(1-1/2^2 )(1-1/3^2 )(1-1/4^2 )…(1-1/2011^2 )
(1-1/2^2)(1-1/3^2)...(1-1/n^2)
=(2^2-1)/2^2 * (3^2-1)/3^2 * ...(n^2-1)/n^2
=(2+1)(2-1)/2^2 * (3+1)(3-1)/3^2 *... (n+1)(n-1)/n^2
=(2+1)(3+1)...(n+1) * (2-1)(3-1)...(n-1)/(1*2*3...n)^2
=((n+1)!/2)(n-1)!/(n!)^2
=(1/2)[(n+1)/(n)] n!*n!/(n1)^2
=(n+1)/(2n)
Therefore
Π(1-1/i^2) i=2 to n
=(n+1)/(2n)
For n=2011,
Π(1-1/i^2) i=2 to 2011
=(2011+1)/(2*2011)
=2012/4022