Question

已知Sn=1+1/2+1/3+...+n/1(n∈N*),并记f(n)=S2n+1-Sn+1

Answer 1

（1）f(n+1)－f(n)=S2n+1-Sn+1-(S2n-1-Sn)=(S2n+1-S2n-1)-(Sn+1-Sn)

=1/(2n+1)+1/2n-1/(n+1)=(2n^2+3n+1)/(2n+1)(2n)(n+1)>0

所以f(n+1)>f(n)

（2）由（1）知f(n)最小值为f(1)=1/3

logm(m-1)+logm-1m=lgm-lg(m-1)+lg(m-1)-lgm=0

[logm(m-1)]^2-11/20[logm-1m]^2＝[logm(m-1)+logm-1m][logm(m-1)-logm-1m]+9/20[logm-1m]^2=9/20[logm-1m]^2