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

Question

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.

Answer 1

Hint :

Write $$a_n b_n \geq a_n L$$

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

Answer 2

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 $a_nb_n\ge a_n\cdot L>K$.

Answer 3

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.