1
$\begingroup$

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 $n=6$.

enter image description here

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 $n=6$.

enter image description here

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 $\lim\limits_{n\to\infty}g(n)=1$.

$\endgroup$
7
  • 1
    $\begingroup$ Just to confirm: here $e = 2.718\ldots$, not some arbitrary number? $\endgroup$
    – Michael Lugo
    Commented May 15 at 14:46
  • 1
    $\begingroup$ @MichaelLugo Correct, $e=2.718\dots$. $\endgroup$
    – Dan
    Commented May 15 at 14:47
  • $\begingroup$ Can you show how you found out the arccos part? It will be very helful $\endgroup$
    – Gwen
    Commented May 15 at 15:16
  • $\begingroup$ @Gwen I used the fact that $A-B=\arccos(\cos(A-B))$ if $0<A-B<\pi$. $\endgroup$
    – Dan
    Commented May 15 at 15:20
  • 1
    $\begingroup$ $$ L = \sqrt[4]{{\frac{{{\rm e}^\gamma }}{{2\pi }}}}\exp \left( {\sum\limits_{n = 2}^\infty {\binom{2n}{n}\frac{{\zeta (n)}}{{2^{2n} \cdot 2n}}} } \right) = \sqrt[4]{{\frac{{8{\rm e}^{\gamma - 1} }}{\pi }}}\exp \left( {\sum\limits_{n = 2}^\infty {\binom{2n}{n}\frac{{\zeta (n) - 1}}{{2^{2n} \cdot 2n}}} } \right) \\= 1.2253517701107728567819327 \ldots $$ $\endgroup$
    – Gary
    Commented May 17 at 1:22

1 Answer 1

Reset to default
4
$\begingroup$

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*}

I am skeptical that this expression has an elementary closed form.

$\endgroup$
2
  • $\begingroup$ 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? $\endgroup$
    – Dan
    Commented May 16 at 22:04
  • 1
    $\begingroup$ @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.) $\endgroup$
    – Sangchul Lee
    Commented May 18 at 10:38

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .