Finding the supremum of the set $A = \left\{\sin({n \pi}/{2})-\frac{1}{\sqrt{n}}: n \in \mathbb{N}\right\}$.

Question

I have been asked to find the supremum of the set $\displaystyle A = \left\{\sin({n \pi}/{2})-\frac{1}{\sqrt{n}}: n \in \mathbb{N}\right\}$.

I've some ideas, but I can't complete a solution. I'd appreciate any in put regarding this problem.

Guessing the supremum: Since $|\sin(n\pi/2)| \le 1$, I've guessed that the $\sup(A) = 1$ by considering when $\sin(n\pi/2)$ is maximum and $\frac{1}{\sqrt{n}}$ is "minimum" (does the word minimum make sense here?).

Proving that $1$ is an upper bound: We have $\displaystyle \sin({n \pi}/{2})-\frac{1}{\sqrt{n}} \le 1-\frac{1}{\sqrt{n}} \le 1$ for all $n \in \mathbb{N}$.

Proving that it's the least upper bound: This is the part I couldn't do. I understand that we show that $\varepsilon < 1$ is not an upper bound by exhibitting $n$ such that the set has elements exceeding $\varepsilon$. However, the inequalities needed to solve this beat me. I've considered taking $\displaystyle n=\max\left\{\bigg\lceil{\frac{1}{(1+\varepsilon)^2}\bigg\rceil},5 \right\}+1$. But I don't think it works for all cases of $\epsilon$ and again the inequalities beat me.

I'd appreciate if someone could outline the reverse process for fixing such $n$. Thanks in advance.

Answer 1

For any $\epsilon>0$, choose an $n_{0}$ such that $\dfrac{1}{\sqrt{n}}<\epsilon$ for all $n\geq n_{0}$, then $4n_{0}+1$ is such that $\sin((4n_{0}+1)\pi/2)-\dfrac{1}{\sqrt{4n_{0}+1}}=\sin(\pi/2)-\dfrac{1}{\sqrt{4n_{0}+1}}=1-\dfrac{1}{\sqrt{4n_{0}+1}}>1-\epsilon$, we are done.

Answer 2

Let's consider the sequence $x_n:\mathbb{N}^{+}\to\mathbb{N}^{+} $ defined by $ x_n = 1+4n $.

Then: $$ \lim_{n\to+\infty} \sin\left ( x_n\frac{\pi}{2} \right ) - \frac{1}{\sqrt{x_n}} = 1 $$

As $|\sin(x)| \leq 1 $ for any $ x \in \mathbb{R} $, then you're sure that the supremum is efectibly 1, because $\sin $ can't be bigger than 1 and $|1/\sqrt{x}|$ can't be smaller than 0.