1-1⼀2+1⼀3-1⼀4+……+1⼀（2n-1）-1⼀2n=1⼀(n+1)+1⼀(n+2)+……1⼀2n 用数学归纳法证明

当n=1时，1-1/2=1/2成立
假设n=k(k≥1)时等式成立，即：
1-1-1/2+1/3-1/4+……+1/（2k-1）-1/2k=1/(k+1)+1/(k+2)+……1/2k
则n=k+1时
左边=(1-1/2+1/3-1/4+……+1/（2k-1）-1/2k)+(1/2k+1)-(1/2k+2)
=(1/(k+1)+1/(k+2)+……1/2k)+(1/2k+1)-(1/2k+2)
=1/(k+2)+……1/2k+【1/(k+1)+(1/2k+1)-(1/2k+2)】
=1/(k+2)+……1/2k+【(1/2k+1)+(1/2k+2)】
所以当n=k+1时等式也成立
综上，等式成立，得证

n=1
1-1/2=1/2
成立

假设n=k成立，k≥1
1-1/2+1/3-1/4+……+1/(2k-1)-1/2k=1/(k+1)+1/(k+2)+……+1/2k
则n=k+1
1-1/2+1/3-1/4+……+1/(2k-1)-1/2k+1/(2k+1)-1/(2k+2)
=1/(k+1)+1/(k+2)+……+1/2k+1/(2k+1)-1/(2k+2)
=1/(k+2)+……+1/2k+1/(2k+1)+[1/(k+1)-1/(2k+2)]
=1/[(k+1)+1]+……+1/2k+1/(2k+1)+[2/(2k+2)-1/(2k+2)]
=1/[(k+1)+1]+……+1/[2(k+1)]
成立

综上
原命题成立