<s>[INST] Convergence of the series for $a \in \mathbb R$ $\sum_{n=1}^\infty\sin\left(\pi \sqrt{n^2+a^2} \right)$ Convergence of the series for $a \in \mathbb R$ $$\sum_{n=1}^\infty\sin\left(\pi \sqrt{n^2+a^2} \right)$$ 
I saw this problem in a calculus book and it gave a hint that says 
HINT First show that $$\sin\left(\pi \sqrt{n^2+a^2} \right)=(-1)^n\sin\frac{\pi a^2}{\sqrt{n^2+a^2}+n}\sim(-1)^n\frac{\pi a^2}{2n}\qquad  (n \to\infty)$$ 
I was able to show that $$\sin\left(\pi \sqrt{n^2+a^2} \right)=\sin\left(\pi \sqrt{n^2+a^2} -\pi n+\pi n\right)=\sin\left(\pi (\sqrt{n^2+a^2}-n)+\pi n \right)=(-1)^n\sin\left(\pi (\sqrt{n^2+a^2}-n) \right)=(-1)^n\sin\frac{\pi a^2}{\sqrt{n^2+a^2}+n}$$
But what I don't understand is how did they come up with that equivalence? First I thought that they used the limit comparison test but now I can see that you can't use that test because we're dealing with alternating series. Did they do a mistake or something?
Can somebody help me understand this hint and how to solve this problem?
 [/INST] Hint:
$$
\begin{align}
\sum_{n=1}^\infty\sin\left(\pi\sqrt{n^2+a^2}\right)
&=\sum_{n=1}^\infty(-1)^n\sin\left(\pi\sqrt{n^2+a^2}-\pi n\right)\\
&=\sum_{n=1}^\infty(-1)^n\sin\left(\frac{\pi a^2}{\sqrt{n^2+a^2}+n}\right)
\end{align}
$$
where $\sin\left(\frac{\pi a^2}{\sqrt{n^2+a^2}+n}\right)$ decreases monotonically to $0$ for $n\ge a^2$. That is, for $n\ge a^2$, $\frac{\pi a^2}{\sqrt{n^2+a^2}+n}\le\frac{\pi a^2}{2n}\le\frac\pi2$ and $\sin(x)$ maps $[0,\pi/2]$ monotonically onto $[0,1]$.
</s>