How do I find supremum of this set?

Question

I had a test today and I had to find $\sup$ and $\inf$ of this set :$$ A=\{\sin(\frac{nπ}{3}+\frac{1}{n}), n \in N\} $$ I calculated the value of $\sin$ for 6 cases (n=6k, n=6k+1, n=6k+2,...) and then said that the minimum value found is the $\inf$ and the maximum value is $\sup$. Is this approach correct? If not, how should this exercise be solved?

Answer 1

Note that ${1 \over n} < {\pi \over 3}$ for all $n$.

Partition $[0, 2 \pi]$ into $6$ intervals $I_k = [0, {\pi \over 3}]+ \{k{\pi \over 3} \}$, $k=0,...,5$.

If we write $n = 6m+k+1$, with $k \in \{0,...,5\}$ and $m \in \{0,1,...\}$ we see that ${n \pi \over 3} + {1 \over n} = {(k+1) \pi \over 3} + {1 \over 6m+k+1} \in I_{k+1}$.

The $\min$ value of $\sin$ on $I_1$ is greater than or equal to the $\max$ value of $\sin$ on $I_0,I_2$, hence we need only consider $k=0$ in order to compute the $\sup$ in the question.

Since $\sin (x) = \cos(x- {\pi \over 2} ) = \cos(|x- {\pi \over 2}| )$, we see that for $x \in I_1$, $\sin$ is a decreasing function of $|x- {\pi \over 2}|$. Hence the $\sup$ will be attained by finding the $\inf_{m \ge 0} |{\pi \over 3} + {1 \over 6m+1}- {\pi \over 2}| $, and a little work shows that this is attained for $m=1$ (this can be shown using the fact that $\pi < {22 \over 7}$), hence $\sup A = \sin( { \pi \over 3} + {1 \over 7})$.

In a similar manner, we need only consider $k=3$ in order to compute the $ \min$. Analogous reasoning shows that $m=0$ attains the $\min$ and so $\inf A = \sin( { 4\pi \over 3} + {1 \over 4})$.