# Homework Help: Infinite Series: ∑ (-1)^(n+1) cos(nx)

1. Oct 18, 2009

### kingwinner

1. The problem statement, all variables and given/known data
Consider the infinite series

∑ (-1)n+1 cos(nx)
n=1
Is there any convergence?

2. Relevant equations/concepts
Infinite series

3. The attempt at a solution
I was looking back in my 1st year calculus textbook, but none of the theorems seem to apply.

Basic divergence test:
If an does not converges to 0 as n->∞, then ∑ an diverges.
[well...here the an's are constants and does not depend on x, so this theorem does not apply in our case]

Secondly, it is not a power series (it has cos(nx) here), so none of the theorems about power series apply here.

Can someone please explain why the above infinite series has no convergence?
Thank you!

2. Oct 18, 2009

### Office_Shredder

Staff Emeritus
Why doesn't this apply? For what possible values of x can the series converge, using this theorem?

3. Oct 18, 2009

### kingwinner

Because it says that the form of the series has to be ∑ an, and I don't see any x in it. (maybe I am wrong?)

When there is a x in the serires, it should be something like ∑ anxn, so the an are just constant coefficients (depends on n only, not on x)...

How can I tell for what possible values of x can the series converge?

4. Oct 18, 2009

### tnutty

You know that cos(nx) will always be less than or equal to 1 , so ...

5. Oct 19, 2009

### kingwinner

But why does this imply that there is no convergence?

6. Oct 19, 2009

### Office_Shredder

Staff Emeritus
You can say: Pretend x is a constant. Let an= (-1)n+1cos(nx). Does ∑an converge? The answer in general will depend on x.

7. Oct 19, 2009

### kingwinner

OK, but how can I possibly test convergence for ALL values of x?

8. Oct 19, 2009

### Petek

Can you find a specific value of x for which the series obviously converges? What about a specific value for which it obviously diverges? If so, then are there other such values of x? Finally, can you generalize?

HTH

Petek

9. Oct 19, 2009

### kingwinner

I can see that for example when x=0, the sequence alternates between -1 and 1, so it does not tend to 0 and so the series diverges.
But there are infinitely many possibilites for x (x can be ANY real number), and I don't see how every possible case can be ruled out...

10. Oct 19, 2009

### Office_Shredder

Staff Emeritus
Think of it the other way: What condition is necessary for the series to converge? Can any x satisfy this condition?

11. Oct 19, 2009

### kingwinner

I am sorry, but I don't get it...can you please give me the names of the theorems that can answer your question?

Here is my try:
In order to converge, the sequence of partial sums must converge to a finite number.
But I don't see how this is going to help.

Once again, the trouble that I am having is that I don't know how to deal with the infinitely many possibilites for x...how can I test convergence in each case? How many cases in total are there?

12. Oct 20, 2009

### Petek

He's referring to the test that you gave in your very first post:

Basic divergence test:
If $a_n$ does not converges to 0 as n->∞, then ∑ $a_n$ diverges.

Next, the answer given above ("no convergence") is incorrect, because the series does converge for some values of x. What happens if cos (nx) = 0 (for all n)? For which values of x is this true?

Two. The one given above (cos (nx) = 0 (for all n)) and ???. For the second case (you supply the condition), use what you called the "Basic divergence test".

Please post again if you would like more help.

Petek

13. Oct 24, 2009

### kingwinner

But I don't think it is possible to have a value of x for which cos(nx)=0 for ALL n. For example, x=pi/2 doesn't work. x=3pi/2 doesn't work as well...

14. Oct 24, 2009

### Dick

No, there isn't such a value. But you are on the right track. Now can you take that one step further and show there is no value of x such that limit n->infinity cos(nx)=0? That's really what you need to show nonconvergence, right? It's pretty easy if x is a rational multiple of pi. If it's not, you need a 'real analysis' type theorem, I think. Like if x is irrational then nx mod 1 is a dense set in the interval [0,1]. Is this a real analysis course? Have you proved anything like that?

Last edited: Oct 24, 2009
15. Oct 25, 2009

### kingwinner

No, it's from an applied partial differential equations course. Please explain in the simplest way since I don't have any background in real analysis so far...

In the text, it is trying to justify why the sin Fourier series for x
i.e. x= ∑ (-1)n+1 (2/n) sin(nx)
cannot be differentiated term-by-term.

So the idea is if

∑ (-1)n+1 cos(nx)
n=1
diverges for all values of x, then the sin Fourier series cannot be differentiated term-by-term, am I right??

16. Oct 25, 2009

### Dick

If x=p*pi/q where p and q are integers (i.e. a rational times pi) then cos(nx)=cos(a multiple of pi) whenever n is divisible by q. That's +1 or -1. In the case where x=r*pi where r is irrational then you can show nx is 'very close' to being a multiple of pi an infinite number of time, so cos(nx) can't approach zero. Maybe they really expect you to prove that, and you can just get a away with stating it and waving your hands around.

17. Oct 26, 2009

### Petek

Quite correct. Sorry for misleading you. To make up for my error, here's a simple proof that cos (nx) doesn't converge to 0 for any value of x:

First, recall the following trig identity:

$$cos (u) - cos (v) = -2 sin \frac{1}{2} (u + v) sin \frac{1}{2} (u - v)$$

Now suppose that cos (nx) converges to 0 for some value of x. Then cos ((n + 2)x) also converges to 0. Hence cos ((n + 2)x) - cos (nx) converges to 0. By the above trig identity,

cos ((n + 2)x) - cos (nx) = -2 sin ((n + 1)x) sin x

Therefore, either sin ((n + 1)x) converges to 0, or sin x = 0. However, since

$$cos^2 ((n + 1)x) + sin^2 ((n + 1)x) = 1$$

we can't have both cos ((n + 1)x) and sin ((n + 1)x) converging to 0. Therefore, sin x = 0. This implies that x = m$\pi$ for some integer m. But then cos (nx) = cos nm$\pi = \pm 1$, contradicting the assumption that cos (nx) converges to 0.

HTH

Petek