Find all real values of $a$ such that $\sum_{n=0}^{\infty} \sin(\pi\sqrt{a^2+n^2})$ diverges

Question

To find all $a$ that makes the series diverge, it is sufficient to make the limit of $\sin(\pi\sqrt{a^2+n^2})$ not exist or exists but not equal to $0$ as $n\to\infty.$ I suspect only $a=0$ would converge the series, and the limit of $\sin(\pi\sqrt{a^2+n^2})$ would not even exist if $a\neq 0.$ .

Answer 1

HINT:

Observe that we can write

$$\begin{align} \sin(\pi\sqrt{a^2+n^2})&=\sin\left(n\pi\sqrt{1+\frac{a^2}{n^2}}\right)\\\\ &=\sin\left(n\pi+\frac{a^2\pi}{2n}+O\left(\frac1{n^3}\right)\right)\\\\ &=(-1)^n\sin\left(\frac{a^2\pi}{2n}+O\left(\frac1{n^3}\right)\right) \end{align}$$

Can you finish now? Is there any real value for $a$ for which this series diverges?

Answer 2

Starting like Mark Viola we have to consider the sum

$$s = \sum _{n=1}^{\infty } \cos (\pi n) \sin \left(\frac{\pi a^2}{2 n}\right)$$

Now, expanding the $\sin()$ and exchanging the order of the summation gives under the k-sum:

$$\frac{(-1)^k \left(\frac{\pi a^2}{2}\right)^{2 k+1} \sum _{n=1}^{\infty } \left(\frac{1}{n}\right)^{2 k+1} \cos (\pi n)}{(2 k+1)!}$$

Doing the n-sum gives for the k-sum:

$$\sum _{k=0}^{\infty } \frac{(-1)^{k+1} 2^{-4 k-1} \left(2^{2 k}-1\right) \pi ^{2 k+1} \left(a^2\right)^{2 k+1} \zeta (2 k+1)}{(2 k+1)!}$$

Doing the k-sum, neglecting the difference betwenn $\zeta$ and unity leads to

$$\sum _{k=0}^{\infty } \frac{(-1)^{k+1} 2^{-4 k-1} \left(2^{2 k}-1\right) \pi ^{2 k+1} \left(a^2\right)^{2 k+1}}{(2 k+1)!} $$

which is explicitly

$$2 \sin \left(\frac{\pi a^2}{4}\right)-\sin \left(\frac{\pi a^2}{2}\right)$$

Hence, contrary to the "guess" in my comment there is no restrictin on "a" for convergence.