Prove that the sequence $\{q_n\}_{n \geq 1}$ diverges to $\infty.$

Question

Let $\{x_n \}_{n \geq 1}$ be a sequence of rational numbers converging to $\pi.$ Let for all $n \geq 1,$ $x_n = \frac {p_n} {q_n},$ where $p_n,q_n \in \Bbb Z$ with $\text {gcd}\ (p_n,q_n) = 1.$ Prove that the sequence $\{q_n \}_{n \geq 1}$ diverges to $\infty.$

How do I prove that? Any help will be highly appreciated.

Thanks in advance.

Oh! I see. Can you please provide the way to approach @Yves Daoust sir? — Anil Bagchi., Nov 09 '20 at 14:51
@Yves Daoust sir can you please be more explicit? What's the lower bound of such approximation? — Anil Bagchi., Nov 09 '20 at 15:00
When you say "diverges to $\infty$" do you mean the sequence is unbounded, or do you mean that $q_n \to +\infty$? The former is what I think you mean but I want to make sure — perpetuallyconfused, Nov 09 '20 at 15:01

score 1 · Accepted Answer · answered Nov 09 '20 at 15:29

1

Suppose that $\{q_n\}$ is bounded, and suppose without loss of generality that $q_n > 0$. Since $q_n \in \mathbb{Z}$ is bounded, define $q = \text{lcm}\{q_n$}. Then we know that there exists $z \in \mathbb{Z}$ such that the following holds:

\begin{equation} \frac{z}{q} < \pi < \frac{z+1}{q}. \end{equation}

Then any $\frac{p_n}{q_n}$ can be written in the form $\frac{z'}{q}$, so cannot improve on the bounds in the above inequality. So we have that $\frac{p_n}{q_n}$ is bounded away from $\pi$.

answered Nov 09 '20 at 15:29

perpetuallyconfused

764

So if we take $\varepsilon = \min \left {\left ( \pi - \dfrac {z} {q} \right ) , \left ( \dfrac {z+1} {q} - \pi \right ) \right }$ then we have $${x_n\ |\ n \in \Bbb N } \cap B \left (\pi, \dfrac {\varepsilon} {2} \right ) = \varnothing.$$ Am I right? But this contradicts the fact that $x_n \to \pi$ as $n \to \infty.$ – Anil Bagchi. Nov 09 '20 at 15:45
That looks correct to me – perpetuallyconfused Nov 09 '20 at 15:49
Elegant proof. +1 Many many thanks for your help. – Anil Bagchi. Nov 09 '20 at 15:51
No problem! I would recommend looking at some of the similar questions that are linked in the comment by @MartinR and on the side, since they'll help with similar questions like proving that the limit is irrational – perpetuallyconfused Nov 09 '20 at 15:53

Derek Luna · Answer 2 · 2020-11-09T16:13:03.453

If the sequence $\{q_{n}\}$ is bounded then clearly $\{p_{n}\}$ must also be bounded as otherwise $\{x_{n}\}$ does not converge. Then $\{x_{n}\}$ must be bounded, and most importantly, finite. Since $\pi$ is irrational, the closest we can get to $\pi$ is obtained by taking the $x_{k}$ (which exists in $\{x_{n}\})$ such that $\inf\{|{x_{n}}-\pi|\}$ is satisfied by $n=k$. Since $x_{k} \neq \pi$, we see we cannot get arbitarily close to $\pi$ using a finite set of fractions. Take $\epsilon < \inf\{|{x_{n}}-\pi|\}$ and we see that $\{x_{n}\}$ does not converge to $\pi$, a contradiction.

Prove that the sequence $\{q_n\}_{n \geq 1}$ diverges to $\infty.$

2 Answers2