# Evaluate $\lim_{n\to\infty}\prod_{k=1}^n \frac{2n}{e}(\arcsin(\frac{k}{n})-\arcsin(\frac{k-1}{n}))$

I'm trying to evaluate $$L=\lim\limits_{n\to\infty}f(n)$$ where

$$f(n)=\prod\limits_{k=1}^n \frac{2n}{e}\left(\arcsin\left(\frac{k}{n}\right)-\arcsin\left(\frac{k-1}{n}\right)\right)$$

We have:
$$f(1)\approx1.15573$$
$$f(10^2)\approx1.22410$$
$$f(10^4)\approx1.22533$$
$$f(10^6)\approx1.22535$$

My attempt

I've only been able to find the following equivalent expressions for $$L$$.

$$L=\lim\limits_{n\to\infty}\exp\left(\sum\limits_{k=1}^n\log\left(\frac{2n}{e}\left(\arcsin\left(\frac{k}{n}\right)-\arcsin\left(\frac{k-1}{n}\right)\right)\right)\right)$$

$$L=\lim\limits_{n\to\infty}\exp\left(\sum\limits_{k=1}^n\log\left(\frac{2n}{e}\arcsin\left(\frac{k\sqrt{n^2-(k-1)^2}-(k-1)\sqrt{n^2-k^2}}{n^2}\right)\right)\right)$$

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

Context

The limit comes from the following geometrical construction that I made up.

In a semicircle with diameter $$na$$ where $$a=\frac{2}{e}$$, draw chords of lengths $$a,2a,3a,\dots,na$$ that meet at one of the endpoints of the semicircle. Here is an example with $$n=6$$.

Consider the arcs between neighboring chord endpoints. The product of the $$n$$ arc lengths is $$f(n)$$ from above.

Why I think $$L$$ might have a closed form

I think $$L$$ might have a closed form, because a similar geometrical construction seems to yield a limit with a closed form.

Instead of starting with a semicircle (which is a cicular segment with central angle $$\pi$$), let's start with a circular segment with central angle $$\pi/3$$, with a bounding chord of length $$nb$$ where $$b=\frac{3\sqrt3}{2e}$$. Draw chords of lengths $$b,2b,3b,\dots,nb$$ that meet at one of the end points of the arc. Here is an example with $$n=6$$.

Consider the arcs between neighboring chord endpoints. The product of the $$n$$ arc lengths is $$g(n)=\prod\limits_{k=1}^n\frac{3\sqrt3n}{e}\left(\arcsin\left(\frac{k}{2n}\right)-\arcsin\left(\frac{k-1}{2n}\right)\right)$$.

Numerical investigation suggests that $$\lim\limits_{n\to\infty}g(n)=1$$.

• Just to confirm: here $e = 2.718\ldots$, not some arbitrary number? Commented May 15 at 14:46
• @MichaelLugo Correct, $e=2.718\dots$.
– Dan
Commented May 15 at 14:47
• Can you show how you found out the arccos part? It will be very helful
– Gwen
Commented May 15 at 15:16
• @Gwen I used the fact that $A-B=\arccos(\cos(A-B))$ if $0<A-B<\pi$.
– Dan
Commented May 15 at 15:20
• $$L = \sqrt[4]{{\frac{{{\rm e}^\gamma }}{{2\pi }}}}\exp \left( {\sum\limits_{n = 2}^\infty {\binom{2n}{n}\frac{{\zeta (n)}}{{2^{2n} \cdot 2n}}} } \right) = \sqrt[4]{{\frac{{8{\rm e}^{\gamma - 1} }}{\pi }}}\exp \left( {\sum\limits_{n = 2}^\infty {\binom{2n}{n}\frac{{\zeta (n) - 1}}{{2^{2n} \cdot 2n}}} } \right) \\= 1.2253517701107728567819327 \ldots$$
– Gary
Commented May 17 at 1:22

Long comment. What I proved in this posting about an asymptotic formula for a related product actually works for $$\alpha, \beta > -1$$ (see the statement of the corollary in the link). The upshot of this asymptotic formula in OP's case is that
\begin{align*} &\prod_{k=1}^{n} \frac{2n}{e} \left[ \arcsin\left(\frac{k}{n}\right) - \arcsin\left(\frac{k-1}{n}\right) \right] \\ &\sim \frac{1}{(2\pi n)^{1/4}} \prod_{k=1}^{n} \frac{2}{1 + \sqrt{1-k^{-1}}} \\ &\sim \sqrt[4]{\frac{e^{\gamma}}{2\pi}} \exp\Biggl[ \sum_{k=1}^{\infty} \log\Biggl( \frac{2}{1 + \sqrt{1-k^{-1}}} \Biggr) -\frac{1}{4k} \Biggr] \\ &\approx 1.2253517701\ldots \end{align*}
• Thanks. In my question, in the section called "Why I think $L$ might have a closed form", I show that there is numerical evidence for $\lim\limits_{n\to\infty}g(n)$ having a closed form. What makes $\lim\limits_{n\to\infty}f(n)$ and $\lim\limits_{n\to\infty}g(n)$ different, in terms of having a closed form?
• @Dan, The limiting constant essentially arises from the product of the form $$\prod_{k=1}^{n}\int_{\frac{k-1}{n}}^{\frac{k}{n}}t^{\alpha}\,\mathrm{d}t,\tag{*}$$ where $\alpha$ comes from the asymptotic behavior of your integrand, $(\arcsin x)'=\frac{1}{1-x^2}$ or $(\arcsin(x/2))'=\frac{1}{4-x^2}$, near the boundary of $[0, 1]$. Now this $\alpha$ is different in the two cases, one admitting telescoping of terms in $\text{(*)}$ to yield a closed form, while the other case not. (I know this comment lacks lots of details, and it would be helpful to give a look on the link in my answer.) Commented May 18 at 10:38