Does $\sum_{n\ge0} \sin (\pi \sqrt{n^2+n+1}) $ converge/diverge? How would you prove convergence/divergence of the following series?
$$\sum_{n\ge0} \sin (\pi \sqrt{n^2+n+1}) $$
I'm interested in more ways of proving convergence/divergence for this series. 
My thoughts
Let
$$u_{n}= \sin (\pi \sqrt{n^2+n+1})$$
trying to bound $$|u_n|\leq |\sin(\pi(n+1) )| $$ since $n^2+n+1\leq n^2+2n+1$ and $\sin$ is decreasing in $(0,\dfrac{\pi}{2} )$
$$\sum_{n\ge0}|u_n|\leq \sum_{n\ge0}|\sin(\pi(n+1) )|$$
or $|\sin(\pi(n+1) )|=0\quad  \forall n\in \mathbb{N}$ then $\sum_{n\ge0}|\sin(\pi(n+1) )|=0$
thus $\sum_{n\ge0} u_n$ is converge absolutely then is converget
any help would be appreciated
 A: Hint: 
A necessary condition for the convergence of an infinite series $\sum\limits_{n=0}^\infty a_n$ is that the limit $\lim\limits_{n \to \infty} a_n$ should exist and be equal $0$.
So for your series investigate the following limit :
$$
\lim_{n\to\infty}\sin\left(\pi\sqrt{n^2+n+1}\right).
$$ 
A: $$\begin{align}\lim_{n\to\infty} \sin(\pi\sqrt{n^2+n+1}) &= \lim_{n\to\infty} \sin\Bigg(\pi \  \sqrt{(n+\frac{1}{2})^2 - \frac{1}{4} + 1}\Bigg) \\&= \lim_{n\to\infty} \sin\Bigg(\pi \  \sqrt{(n+\frac{1}{2})^2 + \frac{3}{4}\Big)}\Bigg) \\&= \lim_{n\to\infty} \sin\Bigg(\pi \  \Big(n+\frac{1}{2}\Big)\sqrt{1 + \frac{3}{4(n+\frac{1}{2})^2}}\Bigg)\end{align}$$ 
then this limit changes between $1$ and $-1$.
Thus you may conclude that the series diverges. 
A: $$\lim_{n\rightarrow\infty}{\sin\left(\pi\cdot\sqrt{n^2+n+1}\right)}=\sin{\lim_{n\rightarrow\infty}\left(\pi\cdot\sqrt{n^2+n+1}\right)}=$$
$$=\sin\left(\pi\cdot\lim_{n\rightarrow\infty}\left(\sqrt{n^2+n+1}-(n+1)+(n+1)\right)\right)=$$
$$=\sin\left(\pi\cdot\left(\lim_{n\rightarrow\infty}\left(\sqrt{n^2+n+1}-(n+1)\right)+\lim_{n\rightarrow\infty}\left(n+1\right)\right)\right)=$$
$$=\sin\left(\pi\cdot\lim_{n\rightarrow\infty}\left(-\frac12+(n+1)\right)\right)=$$
$$=\lim_{n\rightarrow\infty}\sin\left(\pi n+\frac{\pi}{2}\right)=\lim_{n\rightarrow\infty}(-1)^{n}$$
