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.
Find the value of \(A+B+C+D\).
by
Akshat Sharda