1. Apr 19, 2008

### BuBbLeS01

1. The problem statement, all variables and given/known data
Find the sum of the series from n=1 to infinity...
2/(4n^2-1)

2. Relevant equations

3. The attempt at a solution
I want to use the telescoping method....
2/(4n^2) = 2/[(2n-2) * (2n+1)]

I am following an example in my book and this is where I don't know what they did next...but they got
An = 1/(2n-1) - 1/(2n+1)

2. Apr 19, 2008

### Tom Mattson

Staff Emeritus
Correct. And how do you write down the partial fraction decomposition when your denominator has two nonrepeated, linear factors?

3. Apr 19, 2008

### tiny-tim

Hi BuBbLeS01!

I'm confused … 1/(2n - 1) - 1/(2n + 1) is 2/(4n² - 1).

Next step … what is An + An+2 ?

4. Apr 19, 2008

### Tom Mattson

Staff Emeritus
Right, he's asking how the author got that result. So he needs to revisit partial fraction decomposition.

5. Apr 19, 2008

### BuBbLeS01

2/(4n^2) = 2/[(2n-1) * (2n+1)]
So the LCD for the right side would be (2n-1)*(2n+1)

2(2n+1)/(2n-1) + 2(2n-1)/(2n+1)

For some reason that doesn't look right though??

6. Apr 19, 2008

### Tom Mattson

Staff Emeritus
Hold on a second, you dropped a "-1" somewhere along the way. It should be this:

$$\frac{2}{4n^2-1}=\frac{2}{(2n-1)(2n+1)}$$

That's because it's not right, and I don't understand why you're writing it. What you're supposed to do is this:

$$\frac{2}{(2n-1)(2n+1)}=\frac{A}{2n-1}+\frac{B}{2n+1}$$

That ought to jog your memory enough to finish.

7. Apr 19, 2008

### tiny-tim

Hi BuBbLeS01!

That's because you've put a + in the middle instead of a - .

8. Apr 19, 2008

### BuBbLeS01

So is it just....
2 = A(2n+1) + B(2n-1)

9. Apr 19, 2008

### BuBbLeS01

A = 1 and B = -1

10. Apr 19, 2008

### tiny-tim

Yes!

11. Apr 19, 2008

### BuBbLeS01

WOO-HOO LOL Thank you....I think I got it from here :)