It is not

  • mathemachristian[he]
    2•7 months ago
    solution

    Assuming the series converges it converges absolutely. Therefore

    sumn/2(n-1) n >= 1
    = sum(n+1)/2n n >= 0
    = sumn/2n n >= 0 + sum1/2n n >= 0
    = sumn/2n n >= 0 + 2
    = sumn/2n n >= 1 + 2

    =>

    sumn/2(n-1) n >= 1 = sumn/2n n >= 1 + 2

    =>

    2 = sumn/2(n-1) n >= 1 - sumn/2n n >= 1
    = sumn/2(n-1) - n/2n n >= 1
    = sumn/2n n >= 1
    = 1/2 * sumn/2(n-1) n >= 1

    =>

    sumn/2(n-1) n >= 1 = 4