Let $\{a_{n}\}_{n\in\mathbb{N}}$ be a sequence such that $\forall n \in \mathbb{N}, a_{n} > 0$. Let also $k_{1}, k_{2} \in \mathbb{R}$ such that $k_{1} > 0$ and $k_{2}\neq 0$. Assume $\sup\limits_{n\in\mathbb{N}} \bigg[k_{1}a_{n}^{2} - o(1)k_{2}a_{n} \bigg]<\infty$ exists. Show that $\sup\limits_{n\in\mathbb{N}}a_{n}<\infty$!
My attempt so far is to use contradiction by assuming $\sup\limits_{n\in\mathbb{N}}a_{n} = \infty$ so that I have $$k_{1}\sup\limits_{n\in\mathbb{N}}a_{n}^{2} - k_{2}\sup\limits_{n\in\mathbb{N}}o(1)a_{n}<\infty$$But I don't know how to proceed to obtain the contradiction.
Remark : $o(1) \to 0$ as $n\to \infty$.
Any help will be much appreciated! Thank you very much!
Prove by contradiction. We are geiven that $k_1a_n^{2}-o(1)k_2a_n \leq M$ for all $n$ (for some $M <\infty)$. Hence $a_n[k_1a_n-o(1)k_2] \leq M$. If $(a_n)$ is not bounded there is a subsequence which tends to $\infty$. Can you get a contradiction now?.