问题标题：
数学归纳法证明Sn=TnSn=(1-1/2)+(1/3-1/4)+…+(1/(2n-1)-1/(2n))Tn=1/(n+1)+1/(n+2)+1/(n+3)+...+1/2n快啊...
问题描述：
数学归纳法证明Sn=Tn
Sn=(1-1/2)+(1/3-1/4)+…+(1/(2n-1)-1/(2n))
Tn=1/(n+1)+1/(n+2)+1/(n+3)+...+1/2n
快啊...
崔文华回答：
①当n=1时,S1=1/2T1=1/2∴S1=T1②当n=2时,S2=1/2+1/12=7/12T1=1/3+1/4=7/12∴S2=T2③假定Sn=TnS(n+1)=Sn+1/(2n+1)-1/(2n+2)T(n+1)=Tn-1/(n+1)+1/(2n+1)+1/(2n+2)=Tn+1/(2n+1)-[1/(n+1)-1/(2n+2)]=Tn+1/(2n+...
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