Question

Using the definition of convergence, how do you prove that the sequence $\text{limit}\dfrac{\sin n}{n}=0$ converges from n=1 to infinity?

Answer 1

In this problem, we have to prove that the sequence $\text{limit}\dfrac{\sin n}{n}=0$ converges from n=1 to infinity. We know that a sequence of real numbers converges to a real number a, if for every positive number $\varepsilon$, there exist an $N\in N$ such that for all $n\ge N$, $\left| an-a \right| < \varepsilon$. We should know that the sequence converges if the limit as n approaches to infinity, however if the limit goes to infinity, the sequence diverges.

Complete step by step answer:
We should show that,
$\displaystyle \lim_{n \to \infty }\dfrac{\sin n}{n}=0$, converges.
We need to show that, for any positive $\varepsilon$, there is a number M, such that
If n is greater than M, then
$\Rightarrow \left| \dfrac{\sin n}{n} \right| < \varepsilon$
But we are given that $\varepsilon$ is greater than 0.
We can assume that, let M be an integer with,
M > \min \left\\{ 1,\dfrac{1}{\varepsilon } \right\\}
We should also note that, $\dfrac{1}{M} < \varepsilon$.
If $n > M$ then $\dfrac{1}{n} < \dfrac{1}{M}$
We can now write that,
$\Rightarrow \left| \dfrac{\sin n}{n}-0 \right|=\dfrac{\left| \sin n \right|}{n} < \dfrac{1}{n} < \dfrac{1}{M} < \varepsilon$
Therefore, we can say that, for $n > 1$, we have $\left| \sin n \right| < 1$, so $\left| \dfrac{\sin n}{n} \right| < \dfrac{1}{n}$ , the sequence converges.

Note: students should understand the concept of sequence and its convergence to prove these types of problems, we should always remember that, sequence converges if the limit as n approaches to infinity, however if the limit goes to infinity, the sequence diverges and a sequence of real numbers converges to a real number a, if for every positive number $\varepsilon$, there exist an $N\in N$such that for all $n\ge N$, $\left| an-a \right| < \varepsilon$.