1/12+1/22+1/32+------+1/20002

1个回答

  • 1/n2=(n-1)/[n2(n-1)]

    =n/[n2(n-1)]-1/[n2(n-1)]

    =1/[n (n-1)] -1/[n2(n-1)]

    =1/(n-1) -1/n-1/[n2(n-1)]

    当n>2时,总有1/[n2(n-1)]>0

    1/12+1/22+1/32+…+1/20002

    =1/12+1/(2-1) -1/2-1/22(2-1)

    +1/(3-1) -1/3-1/32(3-1)+…

    +1/(2000-1) -1/2000-1/20002(2000-1)

    =1/12+1/(2-1)-1/22(2-1) -1/32(3-1)-…

    -1/2000-1/20002(2000-1)

    =2-1/22(2-1) -1/32(3-1)-…-1/20002(2000-1) -1/2000<2

    证毕

    是1的平方,2的平方,3的平方,2000的平方