n=1時，左=1-1/2=1/2 右面=1/2成立，

假設n=k時，成立：1-1/2+1/3-1/4+...+1/2k-1-1/2k=1/k+1+1/k+2+...+1/k+k

則n=k+1時，

右=1/(k+2)+1/(k+3)+...+1/(k+1+k)+1/(2K+2)

=1/(k+2)+1/k+3)+...+1/(2k+1)+1/(2k+2).........................1

左=[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/(k+k)+1/(2k+1)-1/(2k+2)

=1/(k+2)+1/(k+3)+...+(2k+1)+1/(k+1)-1/(2k+2)

=1/(k+2)+1/(k+3)+...+(2k+1)+(2k+2-k-1)/[(k+1)(2k+2)]

=1/(k+2)+1/(k+3)+...+(2k+1)+(k+1)/[(k+1)(2k+2)]

=1/(k+2)+1/(k+3)+...+(2k+1)+1/(2k+2).............................2

1式=2式

所以n=k+1時也成立，

所以原式成立。