$H_n$ represents the $n\text{th}$ harmonic number. I was messing around with Desmos when I happened to come across this. I typed it into WolframAlpha which confirmed that the limit is equal to $1$ but did not have a step-by-step solution available. I have not formally learned about hyperbolic functions or harmonic numbers. My only idea is to use l'Hopital's rule (I got the derivatives of each function from WolframAlpha):
$$\lim\limits_{n\to\infty}\frac{\cosh^{-1}n}{H_n}=1$$
$$\lim\limits_{n\to\infty}\frac{\frac{1}{\sqrt{n^2-1}}}{\frac{\pi^2}{6}-H^{(2)}_n}=1$$
$$\lim\limits_{n\to\infty}\frac{6}{(\pi^2-6H^{(2)}_n)\sqrt{n^2-1}}=1$$
L'hopital's rule would once again apply but I feel like it would not help.