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If $a_n\to \infty$ and $b_n$ diverges then $a_nb_n$ diverges.

I got this statement and need to show whether it's true or false (by proving it or showing counterexample, respectively).

First of all I don't know for sure if $a_n\to\infty$ means the standard "$a_n\to +\infty$" or just "for any $c>0$ there exists $N$ such that $n\ge N \implies |a_n|>c$". I just try assuming both cases but I'm geting nowhere with any.

If $b_n\to \infty$ or $b_n\to +\infty$ or $b_n\to -\infty$ it's pretty direct in any of the cases. But I don't know how to deal with $b_n$ just assuming that it doesn't converge to any real $a$.

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    $\begingroup$ You are correct about $a_n \to \infty$, that's the definition. I'm glad you're happy if $b_n \to \pm\infty$. If $b_n$ oscillates, without tending to a limit, e.g $-1, 1,-1, 1, \dots$, observe that it is (at least eventually) bounded... $\endgroup$ – bounceback Sep 11 '19 at 6:00
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    $\begingroup$ As soon as $b_n$ is not converge to 0, $a_nb_n$ doesn't converge to any value $\endgroup$ – Zhaohui Du Sep 11 '19 at 6:01
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You don't have $\lim_{n\to\infty}b_n=0$. Therefore, there is some $\varepsilon>0$ such that $\lvert b_n\rvert\geqslant\varepsilon$ infinitely often. In other words, there is a subsequence $(b_{n_k})_{k\in\mathbb N}$ of $(b_n)_{n\in\mathbb N}$ such that$$(\forall k\in\mathbb N):\lvert b_{n_k}\rvert\geqslant\varepsilon.$$But then $(a_{n_k}b_{n_k})_{k\in\mathbb N}\rightarrow\infty$.

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Suppose that $(a_nb_n)$ is convergent. Then $(a_nb_n)$ is bounded, thus there is $c>0$ such that $|a_nb_n| \le c$ for all $n$. Since $a_n \to \infty$, we can assume that $a_n > 0$ for all $n$.

We then get

$$|b_n| = | \frac{a_nb_n}{a_n}| \le \frac{c}{a_n}$$

for all $n$. This shows that $b_n \to 0,$ a contradiction.

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A sequence is called divergent if it is not convergent.

If $(a_nb_n)$ converges then $b_n=\frac {a_n b_n} {a_n} \to 0$ contradicting the hypothesis. Hence $(a_nb_n)$ does not converge.

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  • $\begingroup$ Good morning sir, can we prove this without using contradictory methods? $\endgroup$ – MANI Sep 11 '19 at 6:11
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Suppose:

  • $a_n\to \infty$.$\\[4pt]$
  • $b_n$ diverges.$\\[4pt]$
  • $a_nb_n$ converges.

We can derive a contradiction as follows . . .

Since $a_nb_n$ converges, so does $|a_nb_n|$.

Let $c={\displaystyle{\lim_{n\to\infty}}}|a_nb_n|$.

Let $w > 0$.

Since $a_n\to \infty$ and $|a_nb_n|\to c$, there exists a positive integer $N_w$ such that

    $a_n > w$ and $\Bigl||a_nb_n| - c\Bigr| < 1$

for all $n > N_w$.

Then for $n > N_w$, we have \begin{align*} &\Bigl||a_nb_n| - c\Bigr| < 1\\[4pt] \implies\;&|a_nb_n| - c < 1\;\;\;\;\text{[since $c\ge 0$]}\\[4pt] \implies\;&|a_nb_n| < 1+c\\[4pt] \implies\;&w|b_n| < 1+c\\[4pt] \implies\;&|b_n| < \frac{1+c}{w}\\[4pt] \end{align*} hence, since $w > 0$ is arbitrary, it follows that $b_n \to 0$, contradiction.

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