First of all, I'm aware that this question has been previously asked, (see: show that $\sum \frac {\mu(n)}{10^n}$ is irrational) however I did not find the solutions there particularly useful.
In general, I am comfortable with proving similar statements of this flavour, such as showing that $e$ is irrational, or $sin(1)$ (1 in radians). However this proof is a bit different, since the denominator of interest is not a factorial.
I know that I can think of this as an infinite sequence, and considering $\sum_{n=1}^{\infty} \frac{1+\mu(n)}{10^n}$, I can think of that sum as an infinite ternary sequence of $0,1,2$.
I've tried finding upper and lower bounds for this sum, splitting it into positive and negative sums, and I don't know how to proceed.
The central concern about this problem is to determine whether $1+\mu(n)$ is periodic. If it is periodic, we can soon conclude that this number is rational. We will prove the irrationality by contradicting this statement.
If $1+\mu(n)$ is periodic, then within each period $T$, there should be only a fixed number of occurences of $1+\mu(n)=1$. This indicates that there should be a fixed number of non square-free integers within the period. However, if we were to let $p_k$ be $k$'th prime, we can show that for any $M$>0 the following system of congruences
$$ \begin{cases} n & \equiv0\pmod{p_k^2} \\ n+1 & \equiv0\pmod{p_{k+1}^2} \\ n+2 & \equiv0\pmod{p_{k+2}^2} \\ \vdots & \vdots \\ n+M &\equiv0\pmod{p_{k+M}^2} \end{cases} $$
always has solution by the Chinese remainder theorem, which means that it is possible to have arbitrarily large consecutive non square-free sequence of integers, contradicting our assumption. Therefore, the series
$$ \sum_{n=1}^\infty{\mu(n)\over10^n} $$
converges to an irrational number.