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Let $(a_n)_{n\in\mathbb{N}}$, $(b_n)_{n\in\mathbb{N}}$ two sequences of integers.

Prove that if $u_n=\frac{a_n}{b_n}$ converges to an irrational $x$, then both $a_n$ and $b_n$ diverge to infinity.

I started with a proof by contradiction and proved the first case:

  • If $a_n$ and $b_n$ converge to $l$ and $l'$ respectively, then $x=\frac{l}{l'}$ which is a contraction.

Now, I have to prove it when one or both diverges.

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Math Buster
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    You're nearly there! If $a_n$ diverges and $b_n$ doesn't, what can we say about $ \frac{a_n}{b_n}$? Under what scenario can it converge? – Calvin Lin Dec 08 '19 at 21:48
  • Compare https://math.stackexchange.com/q/1465812/42969. – Martin R Dec 08 '19 at 21:52
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    If $a_n$ doesn't tend to $\infty$, then there is a subsequence $a_{n_k}$ that is bounded. A bounded sequence of integers takes some value infinitely many times. So, we can actually assume that the subsequence is constant. Looking now at the sequence $b_{n_k}$, if is doesn't tend to $\infty$ we can find, as before, a subsequence $b_{n_{m_k}}$ that is constant. Then $u_n=a_{n_{m_k}}/b_{n_{m_k}}$ is constant, and its limit is rational. – conditionalMethod Dec 08 '19 at 21:55
  • @conditionalMethod You should make that $\pm\infty$. – Rushabh Mehta Dec 08 '19 at 21:55
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    @DonThousand What makes you think that it is not already that? – conditionalMethod Dec 08 '19 at 21:56
  • @conditionalMethod What? How do you deal with the sequence $1,-1,2,-2,...$? Do you consider that to "tend towards $\infty$"? That isn't how it is conventionally defined – Rushabh Mehta Dec 08 '19 at 21:57
  • @DonThousand I already said all that needs to be said. You figure it out. – conditionalMethod Dec 08 '19 at 21:58
  • @conditionalMethod Your comment is simply wrong, unless you have a radically different definition of tending to infinity than anything that I've ever seen / can google. – Rushabh Mehta Dec 08 '19 at 21:59
  • I don't understand why you are so unwilling to define the terms you use? Your rude insinuations are quite unnecessary. I just want to understand what you mean. – Rushabh Mehta Dec 08 '19 at 22:00

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