# Knowing that $\lim_{n\to \infty} a_n = 0$ and that $b_n$ is bounded, find $\lim_{n\to \infty} (a_n b_n)$

Knowing that $\lim_{n\to \infty} a_n = 0$ and that $b_n$ is bounded, find $\lim_{n\to \infty} (a_n b_n)$

We know that $$-M \le b_n \le M$$ Now,I don't know if this step is correct at all, but I would like t simply divide both sides of this equation by n, and therefore get: $$-\frac{M}{n} \le \frac{b_n}{n} \le \frac{M}{n}$$ Now, I take the limit as n approaches infinity:
$$0 \le \lim_{n\to \infty} \frac{b_n}{n} \le 0$$ Therefore, $\frac{b_n}{n} \rightarrow 0$.
Now, as we deal with arbitrarily large n-s, I think that any sequence that approaches zero will have the same effect, therefore, I'd like to substitute $1/n$ with $a_n$ and finally get
$$a_n b_n \rightarrow 0$$
I am however afraid that my reasoning is vague or bogus - I'd really appreciate any hints and / or constructive comments.

• Describe what is your $M$, since $b_n$ is bounded, there exists $M>0$, such that $|b_n|\leq M$, and hence $-M \leq b_n \leq M$.

• Rather than multiplying by $\frac1n$, directly multiply by $a_n$,

$$-Ma_n \leq a_nb_n \leq Ma_n$$

Now take limit, by squeeze theorem.

$$\lim_{n \to \infty}-Ma_n \leq \lim_{n \to \infty}a_nb_n \leq \lim_{n \to \infty}Ma_n$$

$$-M\lim_{n \to \infty}a_n \leq \lim_{n \to \infty}a_nb_n \leq M\lim_{n \to \infty}a_n$$

$$0 \leq \lim_{n \to \infty}a_nb_n \leq 0$$

hence the conclusion.

• While working with $\frac1n$ might give you an idea of what is going on. We have to justify more why it works for other sequence that converges to zero. For example knowing that $\sum_{i=1}^\infty \frac1{i^2} < \infty$ doesn't mean the same rule apply for $\sum_{i=1}^\infty \frac1{i}$

Note that there exists $M>0$ such that $$|a_nb_n|\leq M|a_n|\to0$$ since $a_n\to 0$. The result follows.

What you do in your second inequality is multiplying by the sequence $\frac{1}{n}$. Try multiplying by $a_n$...