$ \frac{\pi}{4} = \frac{3}{4} \times \frac{5}{4} \times \frac{7}{8} \times \frac{11}{12} \times \cdots$ (just primes in the numerator) How to prove the following equality?
$$ \frac{\pi}{4} = \frac{3}{4} \times \frac{5}{4} \times \frac{7}{8} \times \frac{11}{12} \times \frac{13}{12} \times \frac{17}{16} \times \frac{19}{20} \times \frac{23}{24} \times \frac{29}{28} \times \frac{31}{32} \times \cdots$$
I've seen this formula on Wikipedia.
The numerators are the odd primes; each denominator is the multiple of four nearest to the numerator.
Thanks.

Your product is nothing but
$$\lim_{n \to \infty}\prod_{\overset{p_k \leq n}{p_k \equiv -1 \pmod4}} \left(1+\dfrac1{p_k}\right)^{-1} \prod_{\overset{p_k \leq n}{p_k \equiv 1 \pmod4}} \left(1-\dfrac1{p_k}\right)^{-1}$$
Now recall that
$$\left(1-\dfrac1r\right)^{-1} = \sum_{k=0}^{\infty} \dfrac1{r^k}$$
Make use of the above to note that
$$\lim_{n \to \infty}\prod_{\overset{p_k \leq n}{p_k \equiv -1 \pmod4}} \left(1+\dfrac1{p_k}\right)^{-1} \prod_{\overset{p_k \leq n}{p_k \equiv 1 \pmod4}} \left(1-\dfrac1{p_k}\right)^{-1} = \lim_{m \to \infty} \sum_{\ell=0}^m \dfrac{(-1)^\ell}{2\ell+1} = \dfrac{\pi}4$$
The last step makes use of the fact that
$$4\ell+1 = \prod_{\overset{q_k \equiv 1 \pmod4}{k=1,2,\ldots,n_1}} q_k^{a_k} \prod_{\overset{p_j \equiv -1 \pmod4}{j=1,2,\ldots,n_2}} p_j^{b_j}$$
such that $\displaystyle \sum_{j=1}^{n_2}b_j = \text{even}$ and 
$$4\ell-1 = \prod_{\overset{q_k \equiv 1 \pmod4}{k=1,2,\ldots,n_1}} q_k^{a_k} \prod_{\overset{p_j \equiv -1 \pmod4}{j=1,2,\ldots,n_2}} p_j^{b_j}$$
such that $\displaystyle \sum_{j=1}^{n_2}b_j = \text{odd}$.
