# Show that:$\prod_{n=1}^{\infty}\left({2n\over 2n-1}\right)^2\left(\cdots\right)={\pi\over2}\cdot{\phi\over5}\cdot\sqrt{\phi\sqrt{5}}$

$\phi$ is the golden ratio

Show that

$$\prod_{n=1}^{\infty}\left({2n\over 2n-1}\right)^2\left(10n-6\over 10n-1\right)\left(10n-4\over 10n+1\right)={\pi\over2}\cdot{\phi\over5}\cdot\sqrt{\phi\sqrt{5}}$$

I try:

$$\prod_{n=1}^{\infty}\left({2n\over 2n-1}\right)^2\left(10n-6\over 10n-1\right)\left(10n-4\over 10n+1\right)={\phi\over5}\cdot\sqrt{\phi\sqrt{5}}\prod_{n=1}^{\infty}\left({2n\over 2n-1}\cdot{2n\over 2n+1}\right)$$

$$\prod_{n=1}^{\infty}\left({2n+1\over 2n-1}\right)\left(10n-6\over 10n-1\right)\left(10n-4\over 10n+1\right)={\phi\over5}\cdot\sqrt{\phi\sqrt{5}}$$

$$\lim_{M\to \infty}(2M+1)\prod_{n=1}^{M}\left(10n-6\over 10n-1\right)\left(10n-4\over 10n+1\right)={\phi\over5}\cdot\sqrt{\phi\sqrt{5}}$$

• Hint: Use general Vieta-Wallis product for $\pi$. Commented Dec 26, 2016 at 20:29
Rewrite this as the product of $$\frac{(2n)^2}{(2n-1)(2n+1)} \cdot \frac{(10n)^2}{(10n-1)(10n+1)} \cdot \frac{(10n/4-1)(10n/4+1)}{(10n/4)^2} \cdot \frac{(2n+1)(10n-6)}{(2n-1)(10n+4)}.$$ Using the infinite product representation of sine, you can show that $$\frac{\pi}{m \sin (\pi / m)} = \prod_{n=1}^{\infty} \frac{(mn)^2}{(mn-1)(mn+1)}$$ which takes care of the first three products. The fourth product telescopes to $4/5$. So the result is $$\frac{\pi}{2\sin(\pi/2)} \cdot \frac{\pi}{10 \sin (\pi/10)} \cdot \frac{5 \sin (2\pi/5)}{2\pi} \cdot \frac{4}{5}$$ in which you can use known values for $\sin(2\pi/5)$ and $\sin(\pi/10).$