Question

如何证明1^2+(1/2)^2+(1/3)^2+……+(1/n)^2<2

Answer 1

原式＜1+1/(1×2)+1/(2×3)+1/(3×4)+……+1/[(n-1)n]
=1+1-1/2+1/2-1/3+1/3-1/4+……+1/(n-1)-1/n
=2-1/n
＜2

追问

麻烦写详细一点，最好讲解一下！

追答

(1/2)^2=1/(2×2)＜1/(1×2)=1-1/2
(1/3)^2=1/(3×3)＜1/(2×3)=1/2-1/3
……
(1/n)^2=1/(n×n)＜1/[(n-1)×n]=1/(n-1)-1/n
∴原式＜1+1/(1×2)+1/(2×3)+1/(3×4)+……+1/[(n-1)n]
=1+1-1/2+1/2-1/3+1/3-1/4+……+1/(n-1)-1/n
=2-1/n
＜2

Answer 2

n^2＞n^2-n=n（n-1）
(1/n)^2＜1/[n（n-1）]=1/（n-1）-1/n
1^2+(1/2)^2+(1/3)^2+……+(1/n)^2 ＜ 1 + 1-1/2 + 1/2-1/3 +~~~~~~~~+1/(n-2)-1/(n-1) + 1/(n-1)-1/n =2-1/n<2