Quote:
Originally Posted by retina
Very good.
If we assign:
A = 1 + 1 + 1 + 1 + 1 + 1 + ...
B = 1 + 2 + 3 + 4 + 5 + 6 + ...
Then B  A = B, since pairwise subtraction gives 0 + 1 + 2 + 3 + ... = 1 + 2 + 3 + ...
Therefore A = 1 + 1 + 1 + 1 + 1 + 1 + ... = 0

Sadly, in the regime of regularized sums, we no longer have 0 + 1 + 2 + 3 + ... = 1 + 2 + 3 + ...