# Proving that the limit of the product of these two sequences is zero

I'm trying to prove this proposition:

Theorem: Suppose $$(a_n)$$ is a sequence that converges to $$0$$, and that $$(b_n)$$ is a bounded sequence. Then the sequence $$(a_n \cdot b_n)$$ is convergent and $$\lim_{n \to \infty} a_n \cdot b_n = 0.$$

Attempt at proof: Because the sequence $$(b_n)$$ is bounded, there exists an $$M > 0$$ such that $$|b_n| \leq M$$ for every $$n \in \mathbb{N}$$. Then we have $$0 \leq | a_n \cdot b_n | \leq M |a_n |.$$ Because $$(a_n)$$ converges to zero, there exists a $$n_0 \in \mathbb{N}$$ such that $$|a_n| < \epsilon$$ for all $$n \geq n_0$$. Since $$M > 0$$ we thus also have $$0 \leq | a_n \cdot b_n | \leq M |a_n | < M \epsilon.$$

Now I'm stuck and don't know how to proceed. Any help would be appreciated.

• You are correct. – sinbadh Jan 20 '16 at 19:07
• Just swap $\epsilon$ for $\frac{\epsilon}{M}$ – Anthony Peter Jan 20 '16 at 19:13
• Let $n>n_0$ imply that $|a_n|<\frac{\epsilon}{M}$. Then, for all $\epsilon>0$, $|a_nb_n|\le M|a_n|<M\frac{\epsilon}{M}=\epsilon$ – Mark Viola Jan 20 '16 at 19:13

You are done. A way to see it better: for any $\varepsilon$ you can find $n_0$ such that $|a_n|<\frac{\varepsilon}{M}$ for $n > n_0$ and so $$0 \leq |a_n \cdot b_n| \leq M |a_n| < M \frac{\varepsilon}{M} = \varepsilon,$$ so $a_n \cdot b_n \to 0$.