How Do You Prove the Summation Formula for 1/(4k^2 - 1)?

[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 ?

\(\displaystyle
\sum 1 / 4k^2 - 1 = n / 2n + 1
\)


Please help

 

[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...

 

[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.