Tricky Sum!

Algebra Level 4

$\sum_{n=2}^\infty \dfrac1{(n^2-1)n^2} = \dfrac{1}{3\cdot4}+\dfrac{1}{8\cdot 9}+\dfrac{1}{15\cdot16}+\dfrac{1}{24\cdot25}+\cdots$

The series above is equal to $\dfrac{A-B\pi^{C}}{D}$ where $A,B,C$ and $D$ are positive integers with $A$ and $B$ coprime.