How do I prove this infinite product?
$${4\over \pi}=\prod_{k=1}^{\infty}\left(1+{1\over 4k}\right)^2\left(2k+1\over 2k+1+(-1)^{k-1}\right)^{(-1)^{k-1}}$$
I try:
$$\prod_{k=1}^{\infty}\left(1+{1\over 4k}\right)^2={5\over 4}\cdot{9\over 8}\cdot{13\over 12}\cdots={144\over 121}$$
$$\prod_{k=1}^{\infty}\left(2k+1\over 2k+1+(-1)^{k-1}\right)^{(-1)^{k-1}}={3\over 4}\cdot{4\over 5}\cdot{7\over 8}\cdot{8\over 9}\cdots$$
Simplified to get
$$\prod_{k=1}^{\infty}\left(2k+1\over 2k+1+(-1)^{k-1}\right)^{(-1)^{k-1}}={3\over 1}\cdot{1\over 5}\cdot{7\over 1}\cdot{1\over 9}\cdot{11\over 1}\cdot{1\over 13}\cdots$$
How do I combine theses two products to show that it is a Wallis's product? This is as far I can go.
Hint. Let's set $$ P_n:=\prod_{k=1}^{n}\left(1+{1\over 4k}\right)^2\left(2k+1\over 2k+1+(-1)^{k-1}\right)^{(-1)^{k-1}}. $$ One may observe that $$ P_{2n+1}=\left(1+{1\over 4(2n+1)}\right)^2\left(4n+3\over 4n+4\right)P_{2n}, \quad n\ge0, $$ since $$ \lim_{n \to \infty}\left(1+{1\over 4(2n+1)}\right)^2\left(4n+3\over 4n+4\right)=1 $$thus, if the limits exist, we have $$ \lim_{n \to \infty}P_{2n+1}=\lim_{n \to \infty}P_{2n}. $$
Then one has, as $n \to \infty$, $$ \begin{align} P_{2n}&=\prod_{k=1}^{2n}\left(1+{1\over 4k}\right)^2\left(2k+1\over 2k+1+(-1)^{k-1}\right)^{(-1)^{k-1}} \\\\&=\prod_{k=1}^{2n}\left(1+{1\over 4k}\right)^2\cdot\prod_{k=1}^{2n}\left(2k+1\over 2k+1+(-1)^{k-1}\right)^{(-1)^{k-1}} \\\\&=\left(\frac{\Gamma\left(2n+\frac{5}{4}\right)}{\Gamma\left(\frac{5}{4}\right) \Gamma\left(2n+1\right)}\right)^2\cdot \prod_{p=1}^{n}\left(4p+1\over 4p\right)^{-1}\prod_{p=1}^{n}\left(4p-1\over 4p\right) \\\\&=\left(\frac{\Gamma\left(2n+\frac{5}{4}\right)}{\Gamma\left(\frac{5}{4}\right) \Gamma\left(2n+1\right)}\right)^2\cdot\frac{\Gamma\left(\frac{5}{4}\right)\Gamma\left(n+\frac{3}{4}\right)}{\Gamma\left(\frac{3}{4}\right)\Gamma\left(n+\frac{5}{4}\right)} \\\\& = \frac{4}\pi+\mathcal{O}\left(\frac1n \right) \end{align} $$ where we have used the generalized Stirling formula.