Question

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

Answer 1

Use the fact that, for `n > 1`, we have `abs sinn < 1`, so `abs (sinn/n) < 1/n`

Explanation:

To show: `lim_(nrarroo)sin n/n = 0`

We need to show that for any positive `epsilon`, there is a number `M`, such that
if `n > M`, then `abs(sin n /n)< epsilon`

Given `epsilon > 0`, Let `M` be an integer with `M > min{1, 1/epsilon}`.
Note that `1/M < epsilon`.

And if `n > M`, then `1/n < 1/M` and
`abs(sin n / n -0) = abs (sin n)/n < 1/n < 1/M < epsilon`