Does the series $\sum_{n=1}^{\infty}\sin\left(2\pi\sqrt{n^2+\alpha^2\sin n+(-1)^n}\right)$ converge?

This series is convergent.

As $n$ tends to $+\infty$, we may write $$ \begin{align} u_n &:=\sin \left( 2\pi \sqrt{n^2+\alpha^2 \sin n+(-1)^n}\right)\\\\ &=\sin \left( 2\pi n \:\sqrt{1+\frac{\alpha^2\sin n}{n^2}+\frac{(-1)^n}{n^2}}\right)\\\\ &=\sin \left( 2\pi n \:\left(1+\frac{\alpha^2\sin n}{2n^2}+\frac{(-1)^n}{2n^2}+\mathcal{O}\left(\frac{1}{n^4}\right)\right)\right)\\\\ &=\sin \left( 2\pi n +\frac{\pi\alpha^2\sin n}{n}+\frac{\pi(-1)^n}{n}+\mathcal{O}\left(\frac{1}{n^3}\right)\right)\\\\ &=\sin \left(\frac{\pi\alpha^2\sin n}{n}+\frac{\pi(-1)^n}{n}+\mathcal{O}\left(\frac{1}{n^3}\right)\right)\\\\ &=\frac{\pi\alpha^2\sin n}{n}+\frac{\pi(-1)^n}{n}+\mathcal{O}\left(\frac{1}{n^3}\right) \end{align} $$ Now recall that $\displaystyle \sum_{n=1}^{+\infty}\frac{\sin n}{n}$ is convergent, moreover $$ \sum_{n=1}^{+\infty}\frac{\sin n}{n}=\Im\sum_{n=1}^{+\infty}\frac{e^{in}}{n}=\Im\left(-\log(1-e^i)\right)=\frac{\pi-1}{2}. $$ Then it is clear that your initial series $\displaystyle \sum_{n=1}^{+\infty} u_n $ is convergent, being the sum of convergent series.