那么当n=k+1时T(k+1)=Tk+b(k+1)=Tk+1/(k+1)²<带陆凯(2k-1)/悉纤k+1/(k+1)²<(2k-1)/k+1/k(k+1)=(2k+1)/(k+1)所蠢唤以n=k+1时不等式也成立,所以Tn<(2n-1)/n包括刚才的回答,哪一步不懂就问我吧