I would like to prove the following:
$$\sin\left(2\pi\sqrt{n^2+(-1)^{n}} \right)=\dfrac{(-1)^{n}\pi}{n}+\mathcal{O}\left( \dfrac{1}{n^2}\right). $$
My attempt:
\begin{align*} \sin\left(2\pi\sqrt{n^2+(-1)^{n}} \right)&=\sin\left(2\pi\sqrt{n^2\left(1+\dfrac{(-1)^{n}}{n^2}\right)} \right) \\ &= \sin\left(2\pi\ n\left(1+\dfrac{(-1)^{n}}{n^2}\right)^{\dfrac{1}{2}} \right) \\ &=\sin\left(2\pi\ n\left(1+\dfrac{(-1)^{n}}{2n^2}+\mathcal{O}\left( \dfrac{1}{n^{4}}\right)\right) \right) \\ &=\sin\left(2\pi\ n+\dfrac{(-1)^{n}\pi}{n}+\mathcal{O}\left( \dfrac{1}{n^{3}}\right)\right) \\ &=(-1)^{2n}\sin\left(\dfrac{(-1)^{n}\pi}{n}+\mathcal{O}\left( \dfrac{1}{n^{3}}\right)\right) \\ &=\dfrac{(-1)^{n}\pi}{n}+\mathcal{O}\left( \dfrac{1}{n^{3}}\right). \\ \end{align*}
I got $\mathcal{O}\left( \dfrac{1}{n^{3}}\right) $ instead of $\mathcal{O}\left( \dfrac{1}{n^{2}}\right). $ Is my proof correct?
Yes, what you have is quite correct and is a more powerful statement than the one they provided. You skipped some details at the end. You in fact need to use $\sin x = x + O(x^3)$ at the last step, and I'm guessing that the original author used only $\sin x = x + O(x^2)$. One gets the stronger result here by using the second-order Taylor polynomial of $\sin$ at $0$.