# If $0<b_n < a_n$ almost for every $n$ then $\lim a_n =\infty$ (given $\lim (a_nb_n) = \infty$ )

Let $$\lim\limits_{n\to\infty }a_n \cdot b_n = \infty$$ , If $$0 almost for every $$n$$ then $$\lim\limits_{n\to\infty }a_n = \infty$$

This is the exact same question here , according to the question asked I know that the statement is true.

This is what I tried and got stuck:

according to the definition of infinite limits , A sequence $$(C_n)$$ tends to infinity if for every $$M \in \Bbb R$$ there exists an $$N \in \Bbb N$$ such that for every $$n \geq N$$ we get $$C_n > M$$.

so according to the given information $$\lim\limits_{n\to\infty }a_n \cdot b_n = \infty$$ there exists an $$N \in \Bbb N$$ such that for every $$n > N$$ we get $$a_n \cdot b_n >M$$

and also from the information $$0 , there exists a $$K \in \Bbb N$$ such that for every $$n > K$$ we get $$0 < b_n < a_n$$ , if we multiply by $$a_n$$ we get $$0

let $$B=max({K,N})$$ then for every $$n>B$$ we get $$M therefore $$M< a_n^2$$

I got stuck here because I cannot get to $$M < a_n$$ from here but I feel like what I did is correct in some way as it only relies on definitions and given information

Thanks for any tips and help!

Trick: Start with $$M^{2}$$, as it is also a positive number, then proceed your argument you will have $$M^{2} and so $$M.
• Actually you can stop at $\sqrt{M}<a_{n}$ and since $\sqrt{M}\rightarrow M$ is one-to-one and onto map on $(0,\infty)$, $\sqrt{M}$ represents every positive real as $M>0$ is varying, but then this may not sound as a precise mathematical reasoning though. Nov 20, 2021 at 13:30