Exercise 7.6.4* (Some irrational numbers are approximable by rational to any order)

Question

Given any constant $c$, prove that there exists an irrational number $\xi$ and infinitely many rational numbers $h/k$ such that
$$\left | \xi - \frac{h}{k} \right | < \frac{1}{k^c}.$$

Richard Ganaye · Accepted Answer

\begin{proof} We can take the same $\xi$ for all constant $c$. Consider the Liouville number $\xi$ given by
$$\xi = \sum_{k=1}^\infty  \frac{1}{10^{k!}} = \frac{1}{10^{1!}} +\frac{1}{10^{2!}}+ \frac{1}{10^{3!}}+ \cdots + \frac{1}{10^{k!}}+ \cdots = 0.1100010\cdots.$$
Consider the partial sums
$$\xi_n = \sum_{k=1}^n  \frac{1}{10^{k!}} = \frac{p_n}{q_n} \qquad (n\geq 1),$$
where $p_n, q_n$ are the integers
\begin{align*}
p_n&= \sum_{k=1}^n  10^{n! - k!},\
q_n &= 10^{n!}.
\end{align*}

Then
\begin{align*}
\left | \xi - \frac{p_n}{q_n} ight | &= \sum_{k=n+1}^\infty 10^{-k!}\
&\leq \sum_{j = (n+1)! }^\infty 10^{-j}\
&=10^{-(n+1)!} \sum_{i=0}^\infty 10^{-i}&(j=(n+1)! - i)\
&=\frac{10}{9} \frac{1}{10^{(n+1)!}}\
&< \frac{2}{q_n^{n+1}}.
\end{align*}
Let $c$ be any real constant. Since $\lim_{n	o \infty} q_n^{n+1-c} = + \infty$, there exists an integer $N$ such that for all $n\geq N$, $2 < q_n^{n+1-c}$, so
$$n\geq N \Rightarrow \left | \xi - \frac{p_n}{q_n} ight | < \frac{2}{q_n^{n+1}} < \frac{1}{q_n^c}.$$
Hence, taking $h = p_n$ and $k = q_n$ for $n\geq N$,  there exist infinitely many rational numbers $h/k$ such that
$$\left | \xi - \frac{h}{k} ight | < \frac{1}{k^c}.$$

\end{proof}

Exercise 7.6.4* (Some irrational numbers are approximable by rational to any order)

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Exercise 7.6.4* (Some irrational numbers are approximable by rational to any order)

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