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?
 A: The first half of the hint is enough to conclude the convergence of the series.
Indeed, notice that $a_n := \pi a^2/(\sqrt{n^2+a^2}+n)$ decreases monotonically to $0$ as $n\to\infty$, and so, $\sin(a_n)$ decreases monotonically to $0$ for large $n$. (For an explicit range of $n$ for which this claim is valid, just pick any $N$ such that $a_N \in [0, \pi/2]$. Then this claim is true for the range $n \geq N$, thanks to the monotonicity of $\sin x$ over $[0, \pi/2]$.)
Then what is the use of the asymptotic formula for $\sin(a_n)$? The advantage is that we can predict the behavior of the series. Indeed, from $\sin(a_n) \sim (-1)^n \pi a^2/ 2n$ we can read out that


*

*$\sum \sin(a_n)$ does not converge absolutely, but 

*somehow alternating series test may be applicable.


So, although the relation cannot be utilized directly, we can set up the direction of our proof.
A: For small $x$ we have that $\sin{(x)} \approx x$. One can prove this from the Taylor series expansion of $\sin{(x)}$. As 
$$\frac{\pi a^2}{\sqrt{n^2+a^2}+n} \approx \frac{\pi a^2}{2n}$$ for large $n$ and $1/n$ is very small for large $n$ we then have
$$(-1)^n\sin\frac{\pi a^2}{\sqrt{n^2+a^2}+n} \approx (-1)^n\frac{\pi a^2}{2n}$$
A: 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 $[0,1]$.
