Characterizing the asymptotic properties of $f(k)>\frac{ak^2}{k-1}$

Context: Let $a>0$ be some given constant. Let $f:\{2,3,\text{...}\}\to\mathbb{R}_+$ be some increasing function. Consider the following inequality:

$$\qquad f(k)> a\frac{k^2}{k-1}. \tag{$*$} $$

I am interested in characterizing the necessary and sufficient conditions on $f$ under which inequality $(*)$ fails to hold beyond a finite value of $x$. I think I was able to show the following:

$$ \exists K \text{ s.t. } (*) \text{ fails to hold } \forall k\geq K \Leftrightarrow \limsup_{k\to\infty} \frac{f(k)}{ak^2/(k-1)}\leq 1. $$

My question has two parts:

1. I would like to double check my proof (I am quite rusty when it comes to $\sup$ and $\limsup$.)
2. What I have proven seems to be obvious and perhaps uninformative. Since $\frac{k^2}{k-1}\sim k+1$, my intuition tells me that $(*)$ holds for arbitrarily large $k$ if and only if $f$ grows faster than linear in some sense (e.g. using one of the Bachmann–Landau notations like big O, little o, etc.). I have tried proving something of this form, but I can't get an if and only if statement to work (only "$\Rightarrow$" or "$\Leftarrow$", but not both). Any guidance towards proving a "tighter"/more informative characterization would be appreciated.

My proof:I first prove the "$\Rightarrow$" part.Suppose that $\exists K\geq 2$ s.t. $f(k)\leq a k^2/(k-1)\ \forall k\geq K.$ It then follows that $\sup_{k\geq K} \frac{f(k)}{a k^2/(k-1)}\leq 1$ by the definition of supremum and because $\frac{f(k)}{a k^2/(k-1)}\leq 1 \Leftrightarrow f(k)\leq ak^2/(k-1)$. It then follows that $$\limsup_{k \to \infty} \frac{f(k)}{a k^2/(k-1)}\leq 1$$ because for any arbitrary sequence $\{y_k\}_{k\in\mathbb{N}}$, $\sup_{k\geq K'+1} (y_{k})\leq \sup_{k\geq K'} (y_{k})$ holds for any $K'\in\mathbb{N}$. This proves the "$\Rightarrow$" part.

I now prove the "$\Leftarrow$" part. Suppose that $\limsup_{k\to\infty} \frac{f(k)}{ak^2/(k-1)}\leq 1.$ It then immediately$^{\color{red}{\text{(right?)}}}$ follows that $\exists K\geq 2$ s.t. $\sup_{k\geq K} \frac{f(k)}{ak^2/(k-1)}\leq 1.$ It then immediately$^{\color{red}{\text{(right?)}}}$ follows that $\frac{f(k)}{ak^2/(k-1)}\leq 1$ holds $\forall k\geq K$. (Do either of my two preceding statements require proof?) Therefore $(*)$ fails to hold $\forall k\geq K$ for some $K\geq 2$. This completes the proof. $\blacksquare$

My thanks: Thank you very much in advance to anyone who considers my question. I sincerely appreciate your help.