I wish to find the infimum of the set $A=\left\{\cos(\pi \sqrt{n}/2)+2/n\right\}$.
I've an attempt below, of which I'm not too confident. I'd appreciate if someone could check it or show me how to do it. Proving there's no greater lower bound (and conversely no lesser upper bound) has been a struggle for me for some reason.
$|\cos(\pi \sqrt{n}/2)| \le 1$ we have $\displaystyle \cos(\pi \sqrt{n}/2)+\dfrac{2}{n} \ge -1+\frac{1}{n} \ge -1. $
So $-1$ is a lower bound for $A$ . To show that it's the greatest lower bound, we need to prove that $\epsilon-1$ is not a lower bound for any $\epsilon > 0$. I've tried what follows, but I'm not sure if it's correct:
Take $\displaystyle k = \lceil \frac{2}{\epsilon-1} \rceil+1$ then for all $n \ge k$, the number $n_0 = 4(2k+1)^2$
$\displaystyle \cos\left(\frac{\pi }{2} \sqrt{n_0}\right)+\frac{2}{n_0} = \cos\left((2k+1)\pi\right)+\frac{2}{4(2k+1)^2} = -1+\frac{2}{4(2k+1)^2} < \frac{2}{4(2k+1)^2} < \frac{2}{k} < \epsilon-1.$
Thus $\epsilon -1$ is not a lower bound of $A$. Hence $\inf(A) = -1$.
Alternative way:
Let $n = (2(2k+1))^2$ where $k \in \mathbb{Z}$,
then $\cos \left( \frac{\pi \sqrt{n}}{2}\right)+\frac{2}{n}= -1+\frac{2}{(2(2k+1))^2}$
$$\lim_{k \to \infty}-1+\frac{2}{(2(2k+1))^2}=-1$$
I have illustrated a subsequence that gets arbitrary close to $-1$, and $-1$ is the infimum.
Remark:
$$1 + \frac{2}{16k^2} < \frac{2}{16k^2}$$ is a false statement.