How can the sum of all naturals numbers up to infinity be -1/12?

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divergent series are not scalar values

>>7920871
Second line is wrong, it should be 0.

>he thinks infinity is a natural number

>>7920868
It isn't, you can't sum that series in a traditional sense because it diverges.

>>7920868
It's not. The sum is undefined, or you could say there is no sum. But you can do some fancy math manipulation to make it equal to -1/12. A kind of fudging in order to use the sum for practical purposes. But in reality, there is no sum (or it's undefined, or whatever they call it).

>>7920868
Stop watching retarded scam garbage like numberphile. That show is literally the quantum autism of the math world

>>7920868
when you prove things like 1=2, at least one>>7921485
of two things is wrong. Either your assumptions before doing the problem are wrong, or you have twisted the mathematical rules in order to arrive at such nonsense.

>>7921402
That's the troll part. It depends on if it ends in -1 or +1, the answer is then 0 or 1, so physicists say the answer is 1/2, or in between the two answers.

>>7921402
>second line is wrong
no, it is true.
proof:
S1 - 1 = -1 + 1 - 1 + 1 - 1 + 1...
=(-1) - (-1) + (-1) - (-1) + (-1) - (-1)...
= - S1

ie S1 - 1 = -S1
ie 2*S1 = 1
ie S1 = 1/2
qed

>>7920868
Actually, [math] \sum_{k=0}^{\infty}k=\frac{e^{i\pi}}{11.999...}[/math] you silly goose
http://fr.scribd.com/doc/233602815/Barnetts-Identity-Pdf1

>>7922861
Wait what did I do wrong with the latex tags? It ruined everything

>>7922868
probably the space at the start

>>7922868
[eqn]\sum_{k=0}^{ \infty } k=\frac{e^{i \pi }}{11.999...}[/eqn]
learn2scitex

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2KB, 136x62px

>>7923021
>>7923107
[math]\sum_{k=0}^{ \infty } k=\frac{e^{i \pi }}{11.999...}[\math]
You know what I'll just keep using pictures

>>7923127
>[math]\sum_{k=0}^{ \infty } k=\frac{e^{i \pi }}{11.999...}[/math]
My bad

>>7923131
Yeah, /sci/'s LaTeX renderer is odd. If you input something right next to a closing bracket, it won't render. But if you add a whitespace between the closing bracket, it will.

t.ex: \sum_{k=0}^{\infty}k=\frac{e^{i \pi }}{11.999...} would not render but \sum_{k=0}^{\infty} k=\frac{e^{i \pi }}{11.999...} will.

>>7923148
I see, that's good to know

>>7920868
it ain't a sum, son

>>7921506
>>7920871


These posts are not true and the actual answer is "more rigorous" and has to do with defining a notion of convergent "summation" based on analytic continuation of the Riemann zeta function in the complex plane.

In fact, Numberphile even did a response video where they point this out. In fact, it's one of Numberphile's best videos, in constrast to the original one (which is rightfully regarded as one of their worst).

>>7920868
The Riemann conditional convergence theorem.
The sum of all natural numbers is also -10. And -pi. Also -123456789. Also exactly 1. Also exactly 0.

>>7923820
I'm amazed someone who knows the Riemann conditional covnergence theorem doesn't know that what OP means by "sum" is not "the limit of partial sums"

>>7923843
Do you really think OP knows the difference?
Also, I'm unsure how exactly you define an infinite sum without using the limit of a sequence of partial sums. To be honest it may be my own ignorance but I would love to hear your explanation.

>>7923820
>The sum of all natural numbers conditionally converges
Fuck off, retard.

>>7923906
>doesn't realize >>7922597 is talking about a conditionally convergent sum.

>>7923906
Same guy you replied to again.
Are you seriously so retarded that you don't realize the "proofs" people are posting in this thread use conditionally convergent series so it might be relevant to the discussion at hand when I mention the RCCT?

>>7920871
Why do they shift the S2 term over on the fourth line in 2S2?