For a positive integer n \geq 1, let a_{n}=\left\lfloor\sqrt[3]{n}+\frac{1}{2}\right\rfloor. Given a positive integer N \geq 1, let \mathcal{F}_{N} denote the set of positive integers n \geq 1 such that a_{n} \leq N. Let S_{N}=\sum\limits_{n \in \mathcal{F}_{N}} \frac{1}{a_{n}^{2}}. As N goes to infinity, the quantity S_{N}-3 N tends to \frac{a \pi^{2}}{b} for relatively prime positive integers a, b. Given that \sum\limits_{k=1}^{\infty} \frac{1}{k^{2}}=\frac{\pi^{2}}{6}, find a+b.