Show that $\prod_{k=1}^\infty \frac{2k+1}{2\pi}\sin{\left(\frac{2\pi}{2k+1}\right)}\sec{\left(\frac{\pi}{k+2}\right)}=\frac{\pi}{2}$

Question

Show that

$$\prod\limits_{k=1}^\infty \frac{2k+1}{2\pi}\sin{\left(\frac{2\pi}{2k+1}\right)}\sec{\left(\frac{\pi}{k+2}\right)}=\frac{\pi}{2}$$

Context:

Inspired by this question, I considered the product of the areas of every regular odd-gon inscribed in a circle of area $1$. This is $\prod\limits_{k=1}^\infty \frac{2k+1}{2\pi} \sin{\frac{2\pi}{2k+1}}=0.18055...$ which I guess does not have a closed form.

On a whim, I divided this number by the Kepler-Bouwkamp constant, $\prod\limits_{k=1}^\infty \cos{\frac{\pi}{k+2}}=0.11494...$, and numerical calculation suggests that the result is $\pi/2$.

My attempt:

I used the double angle formula for sine to get $\prod\limits_{k=1}^\infty \frac{2k+1}{\pi}\sin{\left(\frac{\pi}{2k+1}\right)}\sec{\left(\frac{\pi}{2k+2}\right)}$ but this seems just as intractable as the original form of the product.

Answer 1

Let

$$K = \prod_{k=3}^\infty \frac1{\cos\left(\frac\pi k\right)}$$

As Bouwkamp points out under First method in this paper,

$$\cos(x) = \frac{\sin(2x)}{2x} \cdot \frac x{\sin(x)} \tag{1}$$

Then

$$\begin{align*} \frac1K &= \prod_{k=3}^\infty \cos\left(\frac\pi k\right) \\[1ex] &= \prod_{k=3}^\infty \frac{\sin\left(\frac{2\pi}k\right)}{\frac{2\pi}k} \cdot \prod_{k=3}^\infty \frac{\frac\pi k}{\sin\left(\frac\pi k\right)} \tag{1}\\[1ex] &= \left[\prod_{k=1}^\infty \frac{\sin\left(\frac{2\pi}{2k+1}\right)}{\frac{2\pi}{2k+1}} \cdot \prod_{k=2}^\infty \frac{\sin\left(\frac{2\pi}{2k}\right)}{\frac{2\pi}{2k}}\right] \cdot \prod_{k=3}^\infty \frac{\frac\pi k}{\sin\left(\frac\pi k\right)} \tag{2}\\[1ex] &= \left[\prod_{k=1}^\infty \frac{\sin\left(\frac{2\pi}{2k+1}\right)}{\frac{2\pi}{2k+1}}\right] \cdot \frac2\pi \end{align*}$$

where in $(2)$ we split the product over even and odd indices.

Answer 2

Interesting product. Let us define the Kepler-Bouwkamp constant as $c$. Then

$$c=\prod_{k=1}^\infty \cos\left(\frac{\pi}{k+2}\right)$$

Let $\text{Si}(x)=\frac{\sin(x)}{x}$. Then, $\cos(x)=\frac{\text{Si}(2x)}{\text{Si}(x)}$. Plugging this in, we get

$$c=\prod_{k=1}^\infty\frac{\text{Si}\left(\frac{2\pi}{k+2}\right)}{\text{Si}\left(\frac{\pi}{k+2}\right)}$$

Now, observe that every term in the denominator cancels with a term in the numerator because $\frac{\pi}{k+2}=\frac{2\pi}{2(k+1)}$. All that remains is the second term in the numerator and the odd terms of the numerator, viz

$$c=\prod_{k=1}^\infty \text{Si}\left(\frac{2\pi}{2+2}\right)\text{Si}\left(\frac{2\pi}{2k+1}\right)=\text{Si}\left(\frac{\pi}{2}\right)\prod_{k=1}^\infty \text{Si}\left(\frac{2\pi}{2k+1}\right)$$

Simplifying, and plugging in the actual product expression for $c$, we have

$$\prod_{k=1}^\infty \cos\left(\frac{\pi}{k+2}\right)=\frac{2}{\pi}\prod_{k=1}^\infty\frac{2k+1}{2\pi}\sin\left(\frac{2\pi}{2k+1}\right)$$

Rearranging gives us that $$\frac{\pi}{2}=\prod_{k=1}^\infty\frac{2k+1}{2\pi}\sin\left(\frac{2\pi}{2k+1}\right)\sec\left(\frac{\pi}{k+2}\right)$$

Answer 3

This one seems to be more difficult than the previous and, for the time being, I am stuck.

I did not find where was asked $$P=\prod\limits_{k=1}^\infty \frac{2k+1}{\pi}\sin{\left(\frac{\pi}{2k+1}\right)}$$ for which I post here some ideas. Consider $$S=\sum\limits_{k=1}^\infty \log\Bigg[\frac{2k+1}{\pi}\sin{\left(\frac{\pi}{2k+1}\right)}\Bigg]$$ which, using the series expansion of the logarithm and summing over $k$, write $$S=\frac 14\sum\limits_{n=1}^\infty(-1)^n\frac{ ( 2\pi) ^{2 n}\, B_{2 n}\, \zeta \left(2 n,\frac{3}{2}\right)}{n^2\, \Gamma (2 n)}$$

For the time being, I did not find any closed form but the partial sum converges quite fast as accurately as we wish to $$S=-1.711744530156267356949288\cdots$$ giving $$P=0.1805505418498519239123726\cdots$$