<==>ln[(1/2)(1/2^2)...(1/2^N)] < lne
<==>ln(1/2)+ ln(1/2^2)...+ ln(1/2^N) < 1
对于f x＝LN（1＋X）－AX
f'(x) = 1/(1+x) - A,当x < 0,A <= 1时,有f'(x) > 0
==>A <= 1,x < 0时,f(x)为增函数,
==>此时,f(x) < f(0) = 0,即,ln(1 + x) < Ax(x < 0,A < 1)
==>ln(1/2^i) = ln(1 - (2^i - 1)/2^i) < -A (2^i - 1)/2^i < 1/2^i
==>ln(1/2)+ ln(1/2^2)...+ ln(1/2^N) < 1/2 + ...+1/2^N < 1,证毕
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此题考查的应该是求导的知识,但是这个问题好像有问题
(1/2)(1/2^2)...(1/2^N) < 1*1*1...*1<1所以题目可能应该是(1/2)(1/2^2)...(1/2^N)<1/e,于是
<==>ln(1/2)+ ln(1/2^2)...+ ln(1/2^N) < -1,
==>ln(1/2^i) = ln(1 - (2^i - 1)/2^i) < -A (2^i - 1)/2^i < -1/2^i
==>ln(1/2)+ ln(1/2^2)...+ ln(1/2^N) < -1/2 + ...+（-1/2^N） < -1,证毕