所有奇数的倒数和与所有偶数的倒数和差如何计算？（请不要照搬其他已有问答）

[1+1/3+1/5+……+1/(2k-1)]-[1/2+1/4+1/6+……+1/(2k)]=(1-1/2)+(1/3-1/4)+(1/5-1/6)++[1/(2k-1)-1/(2k)]k=1时，1-1/2=1/2成立k=2时1/2+(1/3-1/4)=1/3+(1/2-1/4)=1/3+1/4k=3时1/3+1/4+(1/5-1/6)=1/4+1/5+(1/3-1/6)=1/4+1/5+1/6可见第一个等号是正确的。证明[1+1/3+1/5+……+1/(2k-1)]-[1/2+1/4+1/6+……+1/(2k)]【减法性质：a-b=(a+c)-(b+c)】=[1+1/2+1/3+1/4+1/5+1/6+……+1/(2k-1)+1/(2k)]-2[1/2+1/4+1/6+……+1/(2k)]=[1+1/2+1/3+……+1/(2k)]-(1+1/2+1/4+……+1/k)=1/(k+1)+1/(k+2)+1/(k+3)+……+1/(2k)【颠倒相加，折半】=[1/(k+1)+1/(2k)]/2+[1/(k+2)+1/(2k-1)]/2+……+[1/(2k)+1/(k+1)]/2【通分】={[(2k+(k+1)]/[(k+1)(2k)]}/2+{[2k-1)+(k+2)]/[(k+2)(2k-1)]}/2+……+[(3k+1)/2]/[(2k)(k+1)]【提公因式】=[(3k+1)/2]*{1/[(k+1)(2k)+1/[(k+2)(2k-1)+……+1/[(2k)(k+1)]}