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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.

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Can I ask you how did you prove the left hand side inequality? – Giovanni De Gaetano Feb 7 '12 at 17:27
$\frac{\log4}{\log3}=\log_{3}{4}>1$ is obvious because $3<4$, the same goes with all $n+1$ logarithms. Then do the sum of all inequalities. – Lazar Ljubenović Feb 7 '12 at 17:30
I´m sorry! I misread the central term, I thought it was the sum over $\frac{log(44...44)}{log(33...33)}$ over all the length from $1$ to $2^n$. Now it makes sense! – Giovanni De Gaetano Feb 7 '12 at 17:35
It's OK. Any ideas for solution? – Lazar Ljubenović Feb 7 '12 at 17:41
Use the fact that for example $\frac{\log(4444)}{\log(3333)}=1+\frac{\log 4-\log 3}{\log 3+\log(1111)}$. And perhaps use the fact that $\log(1111)$ is very close to $\log(10000/9)$. – André Nicolas Feb 7 '12 at 18:33
up vote 6 down vote accepted

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 $\sum_{k=0}^n \frac{1}{\log 3a_{2^k}} < 2$

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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.

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