Let |zr−r| ≤  r,  ∀ r =1,  2,  3,  ...,  n.  Then |∑r=1nzr|  is less than

|∑r=1nzr|  ≤  ∑r=1n|zr|  ≤  ∑r=1nzr-r+r≤  ∑r=1n|zr−r|+∑r=1nr  ≤  2∑r=1nr   =2nn+12=nn+1                      since z1+z2≤z1 +z2