已知数列{an}得通项公式an=1/n+1+1/n+2+1/n+3+...+1/2n(n∈n*).(1)求证:{an}为已知数列{an}得通项公式an=1/(n+1)+1/(n+2)+1/(n+3)+...+1/2n(n∈N*）.（1）求证：{an}为递增数列；（2）若对于一切大于1的自然数n,不等

Question

已知数列{an}得通项公式an=1/n+1+1/n+2+1/n+3+...+1/2n(n∈n*).(1)求证:{an}为
已知数列{an}得通项公式an=1/(n+1)+1/(n+2)+1/(n+3)+...+1/2n(n∈N*）.（1）求证：{an}为递增数列；（2）若对于一切大于1的自然数n,不等式an>1/12log(1/3)a+2/3恒成立,求实数a的取值范围

Answer 1

(1)
a(n)=1/(n+1)+1/(n+2)+1/(n+3)+...+1/2n
a(n+1)=1/(n+2)+1/(n+3)+1/(n+4)+...+1/(2n+2)
a(n+1)-a(n)
=1/(2n+2)+1/(2n+1)-1/(n+1)
>1/(2n+2)+1/(2n+2)-1/(n+1)
=2/(2n+2)-1/(n+1)
=0
所以{an}为递增数列
(2)
an>1/12log(1/3)a+2/3恒成立
因为{an}为递增数列
即需1/12log(1/3)a+2/31/12log(1/3)a+2/3<1/2
1/12log(1/3)alog(1/3)aa>(1/3)^(-2)
a>9