I have a problem:
Let $a \in \mathbb{Z}$, $a \geq 3$, and set $\xi= \sum_{n=0}^\infty
10^{-a^{2n}}>0$. Then the inequality
$$\Big|\,\xi – \dfrac{x}{y}\,\Big| \leq \dfrac{1}{y^a}$$ has infinitely many solutions with $x,y \in \mathbb{Z}$, $y>0$ and $\gcd(x,y)=1$.
I would imitate the proof of Dirichlet's theorem as follow, I claim my lemma
Lemma: Let $\xi=\sum_{n=0}^\infty 10^{-a^{2n}}$, for every integer $Q \geq 2$, there are integers $x,y$ which are not both equal to $0$, such that
$$|x – \xi y| \leq \dfrac{1}{Q^{a-1}}$$
with $0<y \leq Q$ and $\gcd(x,y) =1$
- Try to prove this lemma:
Partition the interval $[0,1]$ into $Q^{a-1}$ subintervals of length $\dfrac{1}{Q^{a-1}}$. Consider $Q^{a-1}+1$ numbers $\xi-[\xi]$,..,$Q^{a-1}\xi-[Q^{a-1}\xi]$ and $1$. By the Dirichlet principle, two among these numbers must lie in the same subinterval of length $\dfrac{1}{Q^{a-1}}$. Hence we can find $x,y \in \mathbb{z}$ such that $|x-y \xi| \leq \frac{1}{Q^{a-1}}$. But now my trouble is $y$ is not smaller than $Q$.
Does anyone have other ideas?
Best Answer
With regards to other ideas, just straight attack. Noting $y=10^{a^{2n}}$ then $$\sum\limits_{k=0}^{n}\frac{1}{10^{a^{2k}}}=\frac{\sum\limits_{k=0}^{n}10^{a^{2n}-a^{2k}}}{y}$$ where $x=\sum\limits_{k=0}^{n}10^{a^{2n}-a^{2k}}=\sum\limits_{k=0}^{\color{red}{n-1}}10^{a^{2n}-a^{2k}}\color{red}{+1}=10\cdot Q\color{red}{+1}$ and $\gcd(x,y)=1$. Now $$\left|\xi -\frac{x}{y}\right|= \sum\limits_{k=n+1}^{\infty}\frac{1}{10^{a^{2k}}}= \frac{1}{10^{a^{2n+2}}}\left(\sum\limits_{k=n+1}^{\infty}\frac{1}{10^{a^{2k}-a^{2n+2}}}\right)< \\ \frac{1}{10^{a^{2n+2}}}\left(\sum\limits_{k=0}^{\infty}\frac{1}{10^{k}}\right)= \frac{1}{10^{a^{2n+2}}}\left(\frac{1}{1-\frac{1}{10}}\right)=\\ \frac{1}{10^{a^{2n}\cdot a^2}}\cdot \frac{10}{9}= \frac{1}{y^{a^2}}\cdot \frac{10}{9}< \frac{1}{y^{a+1}}\cdot \frac{10}{9}=\frac{1}{y^a}\cdot \frac{10}{y\cdot9}<\frac{1}{y^a}$$ for infinitely $n$ and because $y\geq10, \forall n\geq0$ and $a^2 > a+1, \forall a\geq3$.