A Wallis-like formula for $\pi$: $(\frac21)^2(\frac23)^2(\frac43)^2(\frac45)(\frac65)^2(\frac67)^2(\frac87)(\frac89)^2\cdots$

I don't quite understand this and would appreciate a clear explanation, if possible.

If we begin with the Wallis product for $\frac{\pi}{2}$,

$$\prod_{n=1}^\infty \left( \frac{2n}{2n-1} \cdot \frac{2n}{2n+1} \right) = \Big(\frac{2}{1} \cdot \frac{2}{3}\Big) \cdot \Big(\frac{4}{3} \cdot \frac{4}{5}\Big) \cdot \Big(\frac{6}{5} \cdot \frac{6}{7}\Big) \cdot \Big(\frac{8}{7} \cdot \frac{8}{9}\Big) \cdot \; \cdots \\ = \Big(\frac{2}{1}\Big) \cdot \Big( \frac{2}{3}\Big) \cdot \Big(\frac{4}{3}\Big) \cdot \Big(\frac{4}{5}\Big) \cdot \Big(\frac{6}{5}\Big) \cdot \Big(\frac{6}{7}\Big) \cdot \Big(\frac{8}{7}\Big) \cdot \Big(\frac{8}{9}\Big) \cdot \; \cdots \\ = \frac{\pi}{2}$$

By squaring only factors whose numerator and denominator sum to primes, we obtain

$$\Big(\frac{2}{1}\Big)^2 \cdot \Big( \frac{2}{3}\Big)^2 \cdot \Big(\frac{4}{3}\Big)^2 \cdot \Big(\frac{4}{5}\Big) \cdot \Big(\frac{6}{5}\Big)^2 \cdot \Big(\frac{6}{7}\Big)^2 \cdot \Big(\frac{8}{7}\Big) \cdot \Big(\frac{8}{9}\Big)^2 \cdot \; \cdots = \pi\\$$

Why?

Solution 1:

We have to show that the product of those factors is (convergent and) equal to $2$.

The factors are precisely $\frac{p-\chi(p)}{p+\chi(p)}$, where $p$ runs through odd primes, and $$\chi(n)=\begin{cases}(-1)^{(n-1)/2},&n\text{ is odd}\\\hfill 0,\hfill&n\text{ is even}\end{cases}$$ is known as the non-trivial Dirichlet character modulo $4$, and we know that $$\prod_p\big(1-\chi(p)p^{-s}\big)^{-1}=\sum_{n>0}\chi(n)n^{-s}=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^s}\qquad(\Re s>1)$$ since $\chi$ is completely multiplicative. The (harder!) fact that the equality holds at $s=1$ is a particular case of the result discussed here; you may also want to follow the links out there. The sum at $s=1$ is well-known to be $\pi/4$, thus $\prod_p\big(1-\chi(p)/p\big)=4/\pi$. As $$\frac{1-\chi(p)/p}{1+\chi(p)/p}=\frac{\big(1-\chi(p)/p\big)^2}{1-\big(\chi(p)/p\big)^2}=\frac{\big(1-\chi(p)/p\big)^2}{1-1/p^2},$$ and $\prod_p(1-1/p^2)^{-1}=(1-1/4)$$\zeta(2)$$=\pi^2/8$ since the product is over odd primes, we find $$\prod_p\frac{p-\chi(p)}{p+\chi(p)}=\left(\frac{4}{\pi}\right)^2\frac{\pi^2}{8}=2$$ as expected.