Sum of fractions in the form $\frac{1}{n}$ from $\frac{1}{2^n}$ to $\frac{1}{3*2^{(n-1)} - 1}$ less than $\frac{1}{2}$?

Question

https://mathoverflow.net/a/278290/501460

I've been trying to figure out why this works, and why the tiles don't go past the middle, considering all the squares together have an infinite side length.

Is there an easy way to show that the range of fractions in each layer doesn't add up to over $\frac12$ (and similarly, the ones starting from the right that range from, I believe, $\frac{1}{3*2^{(n-1)}}$ to $\frac{1}{2^{(n+1)} - 1}$ don't add up to over $\frac13$)

Answer 1

The function $1/x$ is decreasing.
$$ \sum_{k=2^n}^{3\cdot 2^{n-1}-1} \frac{1}{k} < \int_{2^n-1}^{3\cdot 2^{n-1}-1}\frac{1}{x}\,dx =\log(3\cdot 2^{n-1}-1)-\log(2^n-1) =\log\frac{3\cdot 2^{n-1}-1}{2^n-1} $$ This is decreasing as $n$ increases. For $n \ge 3$ this is $< \frac12$.

Answer 2

Without any serious maths: Count the number of fractions you are adding, then find a trivial upper and lower bound for each fraction, multiply and - voila - you have an upper and lower bound for the sum.

For a very precise estimate: 1/n is the integral of 1/x over the interval from n-1/2 to n+1/2, plus a tiny correction. You can easily sum up integrals where the upper and lower limits match up, and the sum of the corrections is tiny.