Prove $ n+1<\frac{\log 4}{\log3}+\frac{\log 44}{\log33}+\frac{\log4444}{\log3333}+\cdots+\frac{\log 444\ldots444}{\log333\ldots333} Prove that
$$ n+1<\frac{\log 4}{\log3}+\frac{\log 44}{\log33}+\frac{\log4444}{\log3333}+\frac{\log 44444444}{\log33333333}+\cdots+\frac{\log 444\ldots444}{\log333\ldots333} <n+2$$
where last logarithm has $2^n$ digits. 
The left side is easy to prove, but I have no idea where to start for the right one.
PS. This should be proven without using limits.
 A: 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 $\sum_{k=0}^n \frac{1}{\log 3a_{2^k}} < 2$
A: Here is another way:  Suppose we are looking at a term with $2^{k}$  digits, $k\geq1$. Then  $$\frac{\log\left(444\cdots444\right)}{\log\left(333\cdots333\right)}=\frac{\log4+\log\left(111\cdots111\right)}{\log3+\log\left(111\cdots111\right)}=\frac{1+\frac{\log4}{\log\left(111\cdots111\right)}}{1+\frac{\log3}{\log\left(111\cdots111\right)}}$$ 
$$\leq1+\frac{\log4}{\log\left(111\cdots111\right)}\leq1+\frac{\log4}{\log\left(100\cdots000\right)}=1+\frac{\log4}{(2^{k}-1)\log\left(10\right)}.$$ Fortunately this is bounded by a geometric series. Since $$\frac{\log4}{(2^{k}-1)\log\left(10\right)}=\frac{\log4}{2^{k}\log\left(10\right)}+\frac{\log4}{2^{k}(2^{k}-1)\log\left(10\right)}\leq\frac{\log4}{2^{k}\log\left(10\right)}+\frac{2\log4}{2^{2k}\log\left(10\right)},$$ we see that $$\sum_{k=M}^{\infty}\frac{\log4}{(2^{k}-1)\log\left(10\right)}\leq\frac{\log4}{\log\left(10\right)}\sum_{k=M}^{\infty}\left(\frac{1}{2^{k}}+\frac{2}{2^{2k}}\right)=\frac{\log4}{\log\left(10\right)}\left(\frac{1}{2^{M-1}}+\frac{8}{3\cdot2^{2M}}\right).$$ If we take $M=2$, this is $$\frac{2\log4}{3\log10},$$ and hence we may bound the original sum by $$\frac{\log4}{\log3}+\frac{\log44}{\log33}+(n-1)+\frac{2\log4}{3\log10}$$ since there are $n+1$ terms, and we looked at the last $n-1$ leaving the first two alone.  $$Since\frac{\log4}{\log3}+\frac{\log44}{\log33}+\frac{2\log4}{3\log10}-1=1.74550\dots<2$$ the result is proven.
