Finding limit of $a_n = \frac{1}{n|\sin n|}$

Question

Let $a_n = \frac{1}{n|\sin n|}$. Find $\lim_{n\to \infty} a_n$ if it exists. It's known that $b_n = \frac{1}{n\sin n}$ is divergent but what about $a_n = \frac{1}{n|\sin n|}$? I tried to show that $a_n$ converges to $0$ using definition but I am stuck on $\frac{1}{|\sin n|} \lt \epsilon$. Also attempting to show $\sum_{n=1}^\infty a_n$ converges with different tests, like comparison test , was unsuccessful.

Answer 1

We will use the fact $\pi$ is an irrational number.

By Hurwitz's theorem, there are infinite many integer pairs $n,m$ such that

$$\left| \pi - \frac{n}{m} \right| < \frac{1}{\sqrt{5}m^2}$$ It is easy to see the theorem remains valid when we restrict $n, m$ to positive integers.

For such a pair of $n, m$, we have

$$\left|m \pi - n \right| < \frac{1}{\sqrt{5} m} \le \frac{1}{\sqrt{5}n}\left(\pi + \left|\frac{n}{m} - \pi\right|\right) < \frac{\pi + 1}{\sqrt{5} n}$$

This leads to

$$|\sin(n)| = |\sin(m\pi - n)| \le |m\pi - n| < \frac{\pi + 1}{\sqrt{5}n} \implies \frac{1}{n|\sin(n)|} > \frac{\sqrt{5}}{\pi + 1 }$$ Since there are infinitely many such $n$, we can deduce

$$\limsup_{n\to\infty} \frac{1}{n|\sin(n)|} \ge \frac{\sqrt{5}}{\pi + 1 }$$

On the other direction, since $\pi$ is irrational, $n$ will be equidistributed modulo $\pi$. This implies there are infinitely many $n$ such that $|\sin(n)| > \frac12$. For such $n$, we have

$$\frac{1}{n|\sin(n)|} < \frac{2}{n}$$ Since there are infinitely many such $n$, we have

$$\liminf_{n\to\infty} \frac{1}{n|\sin(n)|} \le \liminf_{n\to\infty}\frac{2}{n} = 0$$

Since the limsup and liminf of the sequence $\frac{1}{n|\sin(n)|}$ is different, the limit $\displaystyle\;\lim_{n\to\infty} \frac{1}{n|\sin(n)|}$ doesn't exist.

Answer 2

The limit does not exist in either case. Let us consider the continued fraction of $\pi$, $$ \pi = [3; 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, 14, \ldots]$$ which is infinite since $\pi\not\in\mathbb{Q}$. Let $\frac{p_n}{q_n}$ be its $n$-th convergent. By the properties of continued fractions $$ \left|q_n \pi - p_n\right|\leq \frac{1}{q_n} $$ and by Lagrange's theorem $$ \left|\sin(p_n)-\sin(\pi q_n)\right| \leq |p_n-\pi q_n|\leq \frac{1}{q_n}, $$ so $$ p_n |\sin(p_n)|\leq \frac{p_n}{q_n} $$ and the RHS is $\leq \pi$ if we assume that $n$ has the suitable parity. This implies $$ \limsup_{m\to +\infty} \frac{1}{m|\sin m|}\geq \frac{1}{\pi}. $$ On the other hand $m|\sin m|$ can be pretty large, too. Let us assume that $\frac{P_n}{Q_n}$ is a convergent of $\frac{\pi}{2}$ and $Q_n$ is odd.
$$ \left|\sin(P_n)-\sin\left(\frac{\pi}{2}Q_n\right)\right|\leq \frac{1}{Q_n} $$ implies that $|\sin(Q_n)|\geq 1-\frac{1}{Q_n}$, hence $Q_n|\sin(Q_n)|\geq Q_n-1$ and $$ \liminf_{m\to +\infty}\frac{1}{m|\sin m|}=0$$ contradicts the existence of the limit. The last line can be proved through probabilistic arguments, too: if we consider a "random integer" $M$, the probability that $|\sin M|\geq \frac{1}{\sqrt{2}}$ is $\geq\frac{1}{2}-\varepsilon$, hence $m|\sin m|$ can be as large as $\frac{m}{3}$ infinitely often.