Given $\limsup \frac{a_n}{b_n} < \infty$. Prove there is a constant M such that $a_n \leq Mb_n$

Question

Given sequences $\{a_n\}$ , $\{b_n\}$ of positive real numbers and $\textrm{ limsup } \frac{a_n}{b_n} < \infty$. Prove there is a constant M such that $a_n \leq Mb_n$

Defn: Let $(a_{n})_{n=1}^{\infty}$ be a bounded sequence. Define the sequence $c_n = \sup \{a_{k}: k \geq n\}$ for $n \geq 1$. If the sequence $c_n$ converges, then the value it converges to is the limit superior of $(a_n)$.

Attempt:

So given that the lim sup is finite, it means there exists a value M such that for all $n > k$, $$\frac{a_n}{b_n} < M \\ \Rightarrow \ a_n \leq Mb_n$$

Comment: Surely there is more to it than this and I have missed something.....

Answer 1

It's not quite as simple as you make it in your proof because the sequence may approach the $\lim \sup$ from above. The following proof works, though:

Let $M_1= \lim \sup \frac{a_n}{b_n}$. Then $\exists N~ n \gt N \Rightarrow \frac{a_n}{b_n} \lt M_1+1.$ Let $M_2 = \max \{ \frac{a_n}{b_n} \vert ~n \leq N \}.$ Let $M = \max \{M_1+1, M_2+1 \}$. Then $\forall n \in \Bbb N~\frac{a_n}{b_n} \lt M \text{ so } a_n \lt Mb_n.$