Confirming that the sequence $a_n = \frac{\sqrt{n}\cos{n}}{\sqrt{n^3}-1}$ converges I'm still trying to totally grasp the differences in proofs for sequences and series. I have a sequence $a_n = \frac{\sqrt{n}\cos{n}}{\sqrt{n^3}-1}$
In order to prove that this sequence converges, would it be correct to state that since:
$ -1 \le \cos{n} \le 1$ ,
and for $n \to \infty$ , $\sqrt{n^3}-1 \gt \sqrt{n}$
The denominator will be increasingly larger than the numerator therefore the sequence converges?

To make it more correct, see that
$$0\le|a_n|=\left|\frac{\sqrt n\cos(n)}{\sqrt{n^3}-1}\right|\le\left|\frac{\sqrt n}{\sqrt{n^3}-1}\right|=\left|\frac{n^{0.5}}{n^{1.5}-1}\right|$$
Following your logic, it should be a basic limit concerning rational functions.
You could also go ahead with L'Hospital's rule and squeeze theorem to see that
$$0\le\lim_{n\to\infty}|a_n|\le\lim_{n\to\infty}\left|\frac{n^{0.5}}{n^{1.5}-1}\right|\stackrel{\text{L'Hospital's}}=\lim_{n\to\infty}\left|\frac1{3n}\right|\to0$$
