# Series convergence

1. Nov 20, 2008

### Somefantastik

1. The problem statement, all variables and given/known data
Show uniform convergence

$$\frac{4b}{\pi} \sum^{\inf}_{n=1} \frac{1-(-1)^{n}}{n^{2}}cos(nt)cos(nx)$$

for fixed t

2. Relevant equations

3. The attempt at a solution

$$\left| cos(nt) \right| \leq 1$$

$$\left| cos(nx) \right| \leq 1$$

$$lim \left|\frac{1 - (-1)^{n}}{n^{2}}\right| \ = \ 0$$

What's next?

2. Nov 20, 2008

### Feldoh

Well it's an alternating series isn't it? So... see if it's conditionally convergent and absolutely convergent

3. Nov 20, 2008

### Somefantastik

I'd like to use the Weierstrauss M test. Can someone help me in that direction?

4. Nov 20, 2008

### grief

What's b? It seems like there is a lot of stuff they added here to try to confuse you. Let's ignore the constant 4b/pi for now. Then for any n, for all x the absolute value of the nth term is less than 2/n^2. We know that the series 2/n^2 converges, so this series also converges.

5. Nov 20, 2008

### Dick

Then do so. For each n you want a M_n such the |f_n|<=M_n (where f_n is the nth function in your series) and the series of the M_n converges. It's really pretty straightforward. You've already handled the cos parts. Any suggestions?

6. Nov 20, 2008

### wangchong

This is not an alternative series! The terms with n even are all zeros and terms with n odd are 2/n^2 * cos ( ) * cos (). So take its absolute value and compare it with the series sum(2/n^2, n=1..infty) (I'm ignoring the constant outside the summation. Hence the series is absolutely convergent!

7. Nov 21, 2008

### Somefantastik

Ok thanks everybody! I think I have gleaned enough to get a good answer.