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

Hint: 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 \[n\ge N\], \[\left| an-a \right| < \varepsilon \].