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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.

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  • $\begingroup$ $b_n$ 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? $\endgroup$
    – JonathanZ
    Commented Sep 29, 2020 at 18:10
  • $\begingroup$ $\forall M>0$ one can find a $n_0$ such that if $n\ge n_0$ then $a_n>M/L$, therefore $a_n b_n \ge a_n L > M$. $\endgroup$ Commented Sep 29, 2020 at 18:11
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    $\begingroup$ BTW, in case I didn't emphasize it enough: Your title, as it currently reads, is false. $\endgroup$
    – JonathanZ
    Commented Sep 29, 2020 at 18:20
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    $\begingroup$ 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. $\endgroup$
    – JonathanZ
    Commented Sep 29, 2020 at 18:51
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    $\begingroup$ (The $a_n$ and $b_n$ I gave would be a counterexample to your weaker condition, not to the original statement.) $\endgroup$
    – JonathanZ
    Commented Sep 29, 2020 at 18:55

3 Answers 3

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Hint :

Write $$a_n b_n \geq a_n L$$

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

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

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

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