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.