Show that $\sum \frac {\mu(n)}{10^n}$ is an irrational number.
I had hoped i could use the method that is used to show that e is irrational but it doesnt seem to work.
BWOC: Assume that $\sum \frac {\mu(n)}{10^n}= \frac{p}{q}$ s.t $p,q \in \Bbb N $ and $\operatorname{gcd}(p,q)=1$ then we multiply by $q! $ we get that $(q-1)!p $ is an integer that equals $ q! \sum_{n=1}^{\infty} \frac {\mu(n)}{10^n}$ the problem i am now having is unlike in the proof that $e$ is irrational cause i cant figure out for what value of n that i can throw away till to get the remainder being a decimal.
Consider the number plus $\frac19=\sum \frac{1}{10^n}$.
If $\alpha=\sum \frac{1+\mu(n)}{10^n}$ is irrational, then your number is also irrational.
See this Large gap between two consecutive square-free numbers
Since there are arbitrary long gaps between two square free numbers, the number $\alpha$ has arbitrary long sequence of 1's in the decimal expansion.