# What is the next step of proving that $\lim_{n\to\infty}s_nt_n=0$ if $s_n\to0$ and $t_n$ is bounded? [duplicate]

## Q

There are $t_n$ bounded sequence and $s_n$ tending to zero. Then, prove that $$\lim_{n\to\infty}s_nt_n=0$$

## A

1. Since $t_n$ is bounded, There exists $M>0$ such that $$\left| t_n \right| \le M$$ for all $n$.

2. Since $\displaystyle\lim_{n\to\infty}s_n=0$, for any $\varepsilon_1\gt0$, there exists $N_1$ such that $$\left| s_n \right| \lt \varepsilon_1$$ for $n \ge N_1$.

3. In order to prove this proposition, it is enough to find $N_2$ satisfying that,

for any $\varepsilon_2\gt0$, there exists $N_2$ such that $$\left| s_nt_n \right| \lt \varepsilon_2$$ for $n \ge N_2$

4. If $N_2=\text{_____}$, then $$\left| s_nt_n \right| \le \left| s_n \right| \cdot \left| t_n \right| \lt M \cdot \varepsilon_1 \le \text{_____} \le \varepsilon_2$$ for $n \ge N_2$

I proved until here. Can someone let me know what should be filled in the blank?

## marked as duplicate by Ross Millikan, user91500, user99914, choco_addicted, Claude LeiboviciMar 29 '16 at 5:54

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

## 2 Answers

The point is that in step $2$ you say "for any $\varepsilon_1$ there exists $N_1$ such that $\dots$". That means in $4$ you get to choose $\varepsilon_1$ and get a corresponding $N_1$. Can you find a useful value for $\varepsilon_1$?

• Can you explain a little more detail? I do not understand what you say. Sorry, for my poor English. While it is possible for me to understand the poof of book, I can not solve even this problems which is not difficult for many students. – Danny_Kim Mar 29 '16 at 1:31

Since we have $\mathopen|t_n|\mathclose \ \le M$ for all $n \in \mathbb{N}$ it is sufficient to stick with $N_1$. This is because a choice of $N_1$ gives us the fact that $\mathopen|s_n|\mathclose \ \lt \epsilon$ for any $\epsilon \gt 0$. Since this fact about $s_n$ is true for all $\epsilon \gt 0$, we may choose $\epsilon_1 = \frac{\epsilon}{M}$. Now the proof is easy to finish from what you have. For all $n \ge N_1$ we have $\mathopen|t_n|\mathclose \le M$ and $\mathopen|s_n|\mathclose \lt \frac{\epsilon}{M}$. Following what you already have $|s_nt_n| \le |s_n| \cdot |t_n| \lt M \cdot \frac{\epsilon}{M} = \epsilon$. The key to these types of problems is choosing an $\epsilon$ that fits into what you need for the problem.