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I must proof the following, with $a: \Bbb{N} \to \Bbb{R}$ and $b: \Bbb{N} \to \Bbb{R}$

If $a \to -\infty\ (n\to\infty)$ and $b$ is bounded from below by a constant $k\in\mathbb R^{>0}$, then the $a\cdot b \to -\infty\ (n\to\infty)$

I thinked: If $a \to -\infty\ (n\to\infty)$ then $\forall\ M\in\mathbb R( \exists t\in \Bbb{N}( \forall n \ge t( a_n < -M)))$, therefore also $a_n < -M/k$ and $a_n \cdot k <-M$... but I do not know continue, How can I do?

Thanks in advance!

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All you have to add is that for $n$ large enough, $a_n<0$ and hence $a_n b_n\leq a_n k$ because $b_n\geq k$. –  Michael Greinecker Jan 28 at 0:02
    
@MichaelGreinecker, or Can I study $a_n<-\frac{|M|}{k}$ ?? –  mle Jan 28 at 0:08
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Ultimately, you want to show that for each $N$, $a_n b_n<-N$ for $n$ large enough. Keep the goal in mind. –  Michael Greinecker Jan 28 at 0:15
    
@MichaelGreinecker, I understand.. thanks soo much! :) –  mle Feb 8 at 23:34

1 Answer 1

This question has been answered in comments:

All you have to add is that for $n$ large enough, $a_n<0$ and hence $a_nb_n\leq a_nk$ because $b_n\geq k$.

Ultimately, you want to show that for each $N$, $a_nb_n<−N$ for $n$ large enough. Keep the goal in mind. – Michael Greinecker♦ Jan 28 at 0:15

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