TITLE: Test the convergence of the series $\sum \sin [\pi(\sqrt5+2)^n]$
QUESTION [5 upvotes]: I have just approached the following series
$$\sum_{n=1}^\infty \sin [\pi(\sqrt5+2)^n]$$
And I already have a question. The $\lim_{n \to \infty}\pi(\sqrt5+2)^n=+\infty$. And the $\lim_{n\to \infty} \sin[\pi(\sqrt5+2)^n]$ does not exist.
Can $\sum a_n$ converge if $\lim_{n\to \infty}a_n$ does not exist? I guess the answer is yes because I know that $\sum a_n$ cannot converge if $\lim_{n\to \infty}a_n\neq 0$. Am I right?  
The second part of my question concerns the strategy to use to test convergence of such a series. I can test convergence of series with standard tests. But it seems I need to develop a more sophisticated approach in order to successfully deal with series like this. For example, here, should I use some sort of expansion? If yes, what kind of expansion?
Thank you for your help.

REPLY [5 votes]: As $n$ increases, $(2+\sqrt{5})^n$ gets closer and closer to an integer, since:
$$ (2+\sqrt{5})^n + (2-\sqrt{5})^n \in \mathbb{Z}. $$
If $[x]$ is the distance of $x$ from the closest integer, we have:
$$ [(2+\sqrt{5})^n] = (\sqrt{5}-2)^n =\frac{1}{(2+\sqrt{5})^n}<\frac{1}{4^n},$$
hence:
$$ |\sin(\pi(2+\sqrt{5})^n)| \leq \frac{\pi}{4^n}, $$
giving that the original series is convergent.