# Problem

Show that $\lim\limits_{n \to \infty}\sin n^2$ does not exist, where $n=1,2,\cdots.$

# Proof

Assume that $\lim\limits_{n \to \infty}\sin n^2$ exists, then since $\cos (2n^2)=1-2\sin^2 n^2,$ $\lim\limits_{n \to \infty}\cos(2n^2)$ also exists.

It's easy to know $\lim\limits_{n \to \infty}\sin n^2 \neq 0$, even though the limit exists. Hence, $$\lim\limits_{n \to \infty}\sin (2n)^2=\lim\limits_{n \to \infty}\sin n^2 \neq 0.\tag1$$ Since $$\sin(2n)^2=\sin 2(2n^2)=2\sin(2n^2)\cos(2n^2),\tag2$$ hence $\lim\limits_{n \to \infty}\cos (2n^2) \neq 0.$ Otherwise, if $\lim\limits_{n \to \infty}\cos (2n^2) =0,$ then according to $(2)$, we may obtain $\lim\limits_{n \to \infty}\sin(2n^2) = 0,$ which contradicts $(1)$.

Now, notice that $\sin (2n^2)=\dfrac{\sin(2n)^2}{2\cos(2n^2)}$, for the reason that $\lim\limits_{n \to \infty}\sin n^2 \neq 0$ and $\lim\limits_{n \to \infty}\cos (2n^2) \neq 0$, we may claim that $\lim\limits_{n \to \infty}\sin (2n^2)$ exists but dose not equal $0$.

Since $\sin (2n^2)=2\sin n^2\cos n^2$, hence $\cos n^2=\dfrac{\sin(2n^2)}{2\sin n^2}.$ But we have known that $\lim\limits_{n \to \infty}\sin (2n^2)$ and $\lim\limits_{n \to \infty}\sin n^2$ both exist and $\lim\limits_{n \to \infty}\sin n^2 \neq 0$. Thus, $\lim\limits_{n \to \infty} \cos n^2$ exists. Therefore, $\lim\limits_{n \to \infty} \sin (n+1)^2$ and $\lim\limits_{n \to \infty} \cos (n+1)^2$ both exist as well.

Since $\sin(2n+1)=\sin[(n+1)^2-n^2]=\sin(n+1)^2\cos n^2-\cos(n+1)^2\sin n^2$，hence $\lim\limits_{ n \to \infty}\sin(2n+1)$ exists, which is absurd.

As a result, the assumption at the beginning that $\lim\limits_{n \to \infty}\sin n^2$ exists is false. We are done.

# NOTE

Among the proof above， we apply two facts that $\lim\limits_{n \to \infty} \neq 0,$ and $\lim\limits_{n \to \infty}\sin (2n+1)$ does not exist, which has not been proven. If necessary, I'm pleased to complement the proofs of them.

# Lemma 1

$\lim\limits_{n \to \infty}\sin(2n+1)\neq 0.$

### Proof

Assume that $\lim\limits_{n \to \infty}\sin (2n+1)=0.$ Then $\lim\limits_{n \to \infty}|\sin (2n+1)|=0$ and $\lim\limits_{n \to \infty}|\cos (2n+1)|=1.$ Thus,

\begin{align*} \lim\limits_{n \to \infty}|\sin (2n+1)|&=\lim\limits_{n \to \infty}|\sin (2n+3)|\\ &\geq \lim\limits_{n \to \infty}|\cos (2n+1)\sin 2|- \lim\limits_{n \to \infty}|\sin (2n+1)\cos 2|\\ &= \lim\limits_{n \to \infty}|\cos(2n+1)\sin 2|\\ &=\sin2 \neq 0, \end{align*} which contradicts.

# Lemma 2

$\lim\limits_{n \to \infty}\sin (2n+1)$ does not exist.

### Proof

Assume that $\lim\limits_{n \to \infty}\sin (2n+1)$ exists. Take two arcs with the middle points $\dfrac{\pi}{2},\dfrac{3\pi}{2}$ respectively and the length $\dfrac{2\pi}{3}$ from the unit circle. Notice the common difference of $\{2n+1\}$ is $2$, which is smaller than $\dfrac{2\pi}{3}$. Hence, there are infinitely many terms locating on the arc with the middle point $\dfrac{\pi}{2}$ and the length $\dfrac{2\pi}{3}$. For such terms $\{2n_k+1\}$, we have $$\sin(2n_k+1) \geq \sin \left(\frac{\pi}{2}-\frac{\pi}{3}\right)=\sin\frac{\pi}{6}=\frac{1}{2}.$$ Therefore, $$\lim\limits_{n \to \infty}\sin (2n+1)=\lim\limits_{n \to \infty}\sin (2n_k+1)\geq \frac{1}{2}.\tag3$$ Similarily, there also exists another subsequence $\{2m_k+1\}$ locating on the arc with the middle point $\dfrac{3\pi}{2}$ and the length $\dfrac{2\pi}{3}$. Notice that $$\sin(2m_k+1) \leq \sin \left(\frac{3\pi}{2}+\frac{\pi}{3}\right)=\sin\frac{11\pi}{6}=-\frac{1}{2}.$$ Therefore, $$\lim\limits_{n \to \infty}\sin (2n+1)=\lim\limits_{n \to \infty}\sin (2m_k+1)\leq -\frac{1}{2}.\tag4$$ As a result, $(3)$ and $(4)$ contradict each other.

# Lemma 3

$\lim\limits_{n \to \infty} \sin n^2 \neq 0.$

### Proof

Assume that $\lim\limits_{n \to \infty} \sin n^2 = 0.$ Notice that $\sin(n\pi)=0$ for all $n=1,2,\cdots.$ and the continuity of $y=\sin x$. Thus, for any given $\varepsilon>0$, there exists a $N \in \mathbb{N_+}$, when $k \geq N$, we may choose a positive integer $n_k$ such that $$|k^2-n_k\pi|<\frac{\varepsilon}{2}.$$ Without loss of generality, we may assume $\{n_k\}$ is rigorously increasing. Then

$$|(2k+1)-(n_{k+1}-n_k)\pi|=|(k+1)^2-n_{k+1}\pi-(k^2-n_k\pi)|<\varepsilon.$$ Thus, $$\lim_{k \to \infty}\sin(2k+1)=\lim_{k \to \infty}\sin(n_{k+1}-n_k)\pi=0,$$ which contradicts Lemma 1.

PLEASE CORRECT ME IF I'M WRONG! HOPE TO SEE MORE ELEGANT SOLUTIONS!

• How is it easy to see that $\lim_{n\to\infty}\sin(n^2)\ne0?$ – saulspatz Jul 10 '18 at 17:03
• Caution: when I read $\sin(2n)^2$, I think of $\big(\sin(2n)\big)^2$ and not $\sin\big((2n)^2\big)$ as you meant... – charmd Jul 10 '18 at 17:03
• I (and I guess everyone) guess that here $n$ is an integer... but you should clarify that. – leonbloy Jul 10 '18 at 17:07
• Possibly helpful math.stackexchange.com/questions/2373322/sin-n2-diverges – kingW3 Jul 10 '18 at 17:27
• Why is it absurd that the limit of $\sin(2 n + 1)$ does not exist? (I mean I would prove that particular question in a similar way to what you need to prove.) – Stan Tendijck Jul 10 '18 at 17:43

Suppose $\lim_{n \to \infty} \sin(n^2)$ exists with value $L$.
$\sin((n+1)^2) =\sin(n^2+2n+1) =\sin(n^2)\cos(2n+1)+\cos(n^2)\sin(2n+1)$ so $L \approx L\cos(2n+1)\pm\sqrt{1-L^2}\sin(2n+1)$ or $L(1- \cos(2n+1))\approx\pm\sqrt{1-L^2}\sin(2n+1)$ or $L(2\sin^2(n+1/2)) \approx\pm\sqrt{1-L^2}\sin(2n+1) =\pm\sqrt{1-L^2}2\sin(n+1/2)\cos(n+1/2)$ or $L\sin(n+1/2) \approx\pm\sqrt{1-L^2}\cos(n+1/2)$.
$L = 0$ implies that $\cos(n+1/2) \to 0$, which is false. Therefore $L \ne 0$.
Therefore $\tan(n+1/2) \approx \dfrac{\pm\sqrt{1-L^2}}{L}$. This means that $\tan(n+1/2)$ can have at most two accumulation points.
Since $\pi$ is irrational, $\{(n+1/2)/\pi\}$ (fractional part) is dense in $[0, 1]$, so $\tan(n+1/2) =\tan(\pi\{\frac{n+1/2}{\pi}\})$ has many accumulation points.