<s>[INST] Convergence of $\sin{\pi\sqrt{n}}$ Revising for an exam:

Let $a_n = \sin{(\pi\sqrt{n})}.$ Show that:
(i) $a_{n+1} - a_{n} \rightarrow 0$
(ii) The sequence $(a_n)$ is bounded.
(iii) $(a_n)$ does not converge.

My attempt:
(i) ???
(ii) min($\sin(x)$) = -1, max($\sin{x}$) = 1, so $-1 \leq a_n \leq 1,  \forall n \in \mathbb{N}$. Thus 1 is an upper bound and -1 is a lower bound.
(iii) $a_n$ has a monotonic subsequence which converges to 1 by the Bolanzo-Weierstrass theorem. Note that the subsequence $a_{n^2}$ converges to 0. Since $0 \neq 1$, $a_n$ does not converge.
 [/INST] (i) Since $\left|\sin(x)\right|\le\left|x\right|$
$$
\begin{align}
\left|a_{n+1}-a_n\right|
&=\left|\sin\left(\pi\sqrt{n+1}\right)-\sin\left(\pi\sqrt{n}\right)\right|\\[6pt]
&=2\left|\cos\left(\pi\frac{\sqrt{n+1}+\sqrt{n}}2\right)\sin\left(\pi\frac{\sqrt{n+1}-\sqrt{n}}2\right)\right|\\[3pt]
&\le2\cdot1\cdot\frac\pi2\left(\sqrt{n+1}-\sqrt{n}\right)\\[3pt]
&=\frac\pi{\sqrt{n+1}+\sqrt{n}}\\[3pt]
&\le\frac\pi{2\sqrt{n}}\tag{1}
\end{align}
$$

(ii) Since $\left|\sin(x)\right|\le1$, we have
$$
\begin{align}
\left|a_n\right|
&=\left|\sin\left(\pi\sqrt{n}\right)\right|\\
&\le1\tag{2}
\end{align}
$$

(iii) The limit of the subsequence
$$
\begin{align}
\lim\limits_{n\to\infty}a_{n^2}
&=\lim\limits_{n\to\infty}\sin\left(\pi\sqrt{n^2}\right)\\[3pt]
&=\lim\limits_{n\to\infty}0\\[3pt]
&=0\tag{3}
\end{align}
$$
Since
$$
\begin{align}
\left|\,n+\tfrac12-\sqrt{n^2+n}\,\right|
&=\frac{\frac14}{n+\frac12+\sqrt{n^2+n}}\\
&\le\frac1{8n}\tag{4}
\end{align}
$$
and $\cos(x)\ge1-\frac12x^2$, we have that
$$
\begin{align}
\left|a_{n^2+n}\right|
&=\left|\sin\left(\pi\sqrt{n^2+n}\right)\right|\\
&\ge\left|\cos\left(\pi\left(n+\frac12-\sqrt{n^2+n}\right)\right)\sin\left(\pi\left(n+\frac12\right)\right)\right|\\
&-\left|\sin\left(\pi\left(n+\frac12-\sqrt{n^2+n}\right)\right)\cos\left(\pi\left(n+\frac12\right)\right)\right|\\
&=\left|\cos\left(\pi\left(n+\frac12-\sqrt{n^2+n}\right)\right)\right|\\
&\ge1-\frac12\frac{\pi^2}{64n^2}\\[6pt]
&\gt0.9\tag{5}
\end{align}
$$
for $n\ge1$.
If the sequence converged, then the limit must be the limit of the subsequence computed in $(3)$. However, $(5)$ precludes the limit from being $0$.
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