已知数列an=1/{(n+1)的平方}fn=(1-a1)*(1-a2)*.*(1-an)通过计算f1,f2,f3,的值推测fn的值需证明

令bn=(1-an)=[n/(n+1)][(n+2)/(n+1)]
f(n)=b1b2…bn=[1/2×3/2×2/3×4/3×……(n+2)/(n+1)]=(n+2)/2(n+1)

f1=3/4
f2=2/3
f3=5/8
fn=(1-a1)*(1-a2)*.*(1-an)
=(1-1/2^2)(1-1/3^2)(1-1/4^2)*.*[1-1/(n+1)^2]
=(1-1/2)(1+1/2)(1-1/3)(1+1/3)(1-1/4)(1+1/4)*.*[1-1/(n+1)][1+1/(n+1)
=1/2*3/2*2/3*4/3*3/4*5/4*.*n/(n+1)*(n+2)/(n+1)
=1/2*(n+2)/(n+1)
=(n+2)/(2n+2)