Question

Question #a365c

Answer 1

Expand the sequence, going once around the unit circle. (That is, from `n=0` to `n=11`.)

This allows us to see the sequence as

`1, -1/2, (2+sqrt3)/2, 0, (2+sqrt3)/2, -1/2, 1, -3/2, (2-sqrt3)/2, -2, (2-sqrt3)/2, -3/2`

The sequence then repeats this pattern endlessly.

The limit points are `1, -1/2, (2+sqrt3)/2, 0, -3/2, (2-sqrt3)/2, -2`

The sequence is bounded above by every number greater than or equal to `(2+sqrt3)/2` so the supremum is `(2+sqrt3)/2` and the lim sup is `(2+sqrt3)/2`.

The sequence is bounded below by every number less than or equal to `-2` so the infimum is `-2` and the lim inf is `-2`.