Let $$ \lambda_{n} = \frac{n\pi}{L} - \frac{L}{2n\pi(a-1)} $$ such that $n^{2} > \frac{L^{2}}{2\pi^{2}(a-1)}$, $L > 0$ and $ 1 \neq a$ and $a > 0$.
My question: If $$ B_{n} = \frac{L^{2}}{(n^{2}\pi^{2} - L^{2}\lambda_{n}^{2})} $$ then using asymptotic expansion, we get $$ B_{n} = a- 1 + O(n^{-2}) $$ Why? What does $O(n^{-2})$ mean?
I think: $$ \lambda_{n}^{2} = \bigg(\frac{n\pi}{L} - \frac{L}{2n\pi(a-1)} \bigg)^{2}= \bigg(\frac{2n^{2}\pi^{2}(a-1) - L^{2}}{2n\pi L(a-1)} \bigg)^{2} $$ This is, $$ \lambda_{n}^{2} = \dfrac{(2n^{2}\pi^{2}(a-1))^{2} - 4n^{2}\pi^{2}(a-1)L^{2} + (L^{2})^{2}}{(2n\pi (a-1))^{2}L^{2}} $$ Then, we have $$ n^{2}\pi^{2} - L^{2}\lambda_{n}^{2} = \dfrac{(2n\pi (a-1))^{2}n^{2}\pi^{2} - (2n^{2}\pi^{2}(a-1))^{2} + 4n^{2}\pi^{2}(a-1)L^{2} - (L^{2})^{2}}{(2n\pi (a-1))^{2}} $$ Therefore, we get $$ B_{n} = \dfrac{(2n\pi (a-1))^{2}L^{2}}{(2n\pi (a-1))^{2}n^{2}\pi^{2} - (2n^{2}\pi^{2}(a-1))^{2} + 4n^{2}\pi^{2}(a-1)L^{2} - (L^{2})^{2}}. $$ Unfortunately, I can't continue. Help me, please.