Least upper bound for sequence $s_n= 2^{(-1)^n}\left (1- \frac{1}{n}\right) \sin\frac{nπ}{2}$

Question

Let $\{s_n\}$ be a sequence of reals such that $$s_n= 2^{(-1)^{n}}\left (1- \frac{1}{n}\right) \sin\frac{nπ}{2}.$$ Then find the least upper bound of $s_n$?

Here, $s_n$ will be 0 whenever $n$ is even and when n is odd, then $s_n=\frac{1}{2} \left (1- \frac{1}{n}\right) \sin\frac{nπ}{2}$, also since $|\sin x| \leq 1$ for all $x \in \mathrm{R}$, therefore $$ - \frac{1}{2} \left (1- \frac{1}{n}\right) \leq s_n \leq \frac{1}{2} \left (1- \frac{1}{n}\right) $$ So supremum should be $\frac{1}{2}.$ Is my argument correct?

Answer 1

Your argument is not complete. To say that the supremum is $\frac 1 2$ you also have to show that $s_n$ attains values arbitaririly close to $\frac 1 2$. One way of doing this is to observe that $s_{4m+1} \to \frac 1 2$.

Answer 2

Hint for another approach: examine $n\equiv ?\pmod{4}$ and find $\displaystyle\lim\sup s_n$