This is a continuation of the discussion in the thread titled “(co)sine powers”sine power》).
There, we computed the integrals 
%20%3D%20%5Cint%7B0%7D%5E%7B%5Cpi%2F2%7D%20%5Ccos%5En%20x%20%5C%2C%20dx%20%3D%20%5Cint%7B0%7D%5E%7B%5Cpi%2F2%7D%20%5Csin%5En%20x%20%5C%2C%20d%20x#card=math&code=I%28n%29%20%3D%20%5Cint%7B0%7D%5E%7B%5Cpi%2F2%7D%20%5Ccos%5En%20x%20%5C%2C%20dx%20%3D%20%5Cint%7B0%7D%5E%7B%5Cpi%2F2%7D%20%5Csin%5En%20x%20%5C%2C%20d%20x&id=hrSfS) (which we denoted by 
#card=math&code=C%28n%29&id=I7k0g) then. Now we change the notation to denote them by #card=math&code=I%28n%29&id=AwFjH)). This was considered by John Wallis in 1656. From Wallis, we learn that %2FI(n-2)%20%3D%20(n-1)%2Fn#card=math&code=I%28n%29%2FI%28n-2%29%20%3D%20%28n-1%29%2Fn&id=jiBkl). Depending on the parity of 
, we have
%3D%20%5Cfrac%7B(2n-2)!!%7D%7B(2n-1)!!%7D%2C%5Cquad%20%0AI(2n)%3D%5Cfrac%7B(2n-1)!!%7D%7B(2n)!!%7D%20%5Cfrac%7B%5Cpi%7D%7B2%7D%2C%20%5Cquad%20n%20%3D%201%2C%202%2C%203%2C%20%5Cldots.%20%0A#card=math&code=I%282n-1%29%3D%20%5Cfrac%7B%282n-2%29%21%21%7D%7B%282n-1%29%21%21%7D%2C%5Cquad%20%0AI%282n%29%3D%5Cfrac%7B%282n-1%29%21%21%7D%7B%282n%29%21%21%7D%20%5Cfrac%7B%5Cpi%7D%7B2%7D%2C%20%5Cquad%20n%20%3D%201%2C%202%2C%203%2C%20%5Cldots.%20%0A&id=EzFKp)
Wallis observed that
and hence
%20%5Cleq%20I(2n)%20%5Cleq%20I(2n-1).%0A#card=math&code=I%282n%2B1%29%20%5Cleq%20I%282n%29%20%5Cleq%20I%282n-1%29.%0A&id=YvPXM)
Dividing by #card=math&code=I%282n%2B1%29&id=YeeLP), we get
%7D%7BI(2n%2B1)%7D%20%5Cleq%20%5Cfrac%7BI(2n-1)%7D%7BI(2n%2B1)%7D%20%3D%20%5Cfrac%7B2n%2B1%7D%7B2n%7D.%0A#card=math&code=1%20%5Cleq%20%5Cfrac%7BI%282n%29%7D%7BI%282n%2B1%29%7D%20%5Cleq%20%5Cfrac%7BI%282n-1%29%7D%7BI%282n%2B1%29%7D%20%3D%20%5Cfrac%7B2n%2B1%7D%7B2n%7D.%0A&id=wrHer)
The squeeze theorem tells us that 
%7D%7BI(2n%2B1)%7D%3D1#card=math&code=%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20%5Cfrac%7BI%282n%29%7D%7BI%282n%2B1%29%7D%3D1&id=r7bm5). That is,
%7D%7BI(2n%2B1)%7D%20%3D%20%5Cfrac%7B%5Cpi%7D%7B2%7D%20%5Clim%7Bn%20%5Cto%20%5Cinfty%7D%20%5Cprod%7Bk%3D1%7D%5En%20%5Cbig(%5Cfrac%7B2k-1%7D%7B2k%7D%20%5Ccdot%20%5Cfrac%7B2k%2B1%7D%7B2k%7D%20%5Cbig).%0A#card=math&code=1%20%3D%20%5Clim%7Bn%20%5Cto%20%5Cinfty%7D%20%5Cfrac%7BI%282n%29%7D%7BI%282n%2B1%29%7D%20%3D%20%5Cfrac%7B%5Cpi%7D%7B2%7D%20%5Clim%7Bn%20%5Cto%20%5Cinfty%7D%20%5Cprod_%7Bk%3D1%7D%5En%20%5Cbig%28%5Cfrac%7B2k-1%7D%7B2k%7D%20%5Ccdot%20%5Cfrac%7B2k%2B1%7D%7B2k%7D%20%5Cbig%29.%0A&id=dVjey)
This gives us the Wallis formula
%20%3D%20%5Cfrac%7B2%7D%7B1%7D%5Ccdot%20%5Cfrac%7B2%7D%7B3%7D%5Ccdot%20%5Cfrac%7B4%7D%7B3%7D%5Ccdot%20%5Cfrac%7B4%7D%7B5%7D%5Ccdot%20%5Cfrac%7B6%7D%7B5%7D%5Ccdot%20%5Cfrac%7B6%7D%7B7%7D%5Ccdot%20%5Ccdots%0A#card=math&code=%5Cfrac%7B%5Cpi%7D%7B2%7D%20%3D%20%5Cprod_%7Bk%3D1%7D%5E%7B%5Cinfty%7D%20%28%5Cfrac%7B2k%7D%7B2k-1%7D%5Ccdot%20%5Cfrac%7B2k%7D%7B2k%2B1%7D%29%20%3D%20%5Cfrac%7B2%7D%7B1%7D%5Ccdot%20%5Cfrac%7B2%7D%7B3%7D%5Ccdot%20%5Cfrac%7B4%7D%7B3%7D%5Ccdot%20%5Cfrac%7B4%7D%7B5%7D%5Ccdot%20%5Cfrac%7B6%7D%7B5%7D%5Ccdot%20%5Cfrac%7B6%7D%7B7%7D%5Ccdot%20%5Ccdots%0A&id=MJBy0)
Remark: Using Euler’s infinite product expansion for the sine funciton, it is easy to derive the Wallis formula: since $\frac{\sin x}{x }= \prod_{k=1}^\infty (1-\frac{x2}{k2\pi^2}); \forall x $, putting 
, we get
%20%5CLongleftrightarrow%20%5Cfrac%7B%5Cpi%7D%7B2%7D%20%3D%20%5Cprod%7Bk%3D1%7D%5E%5Cinfty%20(%5Cfrac%7B4k%5E2%7D%7B4k%5E2-1%7D)%20%3D%20%5Cprod%7Bk%3D1%7D%5E%5Cinfty%20(%5Cfrac%7B2k%7D%7B2k-1%7D%5Ccdot%20%5Cfrac%7B2k%7D%7B2k%2B1%7D)%2C%0A#card=math&code=%5Cfrac%7B2%7D%7B%5Cpi%7D%20%3D%20%5Cprod%7Bk%3D1%7D%5E%5Cinfty%20%281-%5Cfrac%7B1%7D%7B4k%5E2%7D%29%20%5CLongleftrightarrow%20%5Cfrac%7B%5Cpi%7D%7B2%7D%20%3D%20%5Cprod%7Bk%3D1%7D%5E%5Cinfty%20%28%5Cfrac%7B4k%5E2%7D%7B4k%5E2-1%7D%29%20%3D%20%5Cprod_%7Bk%3D1%7D%5E%5Cinfty%20%28%5Cfrac%7B2k%7D%7B2k-1%7D%5Ccdot%20%5Cfrac%7B2k%7D%7B2k%2B1%7D%29%2C%0A&id=cFnCd)
as desired.
