1+2+3+4... = -1/12

[Zapmeister]

i still maintain that it doesn't make sense to talk about a divergent series having a finite value as a sum.

let's go back to my earlier example of

Originally Posted by Zapmeister

(1-x)^-2 = 1 + 2x + 3x^2 + 4x^3 + 5x^4 + ...

and the expansion is valid for |x|<1

of course everyone agrees that if you plug x=0.1 in here you get 1+0.2+0.03+0.004= ... = 100/81, because x=0.1 is inside the radius of convergence.

you're saying: if i plug in x=10, then i get a value of the sum 1+20+300+4000+ ... = 1/81 which is justified by analytic continuation, since there is a unique analytic continuation of the sum 1+2x+3x^2+4x^3... outside its radius of convergence, and this analytic continuation just happens to be equal to (1-x)^-2.

here's the thing. if you wanted to do that, you'd need to tell me what the function you're trying to take out of its radius of convergence is. and because the series is divergent, you're not going to get the same answers whichever way you do it. so you're trying to say that 1+2+3+4... represents the analytic continuation of

1^-s + 2^-s + 3^-s + 4^-s + ... | s=-1

at s=-1, which is the zeta function at -1, which is -1/12.

there is no way to determine that when you wrote 1+2+3+4... you're starting from the zeta function and working things out from that. you could be trying to get it from the analytic continuation of another function.

here's another way to write it. how about this?

1+2+3+4+... = 1 + 2x + 3x^2 + 4x^3 + ... | x=1
= (1-x)^-2 | x=1
= 0^-2 = "infinity"

you're still not convinced?

1+2+3+4... = 1 + 2x + 3x^2 + 4x^3 + ... | x=1
= (1-x)^-2 | x=1
= ( (1-x)^-1)^2 | x=1
= (1 + x + x^2 + x^3 + ...)^2 | x=1
= (1 + 1 + 1 + 1 + ...)^2
= (zeta(0))^2
= 1/4

yeah. none of this would be happening if you used a valid x inside the radius of convergence, since everything would agree on a certain value.

this is the point. there may be occasions when you're trying to work out what the analytic continuation of 1^-s + 2^-s + 3^-s + ... is at s=-1. in these cases you'd write it as zeta(-1), because the zeta function is defined to be this analytic continuation. you'd never write it out as 1+2+3+4+... because nobody knows what you mean when you're writing down the sum of a divergent series. it's complete trash. just like your face (oohh burn)