This is a non-proof based only off of notational styles. Let s be a series , with s_n = sum (1 to n, 1) = n if S3 exists (i.e. lim n->infinity s_n exists), the

This is a non-proof based only off of notational styles. Let s be a series , with s_n = sum (1 to n, 1) = n if S3 exists (i.e. lim n->infinity s_n exists), then the series (s-s) (where (s-s)_n = s_n - s_n) converges , and it converges towards S3 - S3 (=0). > = (1 + 1 + 1 + 1 ...) - (0 + 1 + 1 + 1 ...) the series whose sum he's describing here is not (s-s), but another one entirely : u where u_1 = s_1, u_n+1= s_n+1-s_n These are different series, so it is entirely reasonable for them to converge at different values, 1 doesn't equal 0. The "real" reason this result isn't what we think it is(apart from "infinite sums are different" argument, which is a non-answer): >Step 1: Let S = 1 + 2 + 3 + 4 ... What you're doing here firstly, is saying that "I assume that this sum converges, let S be the value it converges to". So you end up proving (if the other steps weren't also flawed) that S = -1/12, all you're saying is that if the sum exists, then it is -1/12. The issue here depends on what sort of thing you're working on. The "classic" definition of a limit (convergence of a series) does not work here, because you can prove that for all n, the sum of n numbers will always be at least -1/12 away from -1/12 (on account of it being positive), so it can't converge to -1/12 (hence S not existing). However, there are techniques for assigning limits to divergent sums. these summability methods will give the same result as the classic analysis for convergent series, but will also give values for some divergent series. It is similar to analytic continuation( f(x) doesn't exist, but a limit exists in x both from the left and to the right (named y), so we sometimes say that f(x)=y ), in that it allows us to extend the resolution domain slightly. But the classic definition no longer works. The one used here is zeta function regularization, which consists in the following: you have a series a , and a function Za(s) = (1/a1^s)+(1/a2^s)+(1/a3^s)+(1/a4^s)+..... Za is only defined for certain values of s depending on the series. But for a serie representing a convergent sum, we know that Za(-1)= a1+a2+a3+... exists. So this method will give the same limit for convergent sums as the classical method. For a series with a diverging sum (a_n)=n, Za(-1) doesn't exist, but the limit in -1 exists from both ends (by analytic continuation), so we extend the domain of Za to -1 by the value of this "limit" : -1/12 The article mentions Ramanujan summation, but Zeta regularization is actually a much more useful tool. In