TITLE: Does $\sum_{n\ge0} \sin (\pi \sqrt{n^2+n+1}) $ converge/diverge?
QUESTION [6 upvotes]: 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

REPLY [3 votes]: $$\begin{align}\lim_{n\to\infty} \sin(\pi\sqrt{n^2+n+1}) &= \lim_{n\to\infty} \sin\Bigg(\pi \  \sqrt{(n+\frac{1}{2})^2 - \frac{1}{4} + 1}\Bigg) \\&= \lim_{n\to\infty} \sin\Bigg(\pi \  \sqrt{(n+\frac{1}{2})^2 + \frac{3}{4}\Big)}\Bigg) \\&= \lim_{n\to\infty} \sin\Bigg(\pi \  \Big(n+\frac{1}{2}\Big)\sqrt{1 + \frac{3}{4(n+\frac{1}{2})^2}}\Bigg)\end{align}$$ 
then this limit changes between $1$ and $-1$.
Thus you may conclude that the series diverges.