Question

已知数列 an=1/（n+1）,若对于任意正整数n,不等式S2n-Sn>m/16恒成立,则常数m所能取得的最大整数为

Answer 1

S(2n)-Sn=[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

∴ {bn}是递增的

则{bn}的最小值是b1

即 b1>m/16