I am trying to show that
$$L=\lim\limits_{n\to\infty}n\left(n\ln{n}+\ln{\sqrt{2}}-n-\sum\limits_{k=1}^n \ln{\left(k-\frac{1}{2}\right)}\right)=\frac{1}{24}$$
Desmos strongly suggests that this is true, but I have not been able to prove this.
Context
A viral question asked to find how many triangles there are in a diagram of this form:
I wasn't really interested in this question; but I did wonder, of course, what happens to the product of the areas of the regions as $n\to\infty$, where $n$ is the number of vertical levels, which equals the number of horizontal intervals. (In the example above, $n=4$.)
It turns out that, assuming the triangle is isosceles, the product of the areas is $\left(\frac{A}{n}\right)^{n^2} \prod\limits_{k=1}^n (2k-1)^n$ where $A$ is the average (arithmetic mean) area of the regions.
Now something interesting happens if we set $A=\frac{e}{2}$: the product seems to approach asymptotically $e^{-1/24}\sqrt2^n$. Proving this amounts to proving that $L=\frac{1}{24}$.
My attempt
I have tried to use the Euler-Maclaurin formula to deal with the $\sum\limits_{k=1}^n \ln{\left(k-\frac{1}{2}\right)}$, but I am not so familiar with this formula and I cannot seem to make it work.
Best Answer
$$L=\lim\limits_{n\to\infty}n\left(n\ln{n}+\ln{\sqrt{2}}-n-\sum\limits_{k=1}^n \ln{\left(k-\frac{1}{2}\right)}\right)$$
$$\sum\limits_{k=1}^n \ln{\left(k-\frac{1}{2}\right)} = \sum\limits_{k=1}^n (\ln{\left(2k-1\right)}-\ln 2) = \ln (1\cdot 3\cdot \ldots \cdot(2n-1)) - n \ln 2 = $$ $$ = \ln\left(\frac{(2n)!}{2^n n!}\right) - n\ln2 = \ln ((2n)!) - \ln n! - 2n\ln2.$$ We know ( https://en.wikipedia.org/wiki/Stirling%27s_approximation ) that $$ \ln n! = n \ln \frac{n}e + \frac{\ln n}2 + \ln \sqrt{2 \pi} + \frac{B_2}{2n} + O(\frac{1}n), $$ where $B_2 = \frac16$.
Hence $$\sum\limits_{k=1}^n \ln{\left(k-\frac{1}{2}\right)} = 2n \ln \frac{2n}e - n \ln \frac{n}e + \frac{\ln 2n}2 - \frac{\ln n}2 + \frac16(\frac1{4n} - \frac1{2n})+O(n^{-1})$$ and now it's easy to find $L$.