Evaluate the series!

$\begin{aligned} S & = \sum_{n-1}^\infty \frac 1{(0.25-n^2)^2} = \sum_{n-1}^\infty \frac 1{\left(n^2-\frac 14\right)^2} = \sum_{n-1}^\infty \frac 1{\left(n-\frac 12 \right)^2\left(n+\frac 12\right)^2} \\ & = \sum_{n-1}^\infty \left(\frac 1{n-\frac 12} - \frac 1{n+\frac 12} \right)^2 \\ & = \sum_{n-1}^\infty \left({\color{#3D99F6}\frac 4{(2n-1)^2}} - {\color{#D61F06}\frac 8{(2n-1)(2n+1)}} + {\color{#3D99F6}\frac 4{(2n+1)^2}} \right) \\ & = {\color{#3D99F6} 8\sum_{n-1}^\infty \frac 1{(2n-1)^2} - 4} - {\color{#D61F06}4 \sum_{n-1}^\infty \left(\frac 1{2n-1} - \frac 1{2n+1} \right)} \end{aligned}$

$\begin{aligned} \ \ \ & = {\color{#3D99F6} 6 \sum_{n-1}^\infty \frac 1{n^2}} - 4 - \color{#D61F06} 4 & \small \color{#3D99F6} \text{See note.} \\ & = 6{\color{#3D99F6} \zeta (2)} - 8 & \small \color{#3D99F6} \text{Riemann zeta function }\zeta (s) = \sum_{k=1}^\infty \frac 1{k^2} \\ & = 6 \times {\color{#3D99F6} \frac {\pi^2}6} - 8 & \small \color{#3D99F6} \text{and }\zeta (2) = \frac {\pi^2}6 \\ & = \pi^2 - 8 \\ & \approx \boxed{1.870} \end{aligned}$

---

Note:

$\begin{aligned} \sum_{n=1}^\infty \frac 1{(2n-1)^2} & = \frac 1{1^2} + \frac 1{3^2} + \frac 1{5^2} + \frac 1{7^2} + \frac 1{9^2} + \cdots \\ & = \frac 1{1^2} + \frac 1{2^2} + \frac 1{3^2} + \frac 1{4^2} + \cdots - \left( \frac 1{2^2} + \frac 1{4^2} + \frac 1{6^2} + \frac 1{8^2} + \cdots \right) \\ & = \frac 1{1^2} + \frac 1{2^2} + \frac 1{3^2} + \frac 1{4^2} + \cdots - \frac 1{2^2} \left( \frac 1{1^2} + \frac 1{2^2} + \frac 1{3^2} + \frac 1{4^2} + \cdots \right) \\ & = \frac 34 \left( \frac 1{1^2} + \frac 1{2^2} + \frac 1{3^2} + \frac 1{4^2} + \cdots \right) \\ & = \frac 34 \sum_{n=1}^\infty \frac 1{n^2} \end{aligned}$

$\sum_{n=1}^ {\infty} \frac{1}{(n^2-0.25)^2}=\sum_{n=1}^ {\infty} \frac{2}{(n+0.5)^2}=\sum_{n=1}^ {\infty} \frac{8}{(2n+1)^2}=8\left(\frac{\pi^2}{8}-1\right)=\pi^2-8\approx \boxed{1.870}$ by the Basel problem.