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.
Best Answer
Hint: (for (i)) $$ \sin a - \sin b = 2\sin\frac{a-b}{2}\cos\frac{a+b}{2} $$ and the product of two sequences, one converging to $0$ and the other bounded, converges to $0$.
In more detail: $$ a_{n+1}-a_n = 2\sin\left(\pi\frac{\sqrt{n+1}-\sqrt{n}}{2}\right)\cos\left(\pi\frac{\sqrt{n+1}+\sqrt{n}}{2}\right) $$ The second factor is bounded as $\cos$ is, and the first goes to $0$ as
$\sqrt{n+1}-\sqrt{n} = \sqrt{n}\left(\sqrt{1+\frac{1}{n}}-1\right) = \frac{1}{2\sqrt{n}}+o\left(\frac{1}{\sqrt{n}}\right)\xrightarrow[n\to\infty]{}0$.
$\sin u\xrightarrow[u\to 0]{}0$
Edit: For part (iii), that I hadn't realized was "still open" as well.
Suppose by contradiction $a_n\to\ell\in\mathbb{R}$.
As you noticed by looking at the subsequence $(a_{n^2})_n$, we necessarily have $\ell=0$.
Now, this implies that $a^2_n \xrightarrow[n\to\infty]{} 0$ as well, and using $\cos^2+\sin^2=1$ we get $\cos(\pi\sqrt{n})^2 \xrightarrow[n\to\infty]{} 1$.
Suppose for now we have shown that $$\cos(\pi\sqrt{n}) \xrightarrow[n\to\infty]{} \beta\in\{-1,1\} \tag{$\dagger$}$$
From the above, we have $$ a_{n+1}-a_n = b_n\cos\left(\pi\frac{\sqrt{n+1}+\sqrt{n}}{2}\right) $$ where $b_n\operatorname*{\sim}_{n\to\infty}\frac{\pi}{\sqrt{n}}$. Let's deal with the other term: as $$ \pi\frac{\sqrt{n+1}+\sqrt{n}}{2} = \pi\sqrt{n}+\frac{\pi}{4\sqrt{n}} + o\left(\frac{1}{\sqrt{n}}\right) $$ we get $$\begin{align} \cos\left(\pi\frac{\sqrt{n+1}+\sqrt{n}}{2}\right) &= \cos\left(\pi\sqrt{n}+\frac{\pi}{4\sqrt{n}} + o\left(\frac{1}{\sqrt{n}}\right)\right)\\ &= \cos \pi\sqrt{n} \cos\left(\frac{\pi}{4\sqrt{n}} + o\left(\frac{1}{\sqrt{n}}\right)\right) - \sin \pi\sqrt{n} \sin\left(\frac{\pi}{4\sqrt{n}} + o\left(\frac{1}{\sqrt{n}}\right)\right)\\ &= \cos \pi\sqrt{n} \cos(o(1)) - \sin \pi\sqrt{n}\sin(o(1)) \\ &\xrightarrow[n\to\infty]{} \beta\cdot 1 - 0\cdot 0 = \beta. \end{align}$$ Putting it all together, this leads to $$ a_{n+1}-a_n \operatorname*{\sim}_{n\to\infty}\frac{\beta\pi}{\sqrt{n}} $$ which by comparison implies that the series $$\sum_{n=0}^{\infty} (a_{n+1}-a_n)$$ diverges to $\infty$ (or $-\infty$, depending on $\beta$). But this is a contradiction, since this is a telescoping series, equal (by assumption on $(a_n)_{n\in\mathbb{N}}$ converging) to $\ell - a_0 = 0$. $\square$
The remaining issue, of course, is that we don't actually have proven ($\dagger$). But it is enough for our purposes (handwaving a bit here, but it's not hard to make it formal) to have either: $$\cos(\pi\sqrt{n}) \xrightarrow[n\to\infty]{} \beta\in\{-1,1\}$$ or two sequances $(k_n)_n$, $(m_n)_n$ partitioning the natural numbers ($\mathbb{N} = \bigcup_n \{k_n\}\cup\{m_n\}$) such that $$\cos(\pi\sqrt{k_n}) \xrightarrow[n\to\infty]{} -1, \qquad \cos(\pi\sqrt{\ell_n}) \xrightarrow[n\to\infty]{} -1$$ which are the only two cases that can happen knowing that $\cos^2(\pi\sqrt{n}) \xrightarrow[n\to\infty]{}1$.
Indeed, the first case we took care of; and for the second case, we can now restrict the above argument to (that's the handwavy part) to either $(a_{k_{n+1}} - a_{k_n})_n$ or $(a_{m_{n+1}} - a_{m_n})_n$, knowing that at least one of the two the series $\sum_{n} \frac{1}{\sqrt{k_n}}$ and $\sum_{n} \frac{1}{\sqrt{m_n}}$ has to diverge.