Telescoping Method & Partial Fractions PLEASE HELP

[BuBbLeS01]

Telescoping Method & Partial Fractions...PLEASE HELP!

Homework Statement

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

Homework Equations

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)

 

[quantumdude]

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?

 

[tiny-tim]

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 ?

 

[quantumdude]

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??

 

[quantumdude]

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.

 

[tiny-tim]

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

 

[tiny-tim]

Yes!
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[BuBbLeS01]

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