Show that $(a_n)$ with $|a_n - a_m|\leq C |\frac{1}{m}-\frac{1}{n}|$ is convergent

Question

Let be $C>0$ a constant such that for all $m,n\in \mathbb{N}$ we have $|a_n - a_m|\leq C |\frac{1}{m}-\frac{1}{n}|$. Show that $(a_n)$ is a convergent sequence.

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If we consider $\mathbb{R}$ as the underlying field, i.e. $(a_n)$ is a real-valued sequence, then the statement follows directly from the fact that $(a_n)$ is a Cauchy-sequence and every Cauchy sequence in $\mathbb{R}$ is convergent.

However, I am wondering if it is possible to show the convergence if we don't know that $(a_n)$ is a real-valued sequence (e.g. assume $\mathbb{Q}$ as the corresponding field)?

Answer 1

No. In fact, the whole reason we need $\mathbb{R}$ is that not all Cauchy sequences in $\mathbb{Q}$ converge to a rational number. That is, $\mathbb{Q}$ is not a complete metric space, while $\mathbb{R}$ is.

Consider, for instance, the sequence defined by $x_n = \sum\limits_{j = 1}^n \frac{1}{10^{(j^2)}}$. This sequence is quite simple to define, and it is clearly a Cauchy sequence. But it does not have a rational limit.

Specifically, WLOG suppose $n > m$. Then we have

$\begin{equation} \begin{split} 0 \leq x_n - x_m &= \sum\limits_{j = m + 1}^n \frac{1}{10^{(j^2)}} \\ &\leq \sum\limits_{j = m + 1}^n \frac{1}{10^{j}} \\ &= \frac{1}{9} (\frac{1}{10^m} - \frac{1}{10^n}) \\ &< \frac{1}{9} \frac{1}{10^m} \\ &< \frac{1}{m(m + 1)} \\ &= \frac{1}{m} - \frac{1}{m + 1} \\ &\leq \frac{1}{m} - \frac{1}{n} \end{split} \end{equation} $

So this is a Cauchy sequence which matches OP's definition. The non-trivial inequality here is that $9 \cdot 10^m \leq m(m + 1)$, which follows induction using the facts that $9 \cdot 10 > 1 \cdot 2$ and that $\frac{m + 2}{m} \leq 3$ for all $n \geq 1$.

In fact, we can go even further; every single real number is the limit of a Cauchy sequence of rational numbers. So every irrational number at all is the limit of some Cauchy sequence of rational numbers. And every Cauchy sequence has a subsequence which satisfies OP's condition.

Answer 2

No. The standard counter-example is to define $\ (a_n)\ $ by choosing, for each $\ n,\ $ that $\ a_n\ $ is a (any) rational number in the interval $\ \left[\ \sqrt{2},\ \sqrt{2}+\frac{1}{n}\ \right].$

Answer 3

In addition to what Mark said: For an arbitrary irrational number $x$ we can construct a sequence $(a_n)$ of rational numbers converging to $x$ and satisfying the given estimate.

For each $n$ choose $a_n \in \Bbb Q$ such that $|a_n - x| < \frac{1}{(n+1)^2}$. For $n < m$ is $$ |a_n - a_m | \le | a_n-x| + |a_m-x| \le \frac{1}{(n+1)^2} + \frac{1}{(m+1)^2} \\ \le \frac{2}{(n+1)^2} \le \frac{2}{n(n+1)} = \frac 2n - \frac{2}{n+1} \le 2 \left( \frac 1n - \frac 1m \right) $$ but the sequence $(a_n)$ is not convergent in $\Bbb Q$.