove if it is uniformly convergent or not.

2. Jan 26, 2010

### Staff: Mentor

No. 1/(n + 2x^2) < 1/n. When the denominator is larger, the overall fraction is smaller. Since the inequality goes the wrong way, you can't conclude that the series diverges.

3. Jan 26, 2010

### Cairo

Stupid me!

I'm at a loss at how to proceed for absolute convergence now. Any ideas?

4. Jan 26, 2010

### ystael

If I write the series as $$\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n + C}$$ where $$C = 2x^2$$ is a constant, does that give you any ideas for what to compare it to?

Remember that convergence or divergence is determined by the tail of the series, and that tail can start as far down as you like.

5. Jan 26, 2010

### dacruick

isnt absolute convergence simply the sum of the absolute values of each term. Or equal to that -1 changing to a 1? And in that case you have a function with the denominator increasing by 1 each time. Which does not converge.

6. Jan 26, 2010

### Cairo

Your comment would suggest comparing it with the Harmonic Series, which diverges.

I have used the Ratio Test and claimed that |a(n+1)/a(n)|<1 for all x in R, as n tends to infinity. Hence converges absolutely.

Is this correct now?

7. Jan 26, 2010

### Cairo

Actually....I think this is wrong since it eventually equals 1, so the test is inconclusive.

8. Jan 26, 2010

### vela

Staff Emeritus
Isn't there a test where you examine the limit of the ratio of the terms of a known converging or diverging series and the series in question?

9. Jan 26, 2010

### l'Hôpital

Here is a hint:

Using Ystael's notation,

$$\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n + C} = \sum_{n=C+1}^\infty \frac{(-1)^{n+1 - C}}{n}$$

10. Jan 26, 2010

### Cairo

The Alternating Series Test reveals that {(-1)^n+1}/n converges.
Since {(-1)^n+1}/n+C < {(-1)^n+1}/n, then the series in question converges by the Comparison Test.

However, this doesn't show absolute convergence.

If I take the modulus of {(-1)^n+1}/n, then I get the Harmonic Series, which diverges.

Confused with where to go with all this!

11. Jan 26, 2010

### vela

Staff Emeritus
To show absolute convergence or divergence, you can use the limit comparison test.

12. Jan 26, 2010

### vela

Staff Emeritus
That only works if C is an integer, which is not guaranteed because C=2x^2 and x can be any real number.

13. Jan 26, 2010

### l'Hôpital

I keep doing this mistake. : /
Good call.

14. Jan 26, 2010

### dacruick

And what is the issue with that? I mean n is to the power of 1, so therefore it must diverge for all values of x except infinity and negative infinity. right?

15. Jan 26, 2010

### Cairo

In a previous post, somebody mentioned that you couldn't determine if the series diverged using this fact, since 1/n+C < 1/n. Wouldn't you need the inequality the other way around?

16. Jan 26, 2010

### vela

Staff Emeritus
17. Jan 26, 2010

### Cairo

I'm so confused now I have no idea what I am trying to prove.

Is the series divergent or absolutely convergent? If I know this, it might give me a starting point for a proof!

18. Jan 26, 2010

### Cairo

Okay...but this doesn't apply since you need ALL positive terms.

19. Jan 26, 2010

### vela

Staff Emeritus
You're testing for absolute convergence, right? So you're no longer looking at the original sequence, but the absolute value of the original sequence.

(By the way, most of the other tests also require a positive or non-negative sequence.)

20. Jan 26, 2010

### Cairo

Okay...So using the Limit Comparison Test I get a value of 1. Since 1/n is divergent, they both diverge, and so my original sequence is not absolutely convergent.

Is that right?