BuBbLeS01

## Homework Statement

Find the sum of the series from n=1 to infinity...
2/(4n^2-1)

## 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)

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Gold Member

## The Attempt at a Solution

I want to use the telescoping method...
2/(4n^2) = 2/[(2n-2) * (2n+1)]

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

Homework Helper
2/(4n^2-1)...
2/(4n^2) = 2/[(2n-2) * (2n+1)]

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

Hi BuBbLeS01!

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

Next step … what is An + An+2 ?

Staff Emeritus
Gold Member
Hi BuBbLeS01!

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

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

BuBbLeS01
Correct. And how do you write down the partial fraction decomposition when your denominator has two nonrepeated, linear factors?

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??

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

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

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

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

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

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.

Homework Helper
Hi BuBbLeS01!

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

BuBbLeS01
So is it just...
2 = A(2n+1) + B(2n-1)

BuBbLeS01
A = 1 and B = -1