# If $(a_n) \rightarrow \infty$ and $(b_n)$ be s.t. inf{bn:n∈N}=L>0. Prove that $\lim_{n\rightarrow\infty}a_n \cdot b_n = \infty$

Let $$(a_n)$$ and $$(b_n)$$ be real sequences such that $$\lim_{n \rightarrow \infty} a_n = \infty$$ and $$\inf \{b_n \colon n \in \mathbb{N}\}=L>0$$. Show that $$\lim_{n \rightarrow \infty} a_nb_n=\infty$$.

I know by definition that for every $$K>0$$, we can find $$n_0 \in \mathbb{N}$$ such that if $$n\geq n_0$$ then $$a_n\geq K$$. And it is given $$\inf \{b_n \colon n \in \mathbb{N}\}=L>0$$, thus implying that $$(b_n)$$ is non-negative. How can I show the $$\lim_{n \rightarrow \infty} a_nb_n$$ diverges to infinity?

I thought of doing by trying to show that $$\forall M > 0$$, one can find a $$n_0$$ such that if $$n\geq n_0$$ then $$a_n b_n > M$$. This $$n_0$$ exists because $$b_n$$ is non-negative and $$a_n \rightarrow \infty$$. But how can I properly show this passage?

Thank you.

• $b_n$ being non-negative is a weaker condition than what you have, and actually isn't enough to reach your desired conclusion. Why don't you find $n_0$ so that $a_n > M/L$ for $a_n \gt n_0$ and see what you can do with that? Commented Sep 29, 2020 at 18:10
• $\forall M>0$ one can find a $n_0$ such that if $n\ge n_0$ then $a_n>M/L$, therefore $a_n b_n \ge a_n L > M$. Commented Sep 29, 2020 at 18:11
• BTW, in case I didn't emphasize it enough: Your title, as it currently reads, is false. Commented Sep 29, 2020 at 18:20
• Consider the (potential) counter-example $a_n = n$ and $b_n = \frac{1}{n^2}$. So "$b_n$ positive", although also true, isn't strong enough to reach your conclusion either. The condition you have is "$b_n$ is bounded away from zero", and some of the other answers/comments show how you can use that to reach your conclusion. Commented Sep 29, 2020 at 18:51
• (The $a_n$ and $b_n$ I gave would be a counterexample to your weaker condition, not to the original statement.) Commented Sep 29, 2020 at 18:55

Hint :

Write $$a_n b_n \geq a_n L$$

(is is true for all $$n$$ ?), and then, let $$n$$ tend to $$+\infty$$.

Since $$\inf\{b_n\}=L>0$$, for each $$n$$ we have $$b_n\ge L$$.

Now, let $$K>0$$ be arbitrary. Then there exists a $$N\in\mathbb N$$ such that for each $$n>N$$, $$a_n>K/L$$. Then $$a_nb_n\ge a_n\cdot L>K$$.

• I am incredibly curious to know why someone would vote this answer down. It is short, rigorous, and straight forward. Commented Sep 30, 2020 at 2:00

If $$a_n\rightarrow\infty$$, then $$a_n > 0$$ for all sufficiently large $$n$$. Then, $$a_nb_n > L\cdot a_n$$ for all sufficiently large $$n$$. The result follows.