We are asked to find the sum of the series $\frac{1}{2^2-1} + \frac{1}{4^2-1} + \frac{1}{6^2-1} + \dots + \frac{1}{(2n)^2-1}$.

1. Problem Description

We are asked to find the sum of the series

$\frac{1}{2^2-1} + \frac{1}{4^2-1} + \frac{1}{6^2-1} + \dots + \frac{1}{(2n)^2-1}$.

2. Solution Steps

The general term of the series is given by

$a_k = \frac{1}{(2k)^2 - 1} = \frac{1}{4k^2 - 1} = \frac{1}{(2k-1)(2k+1)}$.

We can use partial fraction decomposition to rewrite the general term:

$\frac{1}{(2k-1)(2k+1)} = \frac{A}{2k-1} + \frac{B}{2k+1}$

$1 = A(2k+1) + B(2k-1)$

To find $A$, let $2k-1 = 0$, so $k = \frac{1}{2}$. Then,

$1 = A(2(\frac{1}{2}) + 1) + B(0)$

$1 = 2A$

$A = \frac{1}{2}$

To find $B$, let $2k+1 = 0$, so $k = -\frac{1}{2}$. Then,

$1 = A(0) + B(2(-\frac{1}{2}) - 1)$

$1 = -2B$

$B = -\frac{1}{2}$

So, $a_k = \frac{1}{2} (\frac{1}{2k-1} - \frac{1}{2k+1})$.

Now, we want to find the sum $S_n = \sum_{k=1}^{n} a_k$:

$S_n = \sum_{k=1}^{n} \frac{1}{2} (\frac{1}{2k-1} - \frac{1}{2k+1})$

$S_n = \frac{1}{2} \sum_{k=1}^{n} (\frac{1}{2k-1} - \frac{1}{2k+1})$

$S_n = \frac{1}{2} [(\frac{1}{1} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{5}) + (\frac{1}{5} - \frac{1}{7}) + \dots + (\frac{1}{2n-1} - \frac{1}{2n+1})]$

This is a telescoping series. All the terms cancel out except for the first and last terms.

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

$S_n = \frac{n}{2n+1}$

3. Final Answer

$\frac{n}{2n+1}$