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#riemann

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@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
youtube.com/watch?v=YuIIjLr6vU
#maths #AnalyticContinuation #Ramanujan

`Two years later, in 1849, he returned to Göttingen to pursue his PhD with Gauss, completing his thesis in 1851 on the theory of complex variables, the basis for what we now call Riemann surfaces. Gauss described Riemann as having “a gloriously fertile originality” in his report on the thesis, and two years later, when #Riemann was required to give a lecture to land a faculty position at Göttingen, Gauss assigned his star pupil the topic of the foundations of #geometry`

aps.org/archives/publications/

DOMINATED CONVERGENCE THEOREM
Lebesgue's dominated convergence theorem provides sufficient conditions under which pointwise convergence of a sequence of functions implies convergence of the integrals. It's one of the reasons that makes integration more powerful than integration. The theorem an be stated as follows:

Let (fn) be a sequence of measurable functions on a measure space (S,Σ,μ). Suppose that (fn) converges pointwise to a function f and is dominated by some Lebesgue integrable function g, i.e. |fn(x)|g(x) n and xS. Then, f is Lebesgue integrable, and

limnSfn dμ=Sf dμ

Riemann zeta function ζ(s) and n=1n=1+2+3+=112

Have you ever heard that the sum of all natural numbers is 1/12?🤔 Of course not; this doesn't make sense in the usual sum, but using a summation method based on analytic continuation of the Riemann zeta function leads to the following result.

The Riemann zeta function is defined as:
ζ(s)=n=11ns=11s+12s+13s+
for sC such that (s)>1.
It can be extended to a meromorphic function with only a simple pole at s=1, using analytic continuation and the following functional equation:
ζ(1s)=21sπscos(πs2)Γ(s)ζ(s)
For s=2, this gives ζ(1)=n=1n=12π2ζ(2)=12π2π26=112, which is a reason for assigning a finite value to the divergent sum/series (zeta function regularization). That is, n=1n=1+2+3+=112.

The unproved Riemann hypothesis states that the nontrivial zeros of the Riemann zeta function occur only on the critical line z=12+iy. While it is not difficult to understand why these zeros can only occur inside the critical strip 0<Rez<1, the restriction to the critical line is spooky cool.

With an implementation of the zeta function in one has a proof near the origin via . The real part of the function is blue, imaginary red:

mathcell.org/www/riemann-zeta-

Manipulating the imaginary part of the argument along the critical strip shows immediately that zeros only occur on the critical line for an imaginary part of approximately

±14.13, ±21.02, ±25.01, ±30.42, ±32.94, ±37.59, ±40.92, ±43.33, ±48.01, ±49.77

For more context and the relation to the Riemann xi function, visit

analyticphysics.com/Special%20