# Prove that $a_n \times b_n \to 0$ for $n \to \infty$

I want to prove this example:

If $a_n \to 0$ for $n \to \infty$ and $(b_n)_n$ is bounded. Prove that $a_n \times b_n \to 0$ for $n \to \infty$.

My first guess is that I should use the definition of the boundedness and the convergence.

Therefore:

$|a_n| \leq M$ and $|a_n - a |< \epsilon$

My problem is, how to bring this two equations together to prove the theorem?

btw how to code in latex that the $n \to \infty$ is above the $\to$?

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Since $b_n$ are bounded, you have $|b_n| \le B$ for some $B$. Choose $\epsilon>0$ and find $N$ such that $|a_n| < \frac{1}{B} \epsilon$ for $n \ge N$. Then $|a_n b_n| \le B |a_n| < \epsilon$, as required. – copper.hat Aug 29 '13 at 5:04

Hint: Let $M\in \mathbb{R}$ fulfil $|b_n|\leq M$ for all $n$. Then $$|a_n\cdot b_n|\leq |M|\cdot |a_n|$$
Thx for your answer! Ok, then I have $|a_n\cdot b_n|\leq |M\cdot a_n|$ = $|b_n|\leq |M|$ which is true. Is this correct? – Le Chifre Aug 29 '13 at 5:17
@LeChifre no that doesn't help you at all. If you have $|a_n \cdot b_n| \leq M\cdot |a_n|$ take the limit on boths side – Dominic Michaelis Aug 29 '13 at 5:20
ok, so $\lim_{n\to \infty} |a_n \cdot b_n | \leq M \cdot \lim_{n\to \infty} |b_n|$. Therefore, I can argument that $\lim_{n\to \infty} |a_n \cdot b_n |$ goes to $\infty \leq M \cdot \infty$. Therefore, $0 \leq M$ and $a_n \cdot b_n \to 0$. – Le Chifre Aug 29 '13 at 5:34
@LeChifre no we want to use that $b_n$ is bounded – Dominic Michaelis Aug 29 '13 at 6:49