Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.


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

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

I appreciate your answer!!!

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

share|cite|improve this question
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 $$|a_n\cdot b_n|\leq |M|\cdot |a_n|$$

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.