zvavybir 
Math question: Why is there only one way to assign a number to 1 + 2 + 3 + 4 + 5 + …? This sum is always said to either diverge (with the usual definition of equality) or to equal -1/12. Under normal rules it doesn't have a value, but if you ignore that, then – pretty much independent of what you do – you always get out -1/12. Why never any other value? In what way is this genuinely the correct value for this?
ƧƿѦςɛ♏ѦਹѤʞ@zvavybir
It does diverge. It has no sum.
However, the uniquely valued #Riemann #ZetaFunction can be analytically continued into the left half-plane where we find zeta(-1)=-1/12 (which 'looks like' 1+2+...). #Cesàro #summation will get you part of the way there also, and, as you say, yields the same result; presumably due to some ultimate cosmic logical rightness :-)
I very strongly recommend BP's superb exposition of this issue
https://www.youtube.com/watch?v=YuIIjLr6vUA
#maths #AnalyticContinuation #Ramanujan
It does diverge. It has no sum.
However, the uniquely valued #Riemann #ZetaFunction can be analytically continued into the left half-plane where we find zeta(-1)=-1/12 (which 'looks like' 1+2+...). #Cesàro #summation will get you part of the way there also, and, as you say, yields the same result; presumably due to some ultimate cosmic logical rightness :-)
I very strongly recommend BP's superb exposition of this issue
https://www.youtube.com/watch?v=YuIIjLr6vUA
#maths #AnalyticContinuation #Ramanujan
Numberphile v. Math: the truth about 1+2+3+...=-1/12
Adrian Riskin 
@zvavybir That's the value using Cesaro summation but there are very many other methods for assigning values to divergent series. A lot of them are consistent with Cesaro summation, as you say, but not all. E.g. per wiki BGN hyperreal summation yields \frac{\omega}^2}{2} + \frac{\omega}{2}. It seems likely that one could fiddle with parameters in various summation methods to get other values, although I don't know for sure.
https://en.wikipedia.org/wiki/Divergent_series#Miscellaneous_methods
@AdrianRiskin \( \frac{\omega^2}{2} + \frac{\omega}{2} \) isn't a real number though, so it's not really what I care about. (And Cesàro doesn't work.)