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En utilisant \(I_{m+1}\leq J_m\leq I_{m}\), obtenir: \[\frac{2m+1}{2m+2} \frac{(1.3.\cdots.(2m-1))(3.5.\cdots.(2m+1))}{(2.4.\cdots.(2m))^2}\leq \frac2\pi \leq \frac{(1.3.\cdots.(2m-1))(3.5.\cdots.(2m+1))}{(2.4.\cdots.(2m))^2}\] En déduire la formule de Wallis: \[\frac2\pi = \lim_{m\to\infty} \frac{1.3}{2.2}\frac{3.5}{4.4}\cdots\frac{(2m-1).(2m+1)}{(2m).(2m)} = \prod_{m=1}^\infty \left( 1 - \frac1{4m^2}\right)\]