Ad hoc proof of the convergence of $\sum\limits_n\sin\left(\pi\sqrt{n^2+a^2}\right)$ I use the following method to prove that
$$\sum_{n=0}^\infty \sin(\pi\sqrt{n^2+a^2})$$
converges for any $a \in \mathbb{R}$.
First, we can see that
$$\lim_{n\to\infty} \frac{\sqrt{n^2 + a^2}}{n} = 1,$$
which means precisely that $\forall \varepsilon > 0, \exists n_0\ \forall n \ge n_0: \sqrt{n^2+a^2} - n < \varepsilon$. We can therefore find $n_0$ such that $\sqrt{n^2+a^2} - n < \frac\pi2$.
Then we write $$\sum_{n=0}^\infty \sin(\pi\sqrt{n^2+a^2}) = \sum_{n=0}^{n_0} \sin(\pi\sqrt{n^2+a^2})\ + \sum_{n=n_0+1}^\infty \sin(\pi\sqrt{n^2+a^2}),$$
where the first sum contains a finite number of finite terms and evaluates to some $K \in \mathbb{R}$.
The second sum can be rewritten (thanks to the fact that $\sqrt{n^2+a^2} - n < \frac\pi2$) as
$$\sum_{n=n_0+1}^\infty \sin(\pi\sqrt{n^2+a^2}) = \sum_{n=n_0+1}^\infty (-1)^{n+1}|\sin(\pi\sqrt{n^2+a^2})|.$$
We have already shown that $d_n=(\sqrt{n^2+a^2} - n) \to 0$ as $n\to\infty$. $|\sin x|$ has the period $\pi$, and because of that, $|\sin(\pi\sqrt{n^2+a^2})|=|\sin{(n\pi+d_n\pi)}|=|\sin(n\pi + d_n\pi + k\pi)|$ for any $k \in \mathbb{Z}$. Choosing $k=-n$, we get $|\sin(\pi\sqrt{n^2+a^2})| = |\sin{d_n\pi}|$.
As $d_n$ is a monotonic decreasing sequence (we can show easily that $d_{n+1} < d_n\ \forall n$) and $|\sin x|$ is monotonic on $(k\pi,\ k\pi + \frac\pi2)$, we have that $|\sin(\pi\sqrt{n^2+a^2})| \to 0$ is a monotonically decreasing sequence.
The convergence of the sum now follows from the Leibniz's rule.
There is probably a much simpler way. I would like to know whether my solution is correct and if not, why?

Note that $$\sqrt{n^2+a^2}=n\cdot\sqrt{1+\frac{a^2}{n^2}}=n\cdot\left(1+\frac{a^2}{2n^2}+O\left(\frac1{n^4}\right)\right)=n+u_n,$$
where $$u_n=\frac{a^2}{2n}+O\left(\frac1{n^3}\right).$$ Note also that, when $x\to0$, $$\sin x=x+O(x^3),$$ hence, using this for $x=\pi u_n$, one gets $$\sin\left(\pi\sqrt{n^2+a^2}\right)=\sin(n\pi+\pi u_n)=(-1)^n\cdot\sin(\pi u_n)=\frac{(-1)^n}n \frac{\pi a^2}2+O\left(\frac1{n^3}\right).$$
This shows that the series of interest is the sum of an alternating series and of an absolutely convergent series, hence it converges (but not absolutely).
