Write $a_i=1111111...1$ with $i$ digits. Then your sum is:$\sum_{k=0}^n \frac{\log{4a_{2^k}}}{\log{3a_{2^k}}}$
But $\log 4a_i = log 3a_i + \log {\frac 4 3}$ So your sum is:
$\sum_{k=0}^{n} \left(1+ \frac{\log{4/3}}{\log{3a_{2^k}}}\right) = n+1 + \sum_{k=0}^{n} \frac{\log{4/3}}{\log{3a_{2^k}}}$
So you just need to show that $\log\left(\frac4 3\right)\sum_{k=0}^{n} \frac{1}{\log{3a_{2^k}}}<1$. (It's clearly greater than zero, so the first inequality is true.)
So fundamentally, you are trying to come up with a nice lower bound on the term $\log{3a_i}$. But $9a_i$ is $10^i-1$, so you can probably work from there.
Completed proof
One thing to notice is that the base of your logarithm is irrelevant, since $\frac{\log_a b}{\log_a c} = \log_c{b}$ is independent of $a$. So we can use the natural $\log$. Then $\log{4/3} <\frac{1}{3}$, so we only need to show that $\sum_{k=0}^n \frac{1}{\log 3a_{2^k}} < 3$.
Now, $\log 3 a_{2^k} = \log 3 + \log a_{2^k}$. But $a_{2^k} > 10^{2^k-1}$. So $\log a_{2^k} > (2^k-1)\log 10 > 2^k-1$. But $\log 3>1$, so $\log 3a_k > 2^k$.
So $\frac{1}{\log 3a_{2^k}}<2^{-k}$ and therefore $\sum_{k=0}^n \frac{1}{\log 3a_{2^k}} < 2$