n=1时,S1=1-1/2=1/2 ,T1=1/2 成立;假设,n=k 时成立,即Sk=Tk
对于n=k+1 ,S（k+1）=Sk+1/(2k+1)-1/(2k+2)
T(k+1)=Tk+1/(2k+1) +1/(2k+2)-1/k+1=Tk+1/(2k+1)-1/(2k+2)
所以S（k+1）-Sk=T(k+1)-Tk；由于Sk=Tk,所以S（k+1）=T(k+1)
所以Sn =Tn成立.
另给一种直接证法
Sn=1-1/2+1/3-1/4+……+1/2n-1-1/2n=1+(1/2-1)+1/3+(1/4-1/2)+...+1/(2n-1)+[1/(2n)-1/n]
=1+1/2+1/3+1/4+...+1/n+1/(n+1)+1/(n+2)+……+1/2n-(1+1/2+1/3+1/4+...+1/n)
=1/(n+1)+1/(n+2)+……+1/2n=Tn. n=1时,S1=1-1/2=1/2 ,T1=1/2 成立;假设,n=k 时成立,即Sk=Tk
对于n=k+1 ,S（k+1）=Sk+1/(2k+1)-1/(2k+2)
T(k+1)=Tk+1/(2k+1) +1/(2k+2)-1/k+1=Tk+1/(2k+1)-1/(2k+2)
所以S（k+1）-Sk=T(k+1)-Tk；由于Sk=Tk,所以S（k+1）=T(k+1)
所以Sn =Tn成立.
另给一种直接证法
Sn=1-1/2+1/3-1/4+……+1/2n-1-1/2n=1+(1/2-1)+1/3+(1/4-1/2)+...+1/(2n-1)+[1/(2n)-1/n]
=1+1/2+1/3+1/4+...+1/n+1/(n+1)+1/(n+2)+……+1/2n-(1+1/2+1/3+1/4+...+1/n)
=1/(n+1)+1/(n+2)+……+1/2n=Tn.