设:f(n)=[1/(n+1)]+[1/(n+2)]+[1/(n+3)]+…+[1/(3n+1)]
则:f(n+1)=[1/(n+2)]+[1/(n+3)]+[1/(n+4)]+…+[1/(3n+2)]+[1/(3n+3)]+[1/(3n+4)]
两式相减,得:
f(n+1)-f(n)=[1/(3n+2)]+[1/(3n+3)]+[1/(3n+4)]-[1/(n+1)]
=[1/(3n+2)]-[1/(3n+3)]+[1/(3n+4)]-[1/(3n+3)]
=1/[(3n+2)(3n+3)]-1/[(3n+3)(3n+4)]>0
即:f(n+1)>f(n)
从而,f(n)是递增的,即f(n)的最小值是f(1)
则:
a/24