lpetrich
Contributor
That's because of how one gets to the sum 1 + 2 + 3 + ... Done term-by-term, it is convergent, but as a limit of sums of negative powers, it is indeed -1/12, though one needs some mathematical trickery to get there.
That kind of sum is called the
Riemann zeta function, and it is given by
\( \zeta(s) = \sum_{n=0}^\infty \frac{1}{n^s} = \frac{1}{\Gamma(s)} \int_0^\infty \frac{x^{s-1}}{e^x - 1} dx \)
where the Euler gamma function
\( \Gamma(s) = (s-1)! = \int_0^\infty x^{s-1} e^{-x} dx \)
for integer s.
Neither the sum or the integral converge for s <= 1, but evaluating the integral close to s = 1 gives
\( \zeta(s) = \frac{1}{s-1} + O(1) \)
That means that one might be able to define that function for values of s less than 1, and one does that by Analytic continuation, finding some function with the same value for s > 1, but some function that is well-defined by s < 1. In effect, going around s = 1 in the complex plane.
There is, in fact, such an analytic continuation. It uses a variation of the Riemann zeta function called the Dirichlet eta function:
\( \eta(s) = \sum_{n=0}^\infty \frac{(-1)^{n+1}}{n^s} = (1 - 2^{1-s}) \zeta(s) = \frac{1}{\Gamma(s)} \int_0^\infty \frac{x^{s-1}}{e^x + 1} dx \)
The integral here is well-behaved at s = 1, and it can be used to define analytic continuation down to s = 0.
One can go further and get Riemann's functional equation:
\( \zeta(s) = 2^s \pi^{s-1} \sin \left( \frac{\pi s}{2} \right) \Gamma(1-s) \zeta(1-s) \)
This gives us special values
\( \zeta(0) = - \frac{1}{2} ,\ \zeta(-1) = - \frac{1}{12} \)
That is where
1 + 2 + 3 + ... = -1/12
comes from.
It also gives us
1 + 1 + 1 + ... = -1/2
That kind of sum is called the
Riemann zeta function, and it is given by\( \zeta(s) = \sum_{n=0}^\infty \frac{1}{n^s} = \frac{1}{\Gamma(s)} \int_0^\infty \frac{x^{s-1}}{e^x - 1} dx \)
where the
Euler gamma function\( \Gamma(s) = (s-1)! = \int_0^\infty x^{s-1} e^{-x} dx \)
for integer s.
Neither the sum or the integral converge for s <= 1, but evaluating the integral close to s = 1 gives
\( \zeta(s) = \frac{1}{s-1} + O(1) \)
That means that one might be able to define that function for values of s less than 1, and one does that by
Analytic continuation, finding some function with the same value for s > 1, but some function that is well-defined by s < 1. In effect, going around s = 1 in the complex plane.There is, in fact, such an analytic continuation. It uses a variation of the Riemann zeta function called the
Dirichlet eta function:\( \eta(s) = \sum_{n=0}^\infty \frac{(-1)^{n+1}}{n^s} = (1 - 2^{1-s}) \zeta(s) = \frac{1}{\Gamma(s)} \int_0^\infty \frac{x^{s-1}}{e^x + 1} dx \)
The integral here is well-behaved at s = 1, and it can be used to define analytic continuation down to s = 0.
One can go further and get Riemann's functional equation:
\( \zeta(s) = 2^s \pi^{s-1} \sin \left( \frac{\pi s}{2} \right) \Gamma(1-s) \zeta(1-s) \)
This gives us special values
\( \zeta(0) = - \frac{1}{2} ,\ \zeta(-1) = - \frac{1}{12} \)
That is where
1 + 2 + 3 + ... = -1/12
comes from.
It also gives us
1 + 1 + 1 + ... = -1/2