在数列{an}中,a1=1,an+1=(1-1/n+1)an.若对一切n>1的自然数,不等式an+1+an+2+...+a2n>1/12loga (a-1)+2/3恒成立,试求a的取值范围

an+1=(1-1/n+1)an则an+1=（n/n+1）an则an+1=（n/n+1）an=（n/n+1）*（n-1/n）an-1=...=n/n+1*（n-1/n）*..1/2*a1=1/n+1所以an+1=1/n+1则an+1+an+2+...+a2n=1/n+1+1/n+2+..+1/n+n对于1/n+1+1/n+2+..+1/n+n,令Sn=1/n+1...