A friend of mine told me about the following problem:

Let $\{r_n\}$ be a sequence of rational numbers such that $\lim_{n\to\infty}r_n=x\in\Bbb R,$ $r_n\neq x,$ for every $n\in\Bbb N$ and $r_n=\dfrac{a_n}{b_n},$ for each $n\in\Bbb N,$ where $\{a_n\}$ is a sequence of integers and $\{b_n\}$ is a sequence of positive integers. Prove that $\lim_{n\to\infty}b_n=+\infty.$

I proved such result by contradiction:

If $\lim_{n\to\infty}b_n\neq+\infty,$ then $\exists M\in\Bbb R$ such that $\forall N\in\Bbb N,$ there exists some $n\geq N$ such that $b_n<M.$ Therefore, we can construct a subsequence $\{b_{m_k}\}$ of $\{b_n\}$ as follows:

First $\exists m_0\geq0$ such that $b_{m_0}<M.$ Having chosen $m_1,\ldots,m_p,$ let $m_{p+1}$ be such that $m_{p+1}>m_p$ and $b_{m_{p+1}}<M.$

Since $1\leq b_{m_k}<M,$ for every $k\in\Bbb N$ and $\{b_n\}$ is a sequence of positive integers, there must exist some constant subsequence $\{b_{n_k}\}$ of $\{b_n\}.$

Let $b$ be the positive integer such that $b_{n_k}=b$ for every $k\in\Bbb N$ and let $\delta>0$ be arbitrary.

Since $\{r_n\}$ converges to $x,$ then $\{r_{n_k}\}$ converges to $x$ and hence, there is some natural $N_0$ such that $$0<\left|\dfrac{a_{n_k}}{b}-x\right|<\dfrac{\delta}{b},$$ for each $k\geq N_0.$ Then $$0<\left|a_{n_k}-bx\right|<\delta,$$ for every $k\geq N_0.$ Since $\delta$ is arbitrary, this means that $\lim_{k\to\infty}a_{n_k}=bx.$ Therefore $\exists N_1\in\Bbb N$ and $\exists a\in\Bbb Z$ such that $a_{n_k}=a,$ for every $k\geq N_1.$

Therefore, $r_{n_k}=\dfrac{a}{b},$ for sufficiently large $k,$ which contradicts the fact that $r_n\neq x,$ for every $n\in\Bbb N.$

My questions are, is there a direct proof of the exercise? or is there an elegant solution? and, is the idea of my proof correct?

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    $\begingroup$ I like your proof too. You could also argue using the fact that given $N$ a positive integer, there is an open interval containing $x$ that contains no rational $p/q$ with $0<q<N$, except perhaps for $x$. $\endgroup$ – David Mitra Jul 13 '15 at 23:14
  • $\begingroup$ @DavidMitra Nice argument! However, I think the proof of that fact follows in some sense this idea so, it doesn't give precisely a 'direct proof' (in any case, it's more elegant!) $\endgroup$ – Daniel Jul 13 '15 at 23:21

Fix $M$, and consider the all fractions of the form $a/b$ with $b < M$. Although there are infinitely many fractions of this form, only finitely many of them are within distance $1$ of $x$ (since if $|a/b-x| \leq 1$ then and $|a-c| > 2b$ then $|c/b-x| > 1$). Let $r \neq x$ be one of the closest such fractions. Then $|x-a/b| \geq |x-r|$ whenever $b < M$. Since $a_n/b_n \to x$, for large $n$ we must have $|x-a_n/b_n| < |x-r|$, and so $b_n \geq M$. Since this is true for every $M$, we see that $b_n \to \infty$.

When $x$ is rational, say $x = p/q$, then we can see this phenomenon even more vividly: if $b < M$ and $a/b \neq x$ then $$ \left|x - \frac{a}{b}\right| = \left|\frac{pb-aq}{bq}\right| \geq \frac{1}{bq} > \frac{1}{Mq}. $$ So to get within distance $1/(Mq)$, the denominator must be at least $M$.


I like your proof, keep in mind at the and that actually $a=bx$ so that $\frac{a}{b}=x$.

Also, I don't see why do you need the subsequence that you call $\{b_{m_k}\}$, you can 'jump' directly to the constant subsequence $\{b_{n_k}\}$ which is the one you need.

If you're looking for a 'faster' argument, you can begin exactly the same, arguing the existence of a constant subsequence $\{b_{n_k}\}=\{b\}$ of the denominator's sequence, after that, since $\left\{\frac{a_n}{b_n}\right\}$ converges to $x$, $\left\{\frac{a_{n_k}}{b_{n_k}}\right\}$ converges to $x$ too and therefore $\left\{b_{n_k}\cdot\frac{a_{n_k}}{b_{n_k}}\right\}=\left\{a_{n_k}\right\}$ converges to $b\cdot x$, this would simplify your delta argument.

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    $\begingroup$ Thank you, I used the subsequence $\{b_{m_k}\}$ to show that $\{b_n\}$ must necessarily have a constant subsequence, but you are right, I can go directly to $\{b_{n_k}\}.$ $\endgroup$ – CIJ Jul 13 '15 at 23:29

I think that the Cauchy condition can be used here in the proof. I will write the idea briefly

  1. As above assume that $b_k\not\to+\infty$ then there exists a bounded subsequence. WLOG we will call it again $b_k$, i.e. $|b_k|\le M$.

  2. Since $a_k/b_k$ is convergent it is a Cauchy sequence, i.e. $$ \|\frac{a_n}{b_n}-\frac{a_m}{b_m}\|\le\epsilon,\quad\forall n,m\ge N $$ for any $\epsilon>0$. After rewriting we get $$ |a_n b_m-a_m b_n|\le\epsilon|b_n b_m|\le\epsilon M^2. $$

  3. It is possible to pick such $N$ that $\epsilon M^2<1$ which means that $a_n b_m=a_m b_n$, $\forall n,m\ge N$, i.e. $$ \frac{a_n}{b_n}=\frac{a_m}{b_m},\quad \forall n,m\ge N, $$ that is the sequence becomes constant after a while.
  4. Since $a_n/b_n\to x$ it must mean that $a_n/b_n=x$, $\forall n\ge N$, which contradicts the assumption.

How about this: $s_n := r_n-x$ then $\lim_{n\to\infty}s_n=0$ is a null series such that $s_n \neq 0\forall n\in \mathbb{N}$ Because $s_n = \frac{a_n}{b_n}-x = \frac{a_n-xb_n}{b_n} =: \frac{c_n}{b_n}$ and $\lim_{n\to\infty} \frac{c_n}{b_n} = \frac{c}{b}$ as $b_n\neq 0$. This implys that either $\lim_{n\to\infty} c_n = 0$ or $\lim_{n\to\infty} b_n = \infty$. The first case gives a contradiction because $a_n$ and $b_n$ are integers and the series can only go to zero if $a_N = xb_N$ for some $N$, which violates $s_n \neq 0$


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