設(shè)：f(n)=(1+2+3+…+n)(1+ 1/2 + 1/3 +…+ 1/n)-n^2-n+1
f(3)=(1+2+3)(1+ 1/2 + 1/3)-9-3+1=6*11/6-9-3+1=0
f(n+1)-f(n)=(1+2+3+…+n+n+1)[1+ 1/2 + 1/3 +…+ 1/n+1/(n+1)]-(n+1)^2-n
-(1+2+3+…+n)(1+ 1/2 + 1/3 +…+ 1/n)+n^2+n-1
=1+(n+1)(1+ 1/2 + 1/3 +…+ 1/n)+(1+2+3+…+n)(n+1)-2n-2
>1+n+1+(n+1)^2-2n-2>0
f(n)單調(diào)遞增.
f(n)>f(3)≥0
當(dāng)n=2時(shí),1/2^2=1/4=2)時(shí)不等時(shí)成立,那么,對于n=k+1,有
1/2^2+a/3^2+……+1/k^2+1/(k+1)^2