已知不等式1/(n+1)+1/(n+2)+...1/2n>=1/12loga(a-1)+2/3对大于1的整数n恒成立,求a范围

首先对数函数有意义,则a-1>0,即a>1.令an=1/(n+1)+1/(n+2)+...1/2n,则a(n+1)-an=1/(2n+1)+1/(2n+2)-1/(n+1)=1/(2n+1)-1/(2n+2)>0即an是递增数列,1/(n+1)+1/(n+2)+...1/2n>=1/12loga(a-1)+2/3对大于1的整数n恒成立就是...