First, this is literally the most obnoxious maths mene and notation abuse ever devised by humanity.
Second, Numberphile's "proof" mistreats divergent series and basically isn't a proof at all.
You should check out Riemann's functional equation and how it's used to construct Riemann's zeta analytical continuation.
>>8585770
https://www.youtube.com/watch?v=sD0NjbwqlYw
The video is long, but very well made and explain all.
To sum up :
In the domain of real, 1 + 2 + 3 + ... does not equal -1/12. The main argument used by numberphile (that 1 - 1 + 1 - ... = 1/2) is false, because you can just arbitrarily change the order of the terms of a sum that does not converge.
However, the function defined by [math]f(s) = 1 + 1/(2^s) + 1/(3^s) + ... [/math], more known as the Zeta function, can be artificially prolonged outside of where it exists.
When you uses this prolongation at the point (-1), then you have [math]f(-1) = -1/12[/math]
That does not mean : [math] 1 + 2 + ... = -1/12[/math] ; the sum [math] 1 + 2 + ... [/math] does not exist, mathematically speaking. It is infinite. However, we can artificially extends f in a way that f(-1) = -1/12.
>>8585801
>notation abuse
Classical analysis shouldn't have the monopoly on those symbols
It's notation out of the context of more relevant math, but there's no abuse. If I use [math] \omega [/math] to denote both a rotation vector and it's magnitude, then that's abuse of notion. An equation from some context that also has an interpretation in a more common context isn't abuse..
>>8585837
I'd argue the + sign should be used with decimals in traditional sense only, but k lad
Does anyone know when or why it is physically acceptable to use an analytic continuation. To the naive it might appear like a logical fallacy that an extension of a domain would have a physical meaning.
>>8585909
Because it's unique.
>>8585821
Transitivity may not hold, but associativity
>>8585770
The number is so big that it rolls over past infinity and down through the negatives.
>>8585909
b/c it aligns with experiment
Here's your explanation, OP.
Let [math]N \,\in\, \mathbf N[/math].
[eqn]\sum_{n \,=\, 0}^N n \,=\, \frac{N \,\left(N \,+\, 1\right)}{2} \,\xrightarrow[N \,\rightarrow\, \infty]{}\, \infty[/eqn]
Therefore [math]\sum_{n \,=\, 0}^\infty n \,=\, \infty[/math].
It's easy to confuse yourself with this shit but it's quite simple.
All that the ramanujan summation stuff, cutoff and zeta regularization does, is look at the smoothed curve at x = 0.
What sums usually do is look at the value as x->inf.
It's just a unique value you can assign to a sum, really they have many such values.
https://en.wikipedia.org/wiki/File:Sum1234Summary.svg