TITLE: Closed form for an infinite product
QUESTION [15 upvotes]: The following fascinating formula appears in the paper "On gamma quotients and infinite products" by M.Chamberland and A.Straub (see page 9):
$$\prod_{n=2}^\infty\left(1-\frac{1}{n(n-1)}\right)=-\frac1\pi\cos\left(\frac{\sqrt5}{2}\pi\right).$$
Unfortunately, it's given without any reference or explanation. Could anyone give one?

REPLY [4 votes]: We want to compute:
$$S=\sum_{n\geq 1}\log\left(1-\frac{1}{n(n+1)}\right)=\sum_{n\geq 1}\int_{0}^{+\infty}\frac{e^{-(n^2+n-1)x}-e^{-(n^2+n)x}}{x}\,dx$$
where the identity follows from Frullani's theorem. Since $\int_{0}^{+\infty}\frac{f(x)}{x}\,dx=\int_{0}^{+\infty}\mathcal{L}(f)(s)\,ds$,
$$ S = \int_{0}^{+\infty}\sum_{n\geq 1}\left(\frac{1}{n^2+n-1+s}-\frac{1}{n^2+n+s}\right)\,ds $$
that is just:
$$ S = 4\int_{0}^{1}\sum_{n\geq 1}\frac{1}{(2n+1)^2+(s-5)}\,ds=\int_{0}^{1}\left(\frac{1}{1-s}+\frac{\pi\tan\left(\frac{\pi}{2}\sqrt{5-4s}\right)}{\sqrt{5-4s}}\right)\,ds$$
or:
$$ S = \int_{0}^{4}\left(\frac{1}{s}+\frac{\pi\tan\left(\frac{\pi}{2}\sqrt{1+s}\right)}{4\sqrt{1+s}}\right)\,ds=\int_{1}^{\sqrt{5}}\left(\frac{2s}{s^2-1}+\frac{\pi}{2}\,\tan\left(\frac{\pi s}{2}\right)\right)\,ds $$
so:
$$ S = \left. \log(s^2-1)-\log\cos\frac{\pi s}{2}\right|_{1}^{\sqrt{5}} = \log\left(-\pi\sec\frac{\pi\sqrt{5}}{2}\right)$$
and by exponentiating:
$$ \prod_{n\geq 1}\left(1-\frac{1}{n(n+1)}\right) = -\frac{1}{\pi}\,\cos\frac{\pi\sqrt{5}}{2}$$
as wanted.