社数列AN为公差为2,A1=1,前N项和为SN.证2/S1S3+3/S2S4+...+N+1/SNSN+2

题目

社数列AN为公差为2,A1=1,前N项和为SN.证2/S1S3+3/S2S4+...+N+1/SNSN+2<5/16

答案

数列AN为公差为2,A1=1,前n项和为SN.证2/S1S3+3/S2S4+...+(n+1)/SnS(n+2)<5/16
【解】
Sn=n•1+n(n-1)/2•2=n^2.
(n+1)/SnS(n+2)= (n+1)/[n^2•(n+2)^2].
注意到：1/n^2-1/(n+2)^2=[(n+2)^2- n^2]/[ n^2•(n+2)^2]
=(4n+4) /[ n^2•(n+2)^2]=4(n+1)/[n^2•(n+2)^2].
所以(n+1)/[n^2•(n+2)^2]=1/4[1/n^2-1/(n+2)^2],
即(n+1)/SnS(n+2)= 1/4[1/n^2-1/(n+2)^2].
2/S1S3+3/S2S4+...+(n+1)/SnS(n+2)
=1/4[1-1/3^2+1/2^2-1/4^2+1/3^2-1/5^2+……+1/n^2-1/(n+2)^2]
=1/4[1+1/2^2-1/(n+1)^2-1/(n+2)^2]
<1/4[1+1/2^2]=5/16.