what will be the supremum of $a_n $ ?

Question

For $n \ge 1 $ let $ a_n=(1-\frac{1}{n})\sin (\frac{n\pi}{3})$,

what will be the supremum of $a_n $ ?

My attempt : i gots $\frac{\sqrt 3}{2}$ because $\sin\pi/ 3=\frac{\sqrt 3}{2}$

Is its correct ???

Answer 1

Yes, it is correct. Note that $(\forall n\in\mathbb N):\lvert a_n\rvert\leqslant\frac{\sqrt3}2$, which proves that $\sup_na_n\leqslant\frac{\sqrt3}2$. Now, use the fact that $\lim_{n\in\mathbb N}a_{6n+1}=\frac{\sqrt3}2$.

Answer 2

Correct result, incorrect proof. For large enough $n$ the $(1-1/n)$ term gets arbitrarily close to 1, and the second only takes finitely many values of which the max is $\sqrt{3}/2$. So the lowest upper bound is $\sqrt{3}/2$. Note that this is not the maximum, as this value is not attained.