TITLE: Calculate the limit $\lim \limits_{n \to \infty} |\sin(\pi \sqrt{n^2+n+1})|$
QUESTION [11 upvotes]: Calculate 
$$\lim \limits_{n \to \infty} |\sin(\pi \sqrt{n^2+n+1})|$$

REPLY [10 votes]: Note that $$\sqrt{n^2+n+1}-n=\frac{n+1}{\sqrt{n^2+n+1}+n}\to 1/2$$ as $n\to\infty$.
For even $n$, $\sin(\sqrt{n^2+n+1}\pi)=\sin(\sqrt{n^2+n+1}\pi-n\pi)\to \sin(\pi/2)=1$ as $n\to\infty, n$ even.
For odd $n$, $\sin(\sqrt{n^2+n+1}\pi)=-\sin(\sqrt{n^2+n+1}\pi-n\pi)\to -\sin(\pi/2)=-1$ as $n\to\infty, n$ odd.
Therefore, $$|\sin(\sqrt{n^2+n+1}\pi)|\to 1$$ as $n\to\infty$.