f(n)=1+1/2+1/3+...+1/(2^n)-1-n/2g(n)=1+1/2+1/3+...+1/(2^n)-1/2-nf(1)=1+1/2-1-1/2=0若f(n)≥0f(n+1)=1+1/2+1/3+...+1/(2^n)-1-n/2+1+n/2-1-(n+1)/2+1/(2^n +1)+…1/2^(n +1)而f(n)≥01/(2^n +1)+…1/2^(n +1)≥[...