Proving that $\frac{\pi}{2}=\prod_{k=2}^{\infty}\left(1+\frac{(-1)^{(p_{k}-1)/2}}{p_{k}} \right )^{-1}$ an identity of Euler's.
- Decompose the product on the right as $$\prod_{\text{primes}\; p\\ \text{ of the form }4k+1}\left(1+\frac{1}{p}\right)\prod_{\text{primes}\; p\\ \text{ of the form }4k+3}\left(1-\frac{1}{p}\right)$$
- Consider an odd integer $n=2m+1$. It is "easy to see" that if primes of the form $4k+3$ appear in its prime number decomposition an even number of times, then $n$ is of the form $4K+1$ [since $(4k_1+3)(4k_2+3)=1\; \mathrm{mod}\;4$]. If the number of such appearances is odd, then $n$ is of the form $4K+3$.
- Rewrite the right side as $$ \left[\prod_{\mathrm{odd}\;\mathrm{primes}\; p}\left(1-\frac{1}{p^2}\right)\right]^{-1}\times\frac{1}{\left[\prod_{p\\ \text{ of the form }4k+1}\left(1-\frac{1}{p}\right)\prod_{p\\ \text{ of the form }4k+3}\left(1+\frac{1}{p}\right)\right]^{-1}}\qquad\tag{1}$$ Expanding the first factor into geometric series, we find $$ A=\left[\prod_{\mathrm{odd}\;\mathrm{primes}\; p}\left(1-\frac{1}{p^2}\right)\right]^{-1}=\sum_{m=0}^{\infty}\frac{1}{(2m+1)^2}=\frac{\pi^2}{8}$$
- Similarly reexpanding the denominator of the 2nd factor in (1), we have $$ B=\left[\prod_{p\\ \text{ of the form }4k+1}\left(1-\frac{1}{p}\right)\prod_{p\\ \text{ of the form }4k+3}\left(1+\frac{1}{p}\right)\right]^{-1}=\sum_{m=0}^{\infty}\frac{(-1)^{r(m)}}{2m+1},$$ where $r(m)$ counts the number of appearances of primes of the form $4k+3$ in the decomposition of $2m+1$. But then Step 2 allows to write $(-1)^{r(m)}=(-1)^m$, so that $$ B=\sum_{m=0}^{\infty}\frac{(-1)^{m}}{2m+1}=\frac{\pi}{4}$$ and $\displaystyle \frac{A}{B}=\frac{\pi^2/8}{\pi/4}=\frac{\pi}{2}$. $\blacksquare$