Consider the sequence of integrals:
#I_m = int_0^(pi/2) sin^mx dx#
As the integrand is defined and continuous in the whole interval, these integrals exist for every #m in NN#.
Evaluate now the integral by parts:
#I_m = -int_0^(pi/2) sin^(m-1)x d(cosx)#
#I_m = -[sin^(m-1)x cosx]_0^(pi/2) + int_0^(pi/2) (m-1)sin^(m-2)xcos^2xdx#
#I_m = (m-1) int_0^(pi/2) sin^(m-2)x(1-sin^2x)dx#
#I_m = (m-1) I_(m-2) - (m-1)I_m#
and finally we have the recursive formula:
#I_m = (m-1)/m I_(m-2) #
so that:
#I_(m-2)/I_m= m/(m-1)#
and:
#lim_(m->oo) I_(m-2)/I_m= 1#
Given that:
#I_0 = int_0^(pi/2) dx = pi/2#
#I_2 = int_0^(pi/2) sin^2xdx = pi/4#
consider now for every #m# the ratio:
#I_(m-1)/I_(m+1) = I_(m-1)/I_m I_m/I_(m+1)#
and evaluate each ratio using the recursive formula:
#I_(m-1)/I_(m) = ((m-2)/(m-1) I_(m-3))/((m-1)/(m) I_(m-2)#
#I_(m-1)/I_(m)= ((m(m-2))/(m-1)^2 ) I_(m-3)/I_(m-2)#
#I_(m-1)/I_(m) = (((m-1+1)(m-1-1))/(m-1)^2 ) I_(m-3)/I_(m-2)#
#I_(m-1)/I_(m) = (((m-1)^2-1)/(m-1)^2 ) I_(m-3)/I_(m-2)#
#I_(m-1)/I_(m) = (1-1/(m-1)^2 ) I_(m-3)/I_(m-2)#
and:
#I_(m)/I_(m+1)= ((m-1)/(m) I_(m-2))/((m)/(m+1) I_(m-1)#
#I_(m)/I_(m+1)= (((m+1)(m-1))/m^2) I_(m-2)/ I_(m-1)#
#I_(m)/I_(m+1)= ((m^2-1)/m^2) I_(m-2)/ I_(m-1)#
#I_(m)/I_(m+1)= (1-1/m^2) I_(m-2)/ I_(m-1)#
Then:
#I_(m-1)/I_(m+1) = (1-1/m^2) I_(m-2)/ I_(m-1)(1-1/(m-1)^2 ) I_(m-3)/I_(m-2)#
#I_(m-1)/I_(m+1) = (1-1/m^2)(1-1/(m-1)^2 ) I_(m-3)/I_(m-1)#
and iterating:
#I_(m-1)/I_(m+1) = (1-1/m^2)(1-1/(m-1)^2 )(1-1/(m-2)^2)(1-1/(m-3)^2 ) I_(m-5)/I_(m-3)#
or:
#I_(m-1)/I_(m+1) = prod_(j=0)^p (1-1/(m-j)^2) I_(m-p-2)/I_(m-p)#
So for #p=m-2# we have:
#I_(m-1)/I_(m+1) = prod_(j=0)^(m-2) (1-1/(m-j)^2) I_(0)/I_(2)#
#I_(m-1)/I_(m+1) = prod_(j=0)^(m-2) (1-1/(m-j)^2)(pi/2)/(pi/4)#
#I_(m-1)/I_(m+1) = 2 prod_(j=0)^(m-2) (1-1/(m-j)^2)#
Change now the index to #k= m-j#:
#I_(m-1)/I_(m+1) = 2 prod_(k=2)^m (1-1/k^2)#
so:
#prod_(k=2)^m (1-1/k^2) = 1/2 I_(m-1)/I_(m+1)#
and passing to the limit:
#prod_(k=2)^oo (1-1/k^2) = 1/2 lim_(m->oo) I_(m-1)/I_(m+1) = 1/2#