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If $\lim\limits_{n\rightarrow \infty}{a_n}=a<0$ and $\lim\limits_{n\rightarrow \infty}{b_n}=-\infty$, prove that $\lim\limits_{n\rightarrow \infty}{a_nb_n}=+\infty$.

I remember that if the two sequences $a_n$ and $b_n$ are converging and their limits are $a$ and $b$, then $\lim\limits_{n\rightarrow \infty}{a_nb_n} = \lim\limits_{n\rightarrow \infty}{a_n}\lim\limits_{n\rightarrow \infty}{b_n}=a.b$. However, $\lim\limits_{n\rightarrow \infty}{b_n}=-\infty$, not a number. Can I use this theorem or I have to prove the limit in another way?

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  • $\begingroup$ You have to prove that $$\forall M<0, \exists N\in\mathbb N: \forall n\in \mathbb N, n\geq N\implies a_nb_n<M.$$ $\endgroup$ – Surb Feb 14 at 14:15
  • $\begingroup$ You can't use that. What definition do you have for divergence to infinity? Likely something similar to what @Surb has written if I had to guess. $\endgroup$ – Cameron Williams Feb 14 at 14:15
  • $\begingroup$ @CameronWilliams The one Surb posted above. $\endgroup$ – Daniel Halachev Feb 14 at 14:16
  • $\begingroup$ Then that's exactly what you'll need to use. Is there a number directly related to $a$ that you can use to bound the tail of $a_n$ away from $0$? $\endgroup$ – Cameron Williams Feb 14 at 14:17
  • $\begingroup$ No, "a" is a parameter<0. That's all we have. I was think of defining a sequence $c_n=\frac{1}{b_n}$ and then find $\lim{\frac{a_n}{c_n}}$ $\endgroup$ – Daniel Halachev Feb 14 at 14:19
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In order to show that $\lim_{n \to \infty}a_nb_n = +\infty$ you have to show that for any number $M \in \mathbb{R}$ there exists $N \in \mathbb{N}$ such that $a_nb_n > M$ for all $n > N$. So let's start with an arbitrary $M \in \mathbb{R}$ and see if we can show this.

Since $a_n \to a < 0$, we can find $N_a \in \mathbb{N}$ such that $|a-a_n|< \epsilon$ for any $\epsilon > 0$. In a similar way, because $b_n \rightarrow -\infty$, we can find $N_b$ such that $b_n < M_b$ for all $n > N_b$ and any $M_b \in \mathbb{R}$.

Now we need to make some smart choices: For the $\epsilon$ mentioned before, let us take $\epsilon = -\frac{a}{2}$ (which is a positive number since $a<0$), meaning that for any $n > N_a$ we have that $a_n < \frac{a}{2}$. Also, take $N_b$ such that $b_n < \frac{2M}{a}$ for all $n > N_b$. Choose now $N = \max(N_a, N_b)$, so that for all $n > N$ we have:

$$a_nb_n > \frac{a}{2} \cdot \frac{2M}{a} = M \, , $$

which is what we wanted to show.

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Hint.

Let $M>0$. Observe that for sufficiently large $n$, $$ b_n<-M,\quad \frac{3a}{2}<a_n<\frac{a}{2}<0 $$ and thus $$ a_nb_n> -M \cdot \frac{a}{2}>0 $$

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Since you are asking this question, I think you only studied the mentioned theorem in the case of finite limits. If this is the case, you will have to do some work.

One way you can proceed is the following:

By the definition of limit with $\epsilon = \frac{a}{2},\; \exists N > 0,\; \forall n>N, \; a_n < \frac{a}{2}$. Here, I want to use that close to infinity, $a_n$ will be smaller than a fixed negative number, and $\frac{a}{2}$ will work since it is between $a$ and $0$. This is because I know that if $k<0$, $k*b_n$ will go to positive infinity "the same way" $b_n$ goes to negative infinity.

Then, you can use the definition on limit. Let $M>0$. By the definition of the limit of $b_n$, $\exists N'>0, \forall n>N, b_n < \frac{2M}{a}$.

Let $N'' = max(N, N')$ and $n>N''$. We have that $a_n<\frac{a}{2}$ and $b_n<\frac{2M}{a}$, so $a_n b_n > M$. This concludes the proof. Here, I had to use the definition of limit with $\frac{2M}{a}$ to compensate for the $\frac{a}{2}$. As I mentioned before, this is the only thing I had to do with $b_n$, because it $b_n$ and $k*b_n$ diverge pretty similarly.

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