Derivation of Wallis's Formula

Question

Use Euler's product formula $\Gamma (z)={1\over z}\prod_{n=1}^\infty({1+{1\over n}})^z({1+{z\over n}})^{-1}$ the fact that $\Gamma ({1\over 2})=\sqrt\pi$ to prove Wallis's Formula ${\pi\over 4}={2\over 3}\cdot {4\over 3}\cdot{4\over 5}\cdot{6\over 5}\cdots{2n\over 2n+1}\cdot{2n+2\over 2n+1}\cdots$.

Answer 1

Consider Euler's product formula $\Gamma (z)={1\over z}\prod_{n=1}^\infty(1+{1\over n})^z (1+{z\over n})^{-1}$ and $\Gamma ({1\over 2})=\sqrt \pi$. Thus $$ \sqrt \pi = 2\prod_{n=1}^\infty (1+{1\over n})^{1\over 2} ({1+{1\over 2n}})^{-1}. $$ Multiplying both sides of the equation by ${1\over 2}$ and rewriting the right-hand side we obtain $${\sqrt \pi\over 2}=\prod_{n=1}^\infty ({n+1\over n})^{1\over 2}({2n\over 2n+1}).$$ Squaring both sides of the equation gives us $${\pi\over 4}=\prod_{n=1}^\infty({n+1\over n})({4n^2\over (2n+1)^2}).$$ Simplifying the right-hand side we see that $${\pi\over 4}=\prod_{n=1}^\infty({4n^2+4n\over (2n+1)^2}).$$ Thus factoring out $2n$ on the numerator and rewriting the right-hand side we have Wallis's Formula $${\pi\over 4}=\prod_{n=1}^\infty({2n\over 2n+1})({2n+2\over 2n+1}).$$

Answer 2

First of all from the product representation for the $\Gamma$-function it follows that $$ \left( z \Gamma(z) \right)^2 = \prod_{n=1}^\infty \frac{ (n+1)^{2z} }{n^{2z-2} (n+z)^2} $$ substituting $z=\tfrac{1}{2}$ we have: $$ \frac{\pi}{4} = \prod_{n=1}^\infty \frac{2 n}{2n+1} \cdot \frac{2n+2}{2n+1} = \frac{2}{3} \prod_{n=1}^\infty \frac{2 n+2}{2n+3} \cdot \frac{2n+2}{2n+1} $$