Testing convergence of series $\sum_{n=1}^\infty\sin(\pi\sqrt{n^2+k^2})$ Considering

$$\sum_{n=1}^\infty\sin(\pi\sqrt{n^2+k^2})$$

depending on $k$, which can be real. I have absolutely no clue how to proceed. Tried to taylor it, but with no result.
 A: Hint. Your series is convergent.
As $n$ tends to $+\infty$, we may write
$$
\begin{align}
u_n &:=\sin \left( \pi \sqrt{n^2+k^2 }\right)\\
&=\sin \left( \pi n \:\sqrt{1+\frac{k^2}{n^2}}\right)\\
&=\sin \left( \pi n \:\left(1+\frac{k^2}{2n^2}+\mathcal{O}\left(\frac{1}{n^4}\right)\right)\right)\\
&=\sin \left( \pi n +\frac{\pi k^2}{2n}+\mathcal{O}\left(\frac{1}{n^3}\right)\right)\\
&=(-1)^n\sin \left(\frac{\pi k^2}{2n}+\mathcal{O}\left(\frac{1}{n^3}\right)\right)\\
&=\frac{\pi k^2}2\frac{(-1)^n}{n}+\mathcal{O}\left(\frac{1}{n^3}\right)
\end{align}
$$
giving the convergence of the initial series $\displaystyle \sum u_n$. 
A: Notice that
$$\sin ( \pi \sqrt {n^2 + k^2} - n \pi) = \sin ( \pi \sqrt {n^2 + k^2} ) \cos (n \pi) - \cos ( \pi \sqrt {n^2 + k^2} ) \sin (n \pi) = \\
(-1)^n \sin ( \pi \sqrt {n^2 + k^2} ) ,$$
so
$$\sin ( \pi \sqrt {n^2 + k^2} ) = (-1)^n \sin ( \pi \sqrt {n^2 + k^2} - n \pi) = (-1)^n \sin \frac {\pi^2 (\sqrt {n^2 + k^2})^2 - n^2 \pi^2} {\pi \sqrt {n^2 + k^2} + n \pi} = \\
(-1)^n \sin \frac {\pi^2 k^2} {\pi \sqrt {n^2 + k^2} + n \pi} ,$$
therefore your series is
$$\sum \limits _{n \ge 1} (-1)^n \sin \frac {\pi^2 k^2} {\pi \sqrt {n^2 + k^2} + n \pi} .$$
Notice that the fraction inside the sine decreases to $0$ with respect to $n$, so from a certain $n_0$ onwards it will surely be in $[0, \frac \pi 2]$, an interval on which $\sin$ increases, which means that for $n \ge n_0$ the function $n \mapsto \sin \dfrac {\pi^2 k^2} {\pi \sqrt {n^2 + k^2} + n \pi} $ decreases to $0$ (in particular, this also shows that for $n \ge n_0$ the fraction is positive). Since the series is alternating, Leibniz's test tells us that it is convergent.
