Convergence of the series for $a \in \mathbb R$ $\sum_{n=1}^\infty\sin\left(\pi \sqrt{n^2+a^2} \right)$

Question

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?

Answer 1

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

1. $\sum \sin(a_n)$ does not converge absolutely, but
2. somehow alternating series test may be applicable.

So, although the relation cannot be utilized directly, we can set up the direction of our proof.

Answer 2

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 $$(-1)^n\sin\frac{\pi a^2}{\sqrt{n^2+a^2}+n} \approx (-1)^n\frac{\pi a^2}{2n}$$

Answer 3

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 $[0,1]$.