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

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 $$a$$.

• 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... – bounceback Sep 11 '19 at 6:00
• As soon as $b_n$ is not converge to 0, $a_nb_n$ doesn't converge to any value – Zhaohui Du Sep 11 '19 at 6:01

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 $$(a_{n_k}b_{n_k})_{k\in\mathbb N}\rightarrow\infty$$.

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.

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.

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

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.