
The Ramanujan Summation: 1 and 2 and 3 and ⋯ + ∞ = -1/12? - laronian
https://medium.com/cantors-paradise/the-ramanujan-summation-1-2-3-1-12-a8cc23dea793
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matt-attack
So can someone help square this with the obvious fact that the sum should
appear to explode to infinity? I can clearly follow the proof, but I still
cannot reconcile it with basic arithmetic.

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kmill
It certainly does explode to infinity as a series.

What's going on in the factoid (though I'm a bit weak in analysis) is that the
series

    
    
      zeta(s) = sum(k^(-s) for k in 1:Infinity)
    

converges for all complex s whose real part is greater than 1. There is a
concept called _analytic continuation_ where you compute the Taylor series for
a function at a point, find its radius of convergence there, then formally
extend the domain of the original function. You keep doing this to extend the
domain as much as possible. (And if you are lucky, you do not get any
contradictory values when you go around a singularity. Otherwise you'll need
to deal with branch cuts, like with complex logarithms.) This isn't completely
arbitrary: if you have some holomorphic function then restrict its domain to
some small disk, analytic continuation will recover the original function---
that is, even the smallest germ knows everything about the function!

For an example of an analytic continuation, if you were to start with

    
    
      f(z) = sum(z^k for k in 0:Infinity) = 1 + z + z^2 + z^3 + ...,
    

which is a complex function whose domain is the open unit disk centered at 0,
analytic continuation would give f(z)=1/(1-z), which has a much larger domain
---all of C except for 1. We can "make sense of" 1 + 2 + 4 + 8 + 16 + ... by
plugging 2 into the analytic continuation of the original function, which is
1/(1-2) = -1.

Similarly, zeta(-1) would be 1 + 2 + 3 + ..., which doesn't make any sense by
itself, but the analytic continuation of zeta actually takes a value at -1,
which is -1/12\.
[https://www.wolframalpha.com/input/?i=Zeta%5B-1%5D](https://www.wolframalpha.com/input/?i=Zeta%5B-1%5D)

