Relate to Dirichlet's theorem Diophantine approximation

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 $$0 and $$\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?

• It is related to Liouville's theorem, not to Dirichlet's. An idea to follow is at the beginning of the article (it could be put shorter...). – metamorphy Sep 11 at 9:49
• It's quite related but in Liouville's number you have $|\xi -\frac{x}{y}| \leq \frac{1}{y^n}$, meanwhile my problem is $|\xi -\frac{x}{y}| \leq \frac{1}{y^a}$. It's fixed a. – Desunkid Sep 11 at 13:21

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 $$a^2 > a+1, \forall a\geq3$$.