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Problem: Suppose that $a_n$, $b_n$ are sequences of positive numbers such that $\lim_{n \rightarrow \infty} a_n = \infty$ and $\lim_{n \rightarrow \infty} \frac {a_n}{b_n} = \alpha$ for some $\alpha \in \Bbb R$, show that $b_n \rightarrow \infty$.

Thoughts: It seems that I could start with a proof by contradiction of 3 different cases where $b_n$ converges to a real number, or that it converges to negative infinity, and finally that it diverges. This seems like the wrong approach. Any hints much appreciated.

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  • $\begingroup$ Have you tried the epsilon definition of the limits? $\endgroup$ Commented Jan 29, 2017 at 1:52
  • $\begingroup$ @Vik78 sorry my mistake edited. $\endgroup$ Commented Jan 29, 2017 at 2:00
  • $\begingroup$ Why did you accept a faulty answer? $\endgroup$
    – Did
    Commented Sep 11, 2018 at 7:47
  • $\begingroup$ @Did I was just going back to questions that I hadn't accepted and must have been careless. I rescinded the answer sorry about that. $\endgroup$ Commented Sep 11, 2018 at 15:01
  • $\begingroup$ Be aware that $a\in\mathbb{R_+}$ has to hold to deduct $b_n\to \infty$. Otherwise we can e.g. set $b_n = -a_n$. $\endgroup$
    – Sudix
    Commented Sep 11, 2018 at 21:19

1 Answer 1

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if $b_n $ is bounded, given any arbitrary large $\epsilon$ we can find an $N$ such that $\frac{a_n}{b_n} \ge \epsilon$ for some $n \ge N$. after ruling out $b_n$ being bounded, Only case remains is $b_n$ diverging. In this case we can always find a subsequnce of $b_n$ that is bounded, so $\frac{a_n}{b_n}$ contains a subsequence that is going to $\infty$, which is again a contradiction.

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