Question

已知数列{an}的前n项和为Sn,且满足a1=1,an+SnSn-1=0 an=1/n(1-n), Sn=1/n

求证:S1^2+ S2^2+S3^2+……Sn^2 ≤2-1/n
请把详细过程写出来

Answer 1

已知数列{an}的前n项和为Sn,且满足a1=1,an+SnSn-1=0 an=1/n(1-n), Sn=1/n
求证:S1^2+ S2^2+S3^2+……Sn^2 ≤2-1/n
因Sn=1/n,所以
S1^2+ S2^2+S3^2+……Sn^2=
=1+1/2^2+1/3^2+……+1/n^2
因为
1/2^2<1/(1*2)=1-1/2
1/3^2<1/(2*3)=1/2-1/3
…………………………
1/n^2<1/[(n-1)*n]=1/(n-1)-1/n
所以
S1^2+ S2^2+S3^2+……Sn^2=
=1+1/2^2+1/3^2+……+1/n^2<
<1+(1-1/2)+(1/2-1/3)+……+[1/(n-1)-1/n]
=2-1/n