1. Nov 1, 2005

Natasha1

I have to use the partial fraction technique on 1/(4k^2 - 1)...

ANSWER: So far so good and I get 1 / 2(2k-1) - 1 / 2(2k+1), is this correct?

I now need to show that ?

$\sum 1 / 4k^2 - 1 = n / 2n + 1$

Last edited: Nov 1, 2005
2. Nov 2, 2005

pinodk

You have the denominator
$$4k^2-1 = (2k+1)(2k-1)$$
therefore i believe you should get
$$\frac{1}{4k^2-1} = \frac{1}{(2k+1)}+\frac{1}{(2k-1)}$$

Then you know that:
$$\sum_{k=1}^n \frac{1}{4k^2-1} = \sum_{k=1}^n \frac{1}{(2k+1)} + \sum_{k=1}^n \frac{1}{(2k-1)}$$

You should then try to find expressions for the two new summations...

3. Nov 2, 2005

happyg1

hi,
I got what natasha got for the breakdown. You need to look at the sequence of partial sums and see what cancels out. It's easier to do this if you factor your 1/2 out. You should be able to see what's laeft fairly easily. Simplify that and you get your answer.