Prove that ${4\over \pi}=\prod_{k=1}^{\infty}\left(1+{1\over 4k}\right)^2\left(2k+1\over 2k+1+(-1)^{k-1}\right)^{(-1)^{k-1}}$ 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.  
 A: 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.
