Problem
$\sum_{n=4}^{\infty} \frac{6}{n^{2}-1}$
Solution
Step 1 ：Let \(S = \sum_{n=4}^{\infty} \frac{6}{n^{2}-1} = \frac{6}{15} + \frac{6}{23} + \frac{6}{35} + \frac{6}{49} + \dotsb\).
Step 2 ：Then \(n^{2}S = \sum_{n=4}^{\infty} \frac{6n^{2}}{n^{2}-1} = 6 + \frac{6}{2} + \frac{6}{3} + \frac{6}{4} + \dotsb\).
Step 3 ：Subtracting the first equation from the second gives us \(S = 6 + \frac{3}{2} + \frac{2}{3} + \frac{3}{4} + \dots = 6 + \frac{\frac{3}{2}}{1-\frac{1}{2}} = 6 + 3 = \boxed{9}\).
