\rm\sum \limits_(n = 2)^( \infty ) \frac{1}{( {n}^(2) - 1 {)}^(2) } \\​

Question

`\rm\sum \limits_(n = 2)^( \infty ) \frac{1}{( {n}^(2) - 1 {)}^(2) } \\`​

Answer 1

Partial fractions:

`\displaystyle \frac1{(n^2-1)^2} = \frac14 \left(\frac1{n+1} - \frac1{n-1} + \frac1{(n+1)^2} + \frac1{(n-1)^2}\right)`

Now evaluate the sums. The first two terms form a telescoping sum,

`\displaystyle \sum_(n=2)^\infty \left(\frac1{n+1} - \frac1{n-1}\right) = \sum_(n=1)^\infty \left(\frac1{n+2} - \frac1n\right) \\\\ ~~~~~~~~ = \lim_(N\to\infty) \sum_(n=1)^N \left(\frac1{n+2} - \frac1n\right) \\\\ ~~~~~~~~ = \lim_(N\to\infty) \left(\frac1{N+2} + \frac1{N+1} - 1 - \frac12\right) \\\\ ~~~~~~~~= -\frac32`

The others are flavors of
`\zeta(2)`.

`\displaystyle \sum_(n=2)^\infty \frac1{(n+1)^2} = \sum_(n=3)^\infty \frac1{n^2} \\\\ ~~~~~~~~~~~~~~~~~ = \sum_(n=1)^\infty \frac1{n^2} - 1 - \frac1{2^2} \\\\ ~~~~~~~~~~~~~~~~~ = \frac{\pi^2}6 - \frac54`

`\displaystyle \sum_(n=2)^\infty \frac1{(n-1)^2} = \sum_(n=1)^\infty \frac1{n^2} \\\\ ~~~~~~~~~~~~~~~~~ = \frac{\pi^2}6`

It follows that

`\displaystyle \sum_(n=2)^\infty \frac1{(n^2-1)^2} = \frac14 \left(-\frac32 + \left(\frac{\pi^2}6 - \frac54\right) + \frac{\pi^2}6\right) = \boxed{(\pi^2)/(12) - (11)/(16)}`