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Comment by colejohnson66 | Hacker News Reader

Random fact: this only works if the series is finite. Otherwise you end up with sum(1...inf) equaling -1/12.

That's the sum of all integers on the complex plane, solved using an analytical continuation of the Riemann zeta function at -1 (fwiw, by the same definition, the sum of 13, 26, 39...inf is also -1/12).

It's disingenuous to assert that is same as the sum of that infinite series without the associated caveats. By the definitions of infinite series that we all learned in calculus, that is a divergent series with no sum.

Edit: Wolfram Alpha[0] has a good graphic showing why this series converges to -1/12 (the one with the red line, drawing a peach-like shape). It also gives an intuition as to how complex numbers are influencing the results despite being omitted from the equation.

[0] http://mathworld.wolfram.com/RiemannZetaFunction.html

Not true. Infinite sums are not defined for series that don’t converge, and sum(1..n) does not converge as n => infinity.

Yes, you can play a nice trick with that sum if you ignore the fact that inf-inf is meaningless.

It depends on how you define the summation of infinite series. With the standard convergence type definitions you are correct. But it is possible to define this sum in a consistent way (e.g. Ramanujan summation or Riemann Zeta analytic continuation) as shown in Hardy's “Divergent Series”. The cost is that they have properties like rearranging the order of the terms gives a different result. Apparently this sum can come up in Quantum Field Theory when calculating vacuum force between two conducting plates.

This can be disproven by noting that the naturals are closed under addition.

The math involved in reaching this identity is akin to dividing by zero.

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