<s>[INST] Does the series $\sum_{n=1}^{\infty}\sin\left(2\pi\sqrt{n^2+\alpha^2\sin n+(-1)^n}\right)$ converge? Let $\alpha$ be such that $0\leq \alpha \leq 1$. Since $\sin n$ has no limit as $n$ tends to $\infty$, I'm having trouble with finding if the series $$\sum_{n=1}^{\infty}\sin \left(2\pi\sqrt{n^2+\alpha^2\sin n+(-1)^n}\right)$$ is convergent? Thanks.
 [/INST] 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. 
</s>