- Prove Wallis’s product formula
\[ {\frac {\pi}{2}} = {\frac {2 \cdot 2}{1 \cdot 3}} \cdot {\frac {4 \cdot 4}{3 \cdot 5}} \dots {\frac {2 m \cdot 2 m}{(2 m - 1) \cdot (2 m + 1)}} \dots . \]
[Hint: Use the product formula for sin z at \(z = \pi / 2 . ]\)
Solution
The product formula for the sine function is \[ \frac{\sin z}{z}=\prod_{n=1}^{\infty}\left(1-\frac{z^2}{\pi^2n^2}\right). \] Set \(z=\pi/2\). Since \(\sin(\pi/2)=1\), \[ \frac{2}{\pi} =\prod_{n=1}^{\infty}\left(1-\frac1{4n^2}\right) =\prod_{n=1}^{\infty} \frac{(2n-1)(2n+1)}{(2n)^2}. \] Taking reciprocals gives \[ \boxed{ \frac{\pi}{2} =\prod_{n=1}^{\infty} \frac{(2n)^2}{(2n-1)(2n+1)}}. \] This is Wallis’s product formula.