先证明不等式：当x＞0时ln(x+1)＜x令f(x)=x-ln(x+1),则f'(x)=1-1/(x+1)=x/(x+1)＞0,所以f(x)在(0,+∞)上单调递增,于是f(x)＞f(0)=0,即当x＞0时ln(x+1)＜x在不等式中取x为1/n,有当1/n＞0时ln(1/n+1)＜1/n,即n＞0时1/n...