>>11436134Proof:
Assume the result is true for n in N
S(n) where n in N will denote the summation of all natural numbers.
S(n) = -1/12
12S(n) = -1
-12S(n) = 1
Leaving out a negative 1 12 can be introduced into the sum which should be obvious.
-S(n) = 1
S(-n) = 1
S(-n) is the sum of all negative integers Z^- where Z^- := {x < 0}. 1 is not less than 0 so now we've reached a contradiction. The rest is an exercise for the brute.
