This is not the answer to the question of irrationality. This a piece of information about an interesting closed form of the series.
$$q_n=\sum_{k=0}^n 10^k = \frac{10^{n+1}-1}{9}$$
$q_0=1\quad;\quad q_1=11\quad;\quad q_2=111\quad...$
$\sum_{n=0}^m \frac{1}{q_n}=\frac{1}{1}+\frac{1}{11}+\frac{1}{111}+... \quad$ ($m+1$ terms).
$$\sum_{n=0}^m \frac{1}{q_n}=\sum_{n=1}^m \frac{9}{10^{n+1}-1}=
\frac{9}{\ln(10)}\left(\psi_{10}(m+2)-\psi_{10}(1)\right)-9(m+1)$$
$\psi_q(x)$ is the q-digamma function. The leading terms of the asymptotic series are:
$$\psi_{10}(x)\sim (x-\frac{1}{2})\ln(10)-\ln(9)$$
$$\sum_{n=0}^{m\to\infty} \frac{1}{q_n}\sim
\frac{9}{\ln(10)}\left( (m+2-\frac{1}{2})\ln(10)-\ln(9)-\psi_{10}(1)\right)-9(m+1)$$
After simplification :
$$\sum_{n=0}^{\infty} \frac{1}{q_n}=\frac{9}{2}
-\frac{9}{\ln(10)}\left( \ln(9)+\psi_{10}(1)\right)$$
$\psi_{10}(1)\simeq-1.32759401026424207 $
$\frac{9}{2}
-\frac{9}{\ln(10)}\left( \ln(9)+\psi_{10}(1)\right)\simeq 1.100918190836200736 $
Note that $\sum_{n=0}^{\infty} \frac{1}{q_n}$ includes the first term$=1$