Does $\sum_{n\ge1} \sin (\pi \sqrt{n^2+1}) $ converge/diverge?

How would you prove convergence/divergence of the following series?

$$\sum_{n\ge1} \sin (\pi \sqrt{n^2+1}) $$

I'm interested in more ways of proving convergence/divergence for this series. Thanks.

EDIT

I'm going to post the solution I've found here:

$$a_{n}= \sin (\pi \sqrt{n^2+1})=\sin (\pi (\sqrt{n^2+1}-n)+n\pi)=(-1)^n \sin (\pi (\sqrt{n^2+1}-n))=$$ $$ (-1)^n \sin \frac{\pi}{\sqrt{n^2+1}+n}$$ The sequence $b_{n} = \sin \frac{\pi}{\sqrt{n^2+1}+n}$ monotonically decreases to $0$. Since our series is an alternating series then it converges.

$$ \sum_{n\ge1} \sin\left(\pi\sqrt{n^2+1}\right) = \sum_{n\ge1} \pm\sin\left(\pi\left(\sqrt{n^2+1}-n\right)\right) $$ (Trigonometric identity. Later we'll worry about "$\pm$".)

Now $$ \sqrt{n^2+1}-n = \frac{1}{\sqrt{n^2+1}+n} $$ by rationalizing the numerator.

So we have the sum of terms whose absolute values are $$ \left|\sin\left(\frac{\pi}{\sqrt{n^2+1}+n}\right) \right| \le \frac{\pi}{\sqrt{n^2+1}+n} \to0\text{ as }n\to\infty.\tag{1} $$

But the signs alternate and the terms decrease in size, so this converges. (They decrease in size because sine is an increasing function near $0$ and the sequence inside the sine decreases.)

It does not converge absolutely, since $\sin x\ge x/2$ for $x$ small and positive, and the sum of the terms asserted to approach $0$ in $(1)$ above diverges to $\infty$.

It converges as an alternating series. For each $n$, we have $\sin((n+\delta)\cdot\pi)$ for some small $\delta$ which approaches $0$ at the limit and decreases monotonically (as the $1$ in $\sqrt{n^2+1}$ becomes less significant compared to the $n^2$.

For even $n$, this expression will take on smaller and smaller positive values, as $\delta$ shrinks and $(n+\delta)\cdot\pi$ gets closer and closer to a zero at $2m\pi$ for some natural $m$ from the right.

Similarly, it will take on shrinking negative values for odd $n$.

The absolute value of each term tends to $0$ and decreases monotonically, so we have convergence by the alternating series test.