已知数列{an}的通项公式为an=1/(n+1),前n项和为Sn,若对于任意正整数n,不等式S2n-Sn>m/16恒成立,则常数M所能取得饿最大整数为

Answer 1

Sn=a1+a2+a3+.+an
S2n=a1+a2+a3+.+an+a(n+1)+.+a2n
s2n-Sn=a(n+1)+a(n+2)+.+a2n
=[1/2+1/3+1/4+.+1/(n+1)+1/(n+2)+.+1/(2n+1)]-[1/2+1/3+1/4+.+1/(n+1)]
=1/(n+2)+.+1/(2n+1)
设bn=S(2n)-S(n)
则 b(n+1)-b(n)=[1/(n+3)+1/(n+4)+.+1/(2n+1)+1/(2n+2)]+1/(2n+3)-[1/(n+2)+1/(n+3)+.+1/(2n+1)]
=1/(2n+2)+1/(2n+3)-1/(n+2)
>0
所以 {bn}是递增的
则{bn}的最小值是b1
即 b1>m/16
b1=S(2)-S(1)=(a1+a2)-a1=a2=1/3
即1/3>m/16
即 m