证明:1+1/2^2+1/3^2+.+1/n^2〈5/3怎么证?

Question

证明:1+1/2^2+1/3^2+.+1/n^2〈5/3怎么证?

Answer 1

证明:∵1+1/2^2+1/3^2+.+1/11^2+1/12^2+1/13^2
= 1+1/4+1/9+.+1/11^2+1/12^2+1/13^2
＜1+1/3+1/8+.+1/(11^2-1)+1/(12^2-1)+1/(13^2-1)=
1+1/2(1-1/3+1/2-1/4+1/3-1/5+.+1/10-1/12+1/11-1/13+1/12-1/14)=
1+1/2(1+1/2-1/13-1/14)＜5/3
∴当n≤13时,1+1/2^2+1/3^2+.+1/n^2〈5/3总是成立
又∵1+1/2^2+1/3^2+.+1/n^2=1+1/2^2+1/3^2+.+1/(n-3)^2+1/(n-2)^2+1/(n-1)^2+1/n^2＜
1+1/2{1-1/3+1/2-1/4+1/3-1/5+.+1/[(n-3)^2-1]+1/[(n-2)^2-1]+1/[(n-1)^2-1]+1/(n^2-1)}
=1+1/2{1-1/3+1/2-1/4+1/3-1/5+.+1/(n-4)-1/(n-2)+1/(n-3)-1/(n-1)+1/(n-2）-1/n+1/(n-1)- 1/（n+1）}
=1+1/2[1+1/2-1/n-1/(n+1)]
=1+3/4-1/2[1/n+1/(n+1)]
而当n≥13时,（n-6)^2＞42即n^2-12n+36>42
∴n^2-12n-6>0
∴n^2+n>12n+6
∴n(n+1)>6(n+1)+6
∴1/n+1/(n+1)5/3
∴1+1/2{1-1/3+1/2-1/4+1/3-1/5+.+1/(n-4)-1/(n-2)+1/(n-3)-1/(n-1)+1/(n-2）-1/n+1/(n-1)- 1/（n+1）}>5/3
∴1+1/2{1-1/3+1/2-1/4+1/3-1/5+.+1/[(n-3)^2-1]+1/[(n-2)^2-1]+1/[(n-1)^2-1]+1/(n^2-1)}>5/3
∴1+1/2^2+1/3^2+.+1/n^2=1+1/2^2+1/3^2+.+1/(n-3)^2+1/(n-2)^2+1/(n-1)^2+1/n^2>5/3
错了!