Finding a bounded and divergent sequence $(a_n)$ with $\lim_{n\to\infty}a_{n+1}-a_n=0$

I want to find a bounded and divergent sequence $(a_n)$ with $$\lim_{n\to\infty}a_{n+1}-a_n=0,$$and I prefer that $a_n=f(n)$ where the function $f$ is elementary, so that this answer is not relevant. The sequence $a_n=\sin(n)$ is bounded and divergent, but I'm not sure if$$\lim_{n\to\infty} \sin(n+1)-\sin(n)=0.$$ I also came up with $a_n=\sin(\sqrt n),$ and from the inequality $|\sin x-\sin y|\leq |x-y|$ together with (by multiplying by the conjugate) $$\lim_{n\to\infty}\sqrt{n+1}-\sqrt n=0,$$ we get $$\lim_{n\to\infty}\sin(\sqrt{n+1})-\sin(\sqrt n)=0.$$ I'm having trouble showing that $a_n=\sin(\sqrt n)$ diverges.

So I need help with determining if $$\lim_{n\to\infty} \sin(n+1)-\sin(n)=0,$$and also if $a_n=\sin(\sqrt n)$ diverges. Even if it's not true that $$\lim_{n\to\infty} \sin(n+1)-\sin(n)=0,$$ I would still like to know if the sequence $b_n=\sin(n+1)-\sin(n)$ diverges, and if not, what can be said about its limit.