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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:

enter image description here

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.

Dan
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    Extending the argument in this answer we can write $(1/n)\sum_{k=1}^n f(k/n) =\int_0^1 f(x) , dx+\frac{f(1)-f(0)}{2n}+\frac{f'(1)-f'(0)}{12n^2}+o(1/n^2)$. For your problem use $f(x) =\log(1+x)$ to get the limit as $1/24$ (third term of above formula). – Paramanand Singh Oct 27 '22 at 05:05
  • @ParamanandSingh I don't quite understand how to relate the $\sum_{k=1}^n \log{(k-1/2)}$ in my question to your $(1/n)\sum_{k=1}^n \log{(1+k/n)}$. – Dan Oct 27 '22 at 05:20

4 Answers4

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$$L=\lim\limits_{n\to\infty}n\left(n\ln{n}+\ln{\sqrt{2}}-n-\sum\limits_{k=1}^n \ln{\left(k-\frac{1}{2}\right)}\right)$$

$$\sum\limits_{k=1}^n \ln{\left(k-\frac{1}{2}\right)} = \sum\limits_{k=1}^n (\ln{\left(2k-1\right)}-\ln 2) = \ln (1\cdot 3\cdot \ldots \cdot(2n-1)) - n \ln 2 = $$ $$ = \ln\left(\frac{(2n)!}{2^n n!}\right) - n\ln2 = \ln ((2n)!) - \ln n! - 2n\ln2.$$ 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 $L$.

Gary
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Sergei Nikolaev
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  • Thanks, but are you sure your two expressions for $\sum\limits_{k=1}^n \ln{\left(k-\frac{1}{2}\right)}$ are correct? I tested them on desmos, and they don't seem to be correct. – Dan Oct 26 '22 at 15:20
  • @Dan There was a typo, because in fact we have $13\ldots * (2n-1) = \frac{2n!}{2^n n!}$. A made an edit. – Sergei Nikolaev Oct 26 '22 at 15:53
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The exponential of the term in parentheses is $$ \frac{n^n\sqrt 2}{e^{n}\prod_{k=1}^n\left(k-\frac12\right)}.$$ With $$\prod_{k=1}^n\left(k-\frac12\right)=2^{-n}\prod_{k=1}^n(2k-1)= \frac{\prod_{k=1}^{2n}k}{2^n\prod_{k=1}^n(2k)}=\frac{(2n)!}{4^nn!},$$ this becomes $$ \frac{4^nn^n\sqrt 2\,n!}{e^{n}(2n)!}.$$ Now by Stirling, $$n!\approx n^n e^{-n}\sqrt{2\pi n} ,$$ so we get $$\approx \frac{4^nn^n\sqrt 2\,n^ne^{-n}\sqrt{2\pi n}}{e^{n}(2n)^{2n}e^{-2n}\sqrt{4\pi n}} =1. $$ So at least the expression in parentheses converges to $0$.

To find what you are looking for, we need some good enough error bound for the Stirling approximation. Indeed, it is well-known that more precisely, $$ n! = n^ne^{-n}\sqrt{2\pi n}\cdot (1+\frac1{12n}+o(1/n))$$ Then your result follows from $$ \frac{1+\frac1{12n}+o(1/n)}{1+\frac1{24n}+o(1/n)}=1+\frac1{24n}+o(1/n)$$ and $\ln(1+x)=x+o(x)$.

Hagen von Eitzen
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Using Pochhammer symbols $$S_n=\sum_{k=1}^n \log{\left(k-\frac{1}{2}\right)}=\log \left(\left(\frac{1}{2}\right)_n\right)=\log \left(\frac{\Gamma \left(n+\frac{1}{2}\right)}{\sqrt{\pi }}\right)$$ Its expansion is $$S_n=n (\log (n)-1)+\frac{\log (2)}{2}-\frac{1}{24 n}+\frac{7}{2880 n^3}-\frac{31}{40320 n^5}+O\left(\frac{1}{n^7}\right)$$

This makes the expression to be $$\frac{1}{24}-\frac{7}{2880 n^2}+\frac{31}{40320 n^4}+O\left(\frac{1}{n^6}\right)$$

Applied to $n=10$, the truncated expansion gives $$\frac{16790231}{403200000}=\color{red}{0.041642437}996$$

Claude Leibovici
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This is an expansion of my comment to question.

Extending the argument in this answer we can write $$\frac{1}{n}\sum_{k=1}^n f\left(\frac{k}{n}\right) =\int_0^1 f(x) \, dx+\frac{f(1)-f(0)}{2n}+\frac{f'(1)-f'(0)}{12n^2}+o(1/n^2)\tag{1}$$ Next let us observe that the expression $$s_n=n\log n-n +\frac{1}{2}\log 2-\sum_{k=1}^{n}\log\left(k-\frac{1}{2}\right)$$ can be written as $$n\left(\frac{1}{2n}\log 2+\log (4/e) -\frac{1}{n}\log\frac{(2n)!}{n!n^n }\right) $$ which is same as $$n\left(\log(4/e)+\frac{1}{2n}\log 2-\frac{1}{n}\sum_{k=1}^{n}\log\left(1+\frac{k}{n}\right)\right)$$ If $f(x) =\log(1+x)$ then the above expression equals $$s_n=n\left(\int_0^1 f(x) \, dx+\frac{f(1)-f(0)}{2n}-\frac{1}{n}\sum_{k=1}^{n}f\left(\frac{k}{n}\right)\right) $$ Using $(1)$ the expression $ns_n$ tends to $$-\frac{f'(1)-f'(0)}{12}=\frac{1}{24}$$

Paramanand Singh
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  • Thanks. I like all of the answers given so far. I accept yours because it is the most accessible to me. – Dan Oct 27 '22 at 11:52