(1)经归纳,可以猜测 f(n)=1/(n+1).
(2)f(1/n)=1/[(1/n)+1]=n/(n+1)
f(1/2)×f(1/3)×f(1/4)……f(1/99)
=(2/3)(3/4)(4/5)……(99/100)
=2/100
=1/50.
(1)经归纳,可以猜测 f(n)=1/(n+1).
(2)f(1/n)=1/[(1/n)+1]=n/(n+1)
f(1/2)×f(1/3)×f(1/4)……f(1/99)
=(2/3)(3/4)(4/5)……(99/100)
=2/100
=1/50.