4
$\begingroup$

We need to proof that $b_n/a_n\rightarrow0$ for $a_n\rightarrow\pm\infty$ and $(b_n)$ is restricted. But I came to $|b_n| \cdot 1/|a_n| \lt \epsilon$ and now I'm really stuck.. Can someone help me?

  • 2
    hint: Let us suppose for the moment that $a_n\to+\infty$ Use the squeezing theorem. Since $(b_n)$ is bounded, there exists a costant $M>0$ such that $-M$n$ but then $-M/a_n$a_n\to-\infty$ there is a change in the sign of the inequalities, but not much more than that. Can you go on from here? – 2011-09-23

1 Answers 1

4

Let $\varepsilon > 0$. Since $(b_n)$ is bounded there exists a constant $M$ such that $|b_n| < M$ for all $n$. Moreover $|a_n| \to \infty$ hence for some $N$ we get $|a_n| > M/\varepsilon$ for $n > N$. It is equivalent to $|1/a_n| < \varepsilon/M$.

Finally we obtain $$\Bigg|\frac{b_n}{a_n}\Bigg| < \varepsilon$$ for $n > N$ which means that $b_n/a_n \to 0$.

  • 0
    Thanks, but why $|a_n|>M/\epsilon$? I thought $n$ depends on $\epsilon$, so when you take $n$ large enough, $\epsilon$ will become smaller and smaller?2011-09-23
  • 1
    @Kevin: the order is the following. 1. You fix $\varepsilon$, 2. you choose $N'(\varepsilon)$ large enough to ensure the inequality $|a_n|>M/\varepsilon$ for all $n>N'(\varepsilon)$. With increase of $n$, the value of $\varepsilon$ stays unchanged since you fixed it from the beginning. We are only interested that the inequality holds.2011-09-23
  • 0
    @xen: why do you need $N$? If $b$ is bounded, than there is such $M$ that $|b_n|$n$? You're correct, of course, but it may be just a bit confusing for a person who only started to learn limits. – 2011-09-23
  • 0
    @Gortaur: Oh, of course. Thanks.2011-09-23