Prove that $\frac{\sin n}{n}$ is a Cauchy sequence from the definition.

Question

Prove that $\frac{\sin n}{n}$ is a Cauchy sequence from the definition. The following is what I have tried:

Suppose $n>m$ , $$|s_n-s_m|=\frac{\sin n}{n}-\frac{\sin m}{m}<\frac{1}{n}-\frac{1}{m}$$

So here, what $N$ should I choose such that given any $\epsilon>0,n,m\geq N\implies|s_n-s_m|<\epsilon$?

Remark: The sequence I meant here is $$\sin1/1,\sin2/2,\ldots,\sin n/n$$

Answer 1

Let $m>n$. Then:

$$\left| \frac{\sin(n)}{n}-\frac{\sin(m)}{m}\right| \leq\left| \frac{1}{n}-\frac{-1}{m}\right| \leq \frac1n+\frac1m\leq\space\frac2n < \epsilon $$

Thus we can choose $$N(\epsilon)=\left[\frac{2}{\epsilon}\right]$$

Answer 2

Hint: $\lim\limits_{n\to+\infty}\dfrac{m}{n+1}=0$

Answer 3

Since $\sin n$ is bounded, $\frac{\sin n}n$ converges to $0$. Hence, $\frac{\sin n}n$ is a convergent, hence Cauchy sequence.

Answer 4

Let an $\epsilon>0$ be given. There is an $n_0>{2\over\epsilon}$. It follows that $$\left|{\sin n\over n}-{\sin m\over m}\right|\leq{1\over n}+{1\over m}<{\epsilon\over2}+{\epsilon\over2}=\epsilon\qquad \forall\>n, \ m>n_0\ .$$