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A semidisk with diameter $\dfrac{e}{\pi}n$ is divided into $n$ regions of equal area by line segments from a diameter endpoint.

Here is an example with $n=6$.

enter image description here

Consider the $n$ arcs between neighboring line segment endpoints.

Let $P(n)=\text{product of arc lengths}$.

Is the following conjecture true:

Conjecture: $\lim\limits_{n\to\infty}P(n)=\dfrac{\pi}{2}$

In other words, if the average arc length is $\color{red}{\dfrac{e}{2}}$ then the product of arc lengths approaches $\color{red}{\dfrac{\pi}{2}}$.

Evidence for my conjecture

I got the following approximations.

$P(1)\approx 1.359$
$P(2)\approx 1.438$
$P(3)\approx 1.469$
$P(6)\approx 1.507$
$P(12)\approx 1.528$
$P(24)\approx 1.543$
$P(48)\approx 1.552$
$P(96)\approx 1.558$

enter image description here

It seems that if the average arc length is less than $\dfrac{e}{2}$ then the product approaches $0$, and if the average arc length is greater than $\dfrac{e}{2}$ then the product approaches $\infty$.

What makes this difficult

What makes my conjecture difficult for me to prove or disprove, is that I cannot find exact expressions for the arc lengths. For example, in the example shown above with $n=6$, the length of the longest arc is $\dfrac{3ex}{\pi}$ where $x-\sin x=\frac{\pi}{6}$.

I am aware of Kepler's equation, but that doesn't seem to help.

Related questions

This question is essentially the converse of my question, "Product of areas in a disk".

I asked about the product of another kind of arc length related to a circle, in my question, "Another interesting property of $y=2^{n-1}\prod_{k=0}^n \left(x-\cos{\frac{k\pi}{n}}\right)$: product of arc lengths converges, but to what?".

I recently asked a related question, "Product of lengths in a disk of area $\pi/e$".

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    $\begingroup$ For very large $n$ the arc length is inversely proportional to the chord length, since it's basically a triangle. $\endgroup$
    – Trebor
    Commented May 13 at 13:08
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    $\begingroup$ The phrase "if the average arc length is $e/2$" is confusing. I think you are scaling the radius here, so you perhaps mean that "if I choose a radius $r_n$ to scale with $n$ so that average arc length is $e/2$." For a radius $r$ the total length of a half-circle is $\pi r$ and the average arc length over $n$ subdivisions would be $\pi r/n$. So for a fixed radius, the average arc length of course goes to zero. $\endgroup$
    – Michael
    Commented May 13 at 16:12
  • $\begingroup$ I also found the average arc length idea confusing. If I've understood correctly, the radius varies with $n$; $r_n=\frac{ne}{2\pi}$. It might be easier though (especially if you need to vary either quantity you're interested in) if you work in a semicircle with unit radius. If the arc lengths that divide that semicircle into $n$ equal areas (per your diagram) are $x_1,x_2,\ldots x_n$, I think your conjecture is $\prod r_n x_i \to \frac{\pi}{2}$, ie $\left(\frac{ne}{2\pi}\right)^n\prod x_i \to \frac{\pi}{2}$, is that correct? $\endgroup$ Commented May 13 at 16:48
  • $\begingroup$ @Michael Yes, that is correct. I have edited to clarify. $\endgroup$
    – Dan
    Commented May 13 at 19:09
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    $\begingroup$ @ChrisLewis Yes, that is correct. I have edited to clarify. $\endgroup$
    – Dan
    Commented May 13 at 19:09

2 Answers 2

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$\newcommand{\eps}{\varepsilon}$ The answer is in fact not $\pi/2$ but $$\frac{3}{(2\pi)^{1/3}}\prod_{k=1}^\infty\left[1-\big((k+1)^{1/3}-k^{1/3}\big)^3\right]\approx 1.56813\ldots.$$ This product converges since $(k+1)^{1/3}-k^{1/3}\sim \frac13\cdot k^{-2/3}$, so each multiplicand is approximately $1-\frac1{27k^2}$. I believe there is no nicer closed form.


The derivation of this expression will be quite lengthy. We will first express each arc length in terms of a continuous function we define, which does not have a closed-form expression but whose inverse is reasonable. This will give us an expression for the product, of which we take the logarithm. In Part 1, we will come up with an estimate for the length of each arc; if these estimates were exactly correct, the limit in the problem would be $1$. In Part 2, we show that our estimates are extremely close to correct for the arcs near the left side of the picture. In Part 3, we quantify how far off our estimates are for arcs on the extreme right side of the picture. It turns out that our estimate for the first arc is too low by a factor of $3e^{-2/3}\approx 1.54025$, our estimate for the second arc is too low by a factor of $\frac{3\cdot 2^{4/3}e^{-2/3}}{2^{2/3}+2^{1/3}+1}\approx 1.00880$, for the third arc by a factor of $\approx 1.00303$, et cetera. Multiplying these together, which we do in Part 4, will give the claimed result.


Part 0. Setting up an expression for the answer.

Let $f(\theta)$ be the area of the region in a unit circle lying above the line joining the points $(1,0)$ and $(\cos\theta,\sin\theta)$. Subtracting the area of a triangle from that of a sector, we compute that $f(\theta)=\frac{\theta-\sin\theta}2$. This function, considered as $f:[0,\pi]\to[0,\pi/2]$, is increasing; let $g:[0,\pi/2]\to[0,\pi]$ be the inverse function of $f$. The lengths of the $n$ arcs described in the question, counting from the right to the left, are $$\frac{en}{2\pi}g\left(\frac{\pi}{2n}\right),\frac{en}{2\pi}\left(g\left(\frac{2\pi}{2n}\right)-g\left(\frac{\pi}{2n}\right)\right),\ldots,\frac{en}{2\pi}\left(g\left(\frac{n\pi}{2n}\right)-g\left(\frac{(n-1)\pi}{2n}\right)\right).$$ We thus need to find $$X:=\lim_{n\to\infty}\prod_{k=0}^{n-1}\frac{en}{2\pi}\left(g\left(\frac{\pi(k+1)}{2n}\right)-g\left(\frac{\pi k}{2n}\right)\right).$$ For notational simplicity, let $\eps=\frac{\pi}{2n}$. We'll evaluate $$\log X=\lim_{n\to\infty}\sum_{k=0}^{n-1}\log\left(\frac e4\cdot\frac{g((k+1)\eps)-g(k\eps)}{\eps}\right).$$


Part 1. Finding a reasonable first approximation.

This part explains why the normalization of setting the diameter to be $\frac e\pi n$ is the correct one. In the above sum, we can imagine replacing the expression that looks like a derivative with a derivative, and then integrating. Define $$Y:=\int_0^{\pi/2}\log\left(\frac e4g'(t)\right)dt=\sum_{k=0}^{n-1}\frac1\eps\int_{k\eps}^{(k+1)\eps}\log\left(\frac e4g'(t)\right)dt.$$ We'll show that $Y=0$. Indeed, since $f(g(t))=t$, we have $g'(t)=1/f'(g(t))$, so we can substitute $u=g(t)$ to get $$Y=\frac{\pi\log(e/2)}2-\int_0^\pi\log(2f'(u))f'(u)du=\frac{\pi(1-\log 2)}2-\frac12\int_0^\pi(1-\cos(u))\log(1-\cos(u))du.$$ Let $I$ be the remaining integral, so that we wish to show that $I=\pi(1-\log 2)$. We see that $-\cos(u)\log(1-\cos(u))$ has an explicit antiderivative $u + \sin(u) - \sin(u)\log(1-\cos u)$, so the integral of this part is $\pi$. It remains to show that $$\int_0^\pi\log(1-\cos u)du=-\pi\log 2.$$ This is a classical integral; see for example here. We conclude that $Y=0$.

Now, we subtract off this "main term" from our sum. This gives $$\log X=\lim_{n\to\infty}\sum_{k=0}^{n-1}\left[\log\left(\frac{g((k+1)\eps)-g(k\eps)}\eps\right)-\frac1\eps\int_{k\eps}^{(k+1)\eps}\log(g'(t))dt\right].$$ Let $$h_\eps(k)=\log\left(\frac{g(k\eps+\eps)-g(k\eps)}\eps\right)-\frac1\eps\int_{k\eps}^{(k+1)\eps}\log g'(t)dt.$$ We shall show that there exist some absolute constants $C$ and $K$ so that $h_\eps(k)\leq C/k^2$ for any $\eps>0$ and $k\geq K$. We'll also show that, for fixed $k$ with $\eps\to 0$, $$h_\eps(k)=\frac23\left(k\log\left(1+\frac1k\right)-1\right)+\log\left(\frac 3{1+r_k+r_k^2}\right)+o(1),$$ where $r_k=(\frac k{k+1})^{1/3}$, and $k\log(1+1/k)$ is set to zero for $k=0$.


Part 2. Showing that large $k$ don't contribute very much.

This part involves some real analysis. We'll need to understand the behavior of $g$ and its derivatives, but not very finely.

Lemma. There exist absolute constants $C_1,C_2,C_3>0$ so that, for all $x\in[0,\pi/2]$, $$g'(x)\geq C_1x^{-2/3};\qquad |g''(x)|\leq C_2x^{-5/3};\qquad |g'''(x)|\leq C_3x^{-8/3}.$$

Proof. We first see that there exist absolute constants $0<b<B$ so that $bg(x)\leq x^{1/3}\leq Bg(x)$. This is because, letting $y=g(x)$, $$\frac{x^{1/3}}{g(x)}=\frac{f(y)^{1/3}}y=\left(\frac{y-\sin y}{2y^3}\right)^{1/3}$$ is positive and decreasing on $[0,\pi]$. Given this, it suffices to show the existence of constants $A_1,A_2,A_3>0$ so that $$y^2g'(f(y))\geq A_1;\qquad y^5|g''(f(y))|\leq A_2;\qquad y^8|g'''(f(y))|\leq A_3$$ for all $0\leq y\leq\pi$. We may compute by implicit differentiation that $$g'(f(y))=\frac1{f'(y)};\qquad g''(f(y))=-\frac{f''(y)}{f'(y)^3};\qquad g'''(f(y))=\frac{3f''(y)^2-f'(y)f'''(y)}{f'(y)^5}.$$ The first result now follows from the fact that $y^2/f'(y)$ is increasing and tends to $4$ near zero. Using the fact that it evaluates to a constant at $\pi$, the other two bounds follow from upper-bounding $|yf''(y)|$ and $|f'''(y)|$ by constants; in fact, both are bounded by $1$. $\square$

Now, let $x=k\eps$; we seek to show that there exists a constant $C>0$ for which, for sufficiently large $k$, $$\left|\log\left(\frac{g(x+\eps)-g(x)}\eps\right)-\frac1\eps\int_x^{x+\eps}\log g'(t)dt\right|\leq Ck^{-2}.$$ Let $w=x+\eps/2$; we'll show that both terms are within $Ck^{-2}$ of $\log g'(w)$, which will be enough. For $-\eps/2\leq \delta\leq\eps/2$, write $$g'(w+\delta)=g'(w)+\delta g''(w)+\mathrm{err}(\delta),$$ and let $M=\sup_{|\delta|\leq\eps/2}|\mathrm{err}(\delta)|$. Note that $M=\frac{\delta^2}2g'''(z)$ for some $z\in(x,x+\eps)$ and some $\delta$ with $|\delta|\leq\eps/2$, so $$0\leq \frac{M}{g'(w)}\leq \frac{\frac{\eps^2}8 C_3z^{-8/3}}{C_1w^{-2/3}}\leq \frac{C_3\eps^2(2x)^{2/3}}{8C_1z^{8/3}}\leq\frac{C_3}{C_1k^2}.$$ Also, for $|\delta|\leq \eps/2$, $$\left|\frac{\delta g''(w)}{g'(w)}\right|\leq \frac{\eps C_2w^{-5/3}}{2C_1w^{-2/3}}=\frac{\eps C_2}{2C_1 w}\leq \frac{C_2}{2C_1k}.$$ We are now ready to bound the quantities at hand. We first compute \begin{align*} \left|\frac{g(x+\eps)-g(x)}\eps-g'(w)\right| &=\left|\frac1\eps\int_{-\eps/2}^{\eps/2} g'(w+\delta)d\delta-\frac1\eps\int_{-\eps/2}^{\eps/2}g'(w)+\delta g''(w)d\delta\right|\\ &=\left|\frac1\eps\int_{-\eps/2}^{\eps/2}\mathrm{err}(\delta)d\delta\right|\leq M, \end{align*} so that $$\log\left(\frac{g(x+\eps)-g(x)}\eps\right)-\log g'(w)=\log\left(g'(w)+\tilde M\right)-g'(w)=\log\left(1+\frac{\tilde M}{g'(w)}\right)\leq \frac{2C_3}{C_1k^2}$$ for large enough $k$ and for some $|\tilde M|\leq M$. Also, we compute \begin{align*} \left|\frac1\eps\int_x^{x+\eps}\log g'(t)dt-\log g'(w)\right| &=\left|\frac1\eps\int_{-\eps/2}^{\eps/2}\log\big(g'(w)+\delta g''(w)+\mathrm{err}(\delta)\big)d\delta-\frac1\eps\int_{-\eps/2}^{\eps/2}\log g'(w)d\delta\right|\\ &=\left|\frac1\eps\int_{-\eps/2}^{\eps/2}\log\left(1+\frac{\delta g''(w)}{g'(w)}+\frac{\mathrm{err}(\delta)}{g'(w)}\right)d\delta\right|. \end{align*} The $\frac{\delta g''(w)}{g'(w)}$ and $\frac{\mathrm{err}(\delta)}{g'(y)}$ terms are bounded by constants times $1/k$ and $1/k^2$, respectively. So, for large enough $k$, we can replace the integrand by $\frac{\delta g''(w)}{g'(w)}$, and we will incur an error of only a constant times $1/k^2$. After that replacement, this integrates to zero. So, we have $$\left|\frac1\eps\int_x^{x+\eps}\log g'(t)dt-\log g'(w)\right|,\left|\log\left(\frac{g(x+\eps)-g(x)}\eps\right)-\log g'(w)\right|\lesssim \frac C{k^2}$$ for some constant $C$. Hence the difference between these two, which is an upper bound for $h_\eps(k)$, is at most $2C/k^2$.


Part 3. Calculating the contribution of each small $k$.

This part will be quite messy. Fix $k$, and let $x=k\eps$. Write $a=g(k\eps)$ and $b=g((k+1)\eps)$. Since $k$ is fixed and $\eps\to 0$, $a$ and $b$ tend to zero, so we shall think of them as small. We can be slightly more precise; since $f(y)=\frac{y-\sin y}2=\frac{y^3}12(1-o(1))$, we have $g(x)=(12x)^{1/3}(1+o(1))$, so $$a=(12k\eps)^{1/3}(1+o(1))\text{ and }b=(12(k+1)\eps)^{1/3}(1+o(1)).$$ (Note that, if $k=0$, then $a$ is actually zero. We'll occasionally write $\log a$, but only when it is multiplied by a positive power of $a$, in which case the product can be thought of as defined to be zero when $a=0$.) We begin our analysis by massaging the expression of $h_\eps(k)$ into a nicer form: \begin{align*} h_\eps(k) &=\log\left(\frac{g(k\eps+\eps)-g(k\eps)}\eps\right)-\frac1\eps\int_{k\eps}^{(k+1)\eps}\log g'(t)dt\\ &=\log\left(\frac{b-a}\eps\right)-\frac1\eps\int_{f(a)}^{f(b)}\log g'(t)dt\\ &=\log\left(\frac{b-a}\eps\right)-\frac1\eps\int_a^bf'(u)\log\left(\frac1{f'(u)}\right)du. \end{align*} We now estimate the integrand. We have $$f'(u)=\frac{1-\cos(u)}2=\sin^2(u/2)=\frac{u^2}4(1+O(u^2)),$$ so \begin{align*} f'(u)\log\left(\frac1{f'(u)}\right) &=-\frac{u^2}4(1+O(u^2))\log\left(\frac{u^2}4(1+O(u^2))\right)\\ &=-\frac{u^2\log(u^2/4)}4(1+O(u^2))+\frac{u^2}4(1+O(u^2))\log(1+O(u^2))\\ &=-\frac{u^2\log(u^2/4)}4+O\left(u^4|\log u|\right). \end{align*} This gives $$h_\eps(k)=\log\left(\frac{b-a}\eps\right)+\frac1{\eps}\int_a^b\frac{u^2\log(u^2/4)}4du+O\left(\frac{b-a}\eps\max_{a\leq u\leq b}u^4|\log u|\right).$$ The bound $|\log u|<1/u$ (which is valid even very close to zero) suffices to bound the error --- we have $(b-a)b^3\leq b^4=O(\eps^{4/3})$, so the error term here is $o(1)$. We can now explicitly compute the integral in the above: \begin{align*} \int_a^b \frac{u^2\log(u^2/4)}4du &=\left(\frac{u^3\log(u^2/4)}{12}-\frac 1{18}u^3\right)\bigg|_a^b\\ &=\frac{b^3\log(b^2/4)-a^3\log(a^2/4)}{12}-\frac{b^3-a^3}{18}\\ &=\frac{a^3\log(b^2/a^2)}{12}+\frac{(b^3-a^3)\log(b^2/4)}{12}-\frac{b^3-a^3}{18}. \end{align*} We claim that $\frac{b^3-a^3}{12}=\eps\big(1+O(\eps^{2/3})\big)$. Indeed, \begin{align*} \eps=f(b)-f(a) &=\frac{(b-\sin(b))-(a-\sin(a))}2\\ &=\frac{b^3-a^3}{12}+O(b^5)=\frac{b^3-a^3}{12}(1+O(b^2))=\frac{b^3-a^3}{12}\big(1+O(\eps^{2/3})\big). \end{align*} This gives \begin{align*} h_\eps(k) &=\log\left(\frac{b-a}\eps\right)+\frac1{\eps}\int_a^b\frac{u^2\log(u^2/4)}4du+o(1)\\ &=\log\left(\frac{b-a}\eps\right)+\frac{a^3\log(b/a)}{6\eps}+\log(b^2/4)(1+O(\eps^{2/3}))-\frac23(1+o(1))+o(1). \end{align*} The term $\log(b^2/4)O(\eps^{-2/3})$ tends to zero, since $\eps^{2/3}$ goes to zero much more quickly than $|\log b|\sim\log(\eps^{-1})$ grows. We may also compute $$\log\left(\frac{b-a}\eps\right)=\log(b-a)-\log\left(\frac{b^3-a^3}{12}(1+o(1))\right)=\log\left(\frac{12}{a^2+ab+b^2}\right)+o(1).$$ So \begin{align*} h_\eps(k) &=\log\left(\frac{b-a}\eps\right)+\frac{a^3\log(b/a)}{6\eps}+\log(b^2/4)-\frac23+o(1)\\ &=\log\left(\frac{12}{a^2+ab+b^2}\right)+\log\left(\frac{b^2}4\right)+\frac{12k\eps\log\left((\frac{k+1}k)^{1/3}\right)}{6\eps}(1+o(1))+\frac 23+o(1)\\ &=\log\left(\frac{3b^2}{a^2+ab+b^2}\right)-\frac 23k\log\left(1+\frac 1k\right)(1+o(1))-\frac 23+o(1)\\ &=\log\left(\frac{3}{1+(a/b)+(a/b)^2}\right)+\frac 23k\log\left(1+\frac 1k\right)(1+o(1))-\frac 23+o(1)\\ &=\log\left(\frac{3}{1+r_k+r_k^2}\right)+\frac 23\left(k\log\left(1+\frac 1k\right)-1\right)+o(1), \end{align*} where we recall $r_k=(\frac k{k+1})^{1/3}$ and use that $a/b=r_k(1+o(1))$.


Part 4. Finishing up.

From the conclusion of Part 2, we have $$\log X=\lim_{n\to\infty}\sum_{k=0}^{n-1}h_\eps(k),$$ with $\eps=\frac{\pi}{2n}$. For each fixed $t$, as $n\to\infty$, we have \begin{align*} \sum_{k=0}^{n-1}h_\eps(k) &=\sum_{k=0}^{t-1}h_\eps(k)+\sum_{k=t}^{n-1}h_\eps(k)\\ &=\frac{C_t}t+o(1)+\sum_{k=0}^{t-1}\frac 23\left(k\log\left(1+\frac 1k\right)-1\right)+\log\left(\frac{3}{1+r_k+r_k^2}\right) \end{align*} by Parts 3 and 4, where $C_t$ denotes some function of $t$ and $n$ which is bounded in both variables. Taking $t\to\infty$ at some rate, we obtain $$\log X=\sum_{k=0}^\infty\frac 23\left(k\log\left(1+\frac 1k\right)-1\right)+\log\left(\frac{3}{1+r_k+r_k^2}\right).$$ We need only justify that this series actually converges, and express it in the form given at the beginning of the answer. We first pull out the $k=0$ term, as $r_0=0$, this is $\log 3-\frac 23$. For larger $k$, call this expression $T_k$, and let $a_k=k^{1/3}$ and $b_k=(k+1)^{1/3}$. We simplify \begin{align*} T_k&=\frac 23\left(k\log\left(1+\frac 1k\right)-1\right)+\log\left(\frac{3}{1+r_k+r_k^2}\right)\\ &=\frac 23\left(3k\log\left(\frac{b_k}{a_k}\right)-1\right)+\log\left(\frac{3b_k^2}{a_k^2+a_kb_k+b_k^2}\right)\\ &=2k\log\left(\frac{b_k}{a_k}\right)-\frac 23+\log\left(3b_k^2(b_k-a_k)\right)\\ &=(2k+1)\log\left(\frac{b_k}{a_k}\right)-\frac 23+\log\left(3a_kb_k(b_k-a_k)\right). \end{align*} We have $$1-(b_k-a_k)^3=b_k^3-a_k^3-(b_k-a_k)^3=3a_kb_k(b_k-a_k),$$ so we can simplify the term inside the sum. We also compute $$\sum_{k=1}^{N-1}(2k+1)\log\left(\frac{b_k}{a_k}\right)=\sum_{k=1}^{N-1}\frac{(2k+1)\log(k+1)-(2k+1)\log k}3=\frac{2N+1}3\log N-\frac 23\log(N!),$$ and so this term is, by Stirling's formula, $$\frac{2N+1}3\log N-\frac 23\left(N\log N-N+\frac12\log N+\log\sqrt{2\pi}+o(1)\right)=\frac{2N}3-\frac{\log2\pi}3+o(1).$$ This implies that $$\sum_{k=1}^{N-1}T_k=\frac23-\frac{\log2\pi}3+o(1)+\sum_{k=1}^{N-1}\log\left[1-\left((k+1)^{1/3}-k^{1/3}\right)^3\right].$$ Taking $N\to\infty$ and adding on the term from $k=0$ gives $$\log X=\log 3-\frac{\log2\pi}3+\sum_{k=1}^\infty\log\left[1-\left((k+1)^{1/3}-k^{1/3}\right)^3\right].$$ Exponentiating gives the result claimed in the introduction.

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  • $\begingroup$ Brilliant, thank you! Interesting method, and very nice explanations. Looking back, I'm not surprised that the limit does not seem to have a closed form. I wonder if the core reason for not having a closed form, is that $x-\sin x=c$ has no closed form solution for $x$. $\endgroup$
    – Dan
    Commented May 15 at 13:25
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    $\begingroup$ @Dan I don't think that's the reason. It certainly made the analysis more messy, but I suspect that the same sort of expression would pop out if $\frac{x-\sin x}2$ were replaced by any function which grows like $x^3$ near zero. The arguments in Part 2 can be made to show that, if we had a function $f$ whose derivative was bounded below, the answer would just be $1$ (regardless of if we had a closed-form expression for $g$). I believe the ugly expression for the answer comes from the fact that the "discretization" which is happening here is a bit unnatural, in terms of where $\log$ shows up. $\endgroup$ Commented May 15 at 20:38
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    $\begingroup$ I've asked about a closed form expression for the answer at MO. $\endgroup$
    – Dan
    Commented May 16 at 22:22
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    $\begingroup$ This has to be one of the best answers I've ever seen on MSE. Amazing! $\endgroup$
    – KStarGamer
    Commented May 16 at 23:49
  • $\begingroup$ This is more than impressive ! Thaks for posting such kind of answers. Cheers :-) $\endgroup$ Commented May 21 at 14:39
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Too long for a comment.

Starting from the first line of @Carl Schildkraut's more than impressive solution $$A=\frac{3}{(2\pi)^{1/3}}\prod_{k=1}^\infty\left[1-\big((k+1)^{1/3}-k^{1/3}\big)^3\right]$$

Taking logarithms, consider $$\sum_{k=1}^\infty \log\left[1-\big((k+1)^{1/3}-k^{1/3}\big)^3\right]=\sum_{k=1}^\infty a_k$$

The convergence has to be quite slow since $$\frac {a_{k+1}}{a_k}=1-\frac{2}{k}+\frac{4}{k^2}+O\left(\frac{1}{k^3}\right)$$

Expanding the summand as Taylor series, it is $$ \log\left[1-\big((k+1)^{1/3}-k^{1/3}\big)^3\right]=-\frac{1}{27 k^2}\sum_{n=0}^\infty (-1)^n \,\,\frac{b_n}{k^n}$$ where the first coefficients are $$\left\{1,1,\frac{49}{54},\frac{22}{27},\frac{3215}{4374} ,\frac{487}{729},\frac{16055}{26244},\frac{3701}{6561 },\frac{1390556}{2657205},\frac{259391}{531441},\frac {78754373}{172186884},\cdots\right\}$$ Even using many terms, the convergence is very slow.

At the opposite, making the series an $[n,n+2]$ Padé approximant $P_n$, the simplest being $$P_2=-\frac{324 k^2}{8748 k^4+8748 k^3+810 k^2-37}$$ we have explicit results for the infinite summation in terms of polygamma functions.

To give an idea of the accuracy of these approximations, conisder the infinite norm $$\Phi_n=\int_1^\infty \Big(\log\left[1-\big((k+1)^{1/3}-k^{1/3}\big)^3\right] -P_n\Big)^2\,dk$$ we have $\Phi_2=3.61\times 10^{-11}$, $\Phi_3=1.75\times 10^{-15}$, $\Phi_4=4.39\times 10^{-17}$.

Compared to the Taylor series, the error of these approximants is $$\Delta_n \sim \frac {e^{-(3n+4)}}{k^{2n+3}}$$

For a few values of $n$, the results

$$\left( \begin{array}{cc} n & A_{(n)} \\ 2 & 1.5681035248415 \\ 3 & 1.5681341279698 \\ 4 & 1.5681338490782 \\ 5 & 1.5681338908977 \\ 6 & 1.5681338907207 \\ 7 & 1.5681338907985 \\ 8 & 1.5681338907980 \\ \end{array} \right)$$

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