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Problem: Given that $ \lim\limits_{n \to \infty} a_n = - \infty $ and $ \lim\limits_{n \to \infty} b_n = b <0 $, show $ a_n b_n \longrightarrow +\infty $
Note: there's an answer to the problem here Prove that $\lim\limits_{n\rightarrow \infty}{a_nb_n}=+\infty$ , but I tried proving the theorem and erroneously got that $ a_n b_n \longrightarrow -\infty $, I can't figure out where is the error in my attempt of proof, maybe you can help please?
My attempt: Suppose $ \lim\limits_{n \to \infty} a_n = - \infty $ and $ \lim\limits_{n \to \infty} b_n = b <0 $. Meaning,
For every $ \epsilon > 0$ exists $ N_1 \in \mathbb{N} $ s.t. $ |b_n - b| < \epsilon \iff b-\epsilon < b_n < b + \epsilon $.
For every $ M' > 0 $ exists $ N_2 \in \mathbb{N} $ s.t. $ a_n < -M' $.
Let $ M > 0 $. Choose $ \epsilon = -\frac{5b}{2} = \frac{5|b|}{2} $ ( since $ b<0 $ ). Choose $ M' = \frac{2M}{3|b|} $. Thus, there exists $ N_1, N_2 \in \mathbb{N} $, denote $ N = max\{N_1,N_2\} $ and we have for all $ n > N $, $ -\frac{7|b|}{2}<b_n < \frac{3|b|}{2} $ and $ a_n < -\frac{2M}{3|b|} $, Hence:
$ a_n b_n < a_n \cdot \frac{ 3|b|}{2} < (-\frac{2M}{3|b|}) \cdot (\frac{3|b|}{2} ) = -M $, Therefore since $ M>0 $ was arbitrary, we have $ a_n b_n \longrightarrow -\infty $

hazelnut_116
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    You could take a simple concrete example, e.g. $a_n=-n$ and $b_n=-1$, and check where your proof goes wrong. – Martin R Apr 02 '21 at 14:54

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The error lies in taking $\varepsilon=\frac{5b}2$. You cannot do that, since $\frac{5b}2<0$.

Taking $\varepsilon=-\frac b2>0$, you will easily prove that $\lim_{n\to\infty}a_nb_n=\infty$.

José Carlos Santos
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  • You are right, I mistakenly copied the $ \epsilon $ wrongly from a draft paper, can you please look again at the proof under the new $\epsilon $? – hazelnut_116 Apr 02 '21 at 14:53
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    It is still wrong: from $a_n<-\frac{2M}{3|b|}$ and $b_n<\frac{3|b|}2$, what you get is that $a_nb_n\color{red}>M$, since $a_n<0$. – José Carlos Santos Apr 02 '21 at 14:59
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    @JoséCarlosSantos In fact, $a_n b_n>M$ might not be right. Just like $(-2)\times 2<(-1)\times 3$. – Yan Peng Apr 02 '21 at 15:17
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    @YanPeng You are right. What I should have said was that from $a_n<-\frac{2M}{3|b|}$ and $b_n<\frac{3|b|}2$ it cannot be deduced that $a_nb_n<M$. – José Carlos Santos Apr 02 '21 at 15:19
  • @JoséCarlosSantos so I cannot a-priori multiply $ a_n $ with $ b_n $ and deduce facts about the inequality $ a_n b_n < ... $ ? , Must I reach some inequalities first involving $ a_n $ and $ b_n $ separately and only then reach an inequality involving their multiplication? ( for example : $ b_n<-M,\quad \frac{3a}{2}<a_n<\frac{a}{2}<0 $ , for $ a<0 $ , then the inequality for $ a_n b_n $ is $ a_nb_n> -M \cdot \frac{a}{2}>0 $) – hazelnut_116 Apr 03 '21 at 10:09
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    The problem is is that if you do not know the sign of $a_n$ or of $b_n$, then you do not know whether the equality will be a smaller than or a greater than. But after you know their signs, there is no problem. – José Carlos Santos Apr 03 '21 at 10:12