Simplification of $\sum_{k=1}^n \frac{1}{4k^2-1}$

Question

So I want to simplify this expression: $$\sum_{k=1}^n \frac{1}{4k^2-1}$$ and Wolfram Alpha tells me it can be simplified to two forms: $$\frac{n}{2n+1}$$ and $$\frac{1}{2}-\frac{1}{2(2n+1)}$$ The problem being that I don't know how to do so and I cannot seem to, after sincerely trying, find any helpful information in my text book nor in Google. Thanks.

Answer 1

Hint: $$\frac 1{4k^2 - 1} = \frac 1{(2k + 1)(2k - 1)} = \frac 12\left(\frac{1}{2k - 1} - \frac{1}{2k + 1}\right)$$ So now your sum can collapse to contain only some initial and terminal terms.

Answer 2

There is a problem in what you say.

Hint

$$\frac{1}{4k^2-1}=\frac{1}{(2k-1)(2k+1)}.$$ Then, find $A,B$ such that $$\frac{1}{4k^2-1}=\frac{A}{2k-1}+\frac{B}{2k+1},$$ and your sum will follow...