# Using the definition of convergence, how do you prove that the sequence limit (sin n)/ (n) = 0 converges from n=1 to infinity?

Oct 30, 2015

Use the fact that, for $n > 1$, we have $\left\mid \sin \right\mid n < 1$, so $\left\mid \sin \frac{n}{n} \right\mid < \frac{1}{n}$

#### Explanation:

To show: ${\lim}_{n \rightarrow \infty} \sin \frac{n}{n} = 0$

We need to show that for any positive $\epsilon$, there is a number $M$, such that
if $n > M$, then $\left\mid \sin \frac{n}{n} \right\mid < \epsilon$

Given $\epsilon > 0$, Let $M$ be an integer with $M > \min \left\{1 , \frac{1}{\epsilon}\right\}$.
Note that $\frac{1}{M} < \epsilon$.

And if $n > M$, then $\frac{1}{n} < \frac{1}{M}$ and
$\left\mid \sin \frac{n}{n} - 0 \right\mid = \frac{\left\mid \sin n \right\mid}{n} < \frac{1}{n} < \frac{1}{M} < \epsilon$