猜想：Sn=Tn .
证明：（1）当n=1时,S1=1-1/2=1/2 ,T1=1/2 ,因此 S1=T1 ,命题成立.
（2）设当 n=k（k>=1 ,为正整数）时 Sk=Tk ,
两边同时加上 1/(2k+1)-1/(2k+2) ,得
Sk+1/(2k+1)-1/(2k+2)=Tk+1/(2k+1)-1/(2k+2) ,
上式左边=S(k+1) ,
右边=1/(k+1)+1/(k+2)+.+1/(2k)+1/(2k+1)-1/(2k+2)
=[1/(k+2)+1/(k+3)+.+1/(2k)+1/(2k+1)+1/(2k+2)]+[1/(k+1)-2/(2k+2)]
=1/(k+2)+1/(k+3)+.+1/(2k)+1/(2k+1)+1/(2k+2)
=T(k+1) ,
因此命题对n=k+1也成立,
由（1）（2）可得,对所有正整数 n ,有 Sn=Tn .