Do you guys believe this,if not demonstrate why not

If S is this sum, you can show that [math]0 \,=\, S \,-\, 2\,S \,+\, S \,=\, 1[/math]. Stop falling for Numbermemes.

Prove it.

...

Its true in the right context

Here's your spoonfeeding, Pajeet.

The result is a regularization.

Note [math]\sum\limits_{n = 1}^\infty {n{e^{ - \varepsilon n}}} = \frac{{{e^\varepsilon }}}{{{{\left( {{e^\varepsilon } - 1} \right)}^2}}} = \frac{1}{{{\varepsilon ^2}}} - \frac{1}{{12}} + O\left( {{\varepsilon ^2}} \right)[/math].

Just to flesh this out a bit:

youtube.com/watch?v=jcKRGpMiVTw

This video makes it much more clear what people mean when they say "the sum of the natural numbers is -1/12". It's a distorted statement and very clickbaity; I refuse to watch any channels that make this claim without qualifying the statement with some amount of exploration of the Riemann Zeta function.

>ITT: confusing a function for the analytic continuation of that function

yeah i saw that video too

This is a twisted conception of math.

If we agree that S is equal to S (damn I can't believe I have to explain this) as long as the sum of numbers reaches a certain number k, the theory this video shows goes to shit because:

1+2+3+4+....+ a + a+1 + ....+ k
0 -2 -4 -6 -....- 2(a-1)- 2*a -....- 2*(k -1) - 2*k
0+0+1+2+...+ a-2 + a-1 +... + k -2 + k-1 + k
= 1+0+0+0+...+ 0 + 0 +........+ 0 - k -1 + k = 1 -1 -k +k = 0

See? It's stupid to think the result is 1.
You don't handle infinites that way.
Much less in a case like this where the sums are the same so therefore they "stop" when n = k, and if they're infinite a higher k is reached every time and the whole thing cancels itself with the "last" step.

Besides I hate asian faces so I get all the cringes when they try to explain things.

That is evidence that QFT is on the wrong track. Obviously any math that doesn't correspond to reality should be thrown out, because the point of math is to describe/predict/model reality. It's obvious that in reality if you have an uncountable positive number of something, that is not the same as having negative one twelfth of that thing.

terrytao.wordpress.com/2010/04/10/the-euler-maclaurin-formula-bernoulli-numbers-the-zeta-function-and-real-variable-analytic-continuation/

QFT is not like String Theory. Things like this aren't just used because they give mathematically "nice" results, but because they result in predictions that align with experimental values.

That approaches infinity as epsilon tents to infinity, fool.

>That is evidence that QFT is on the wrong track

Actually it's evidence of the opposite. If the predictions made by QFT using this formula weren't vindicated, then then formula would have been thrown out. The formula in fact does describe/predict/model reality quite well in the appropriate context, which is why it continues to be used in the theory.

en.wikipedia.org/wiki/Casimir_effect

Yes it does. But in QFT you deal with stuff like that all the time by setting bounds like [math]\frac{1}{{{\varepsilon ^2}}} < \Lambda [/math], determined by experiment. This then allows you to follow through and make calculations.

>is given proof that you can't sum naturals in a linear and stable fashion
>rejects the proof by summing in a completely different way (the correct one)
Are you deeply retarded or have you been half-asleep each time you've been exposed to Numberphile memes?

The summation index is countable though.

It's not true.
Proof :
By definition, [math]\zeta(s)[/math] is only defined for [math]\mathrm{Re}(s) > 1[/math]. The sum [eqn]1 + 2 + 3 + ... = \frac{-1}{12}[/eqn]
is obtained by computing [math]\zeta(-1)[/math] . Since [math] ]\mathrm{Re}(-1) = -1 \ngtr 1[/math], using the zeta function to compute this sum is a mistake.
A funny mistake, but a mistake.

THAT WAS THE JOKE MADE YOU FUCKING FAGGOT.
REEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE

>[math]\ngtr[/math]

good reaction image

Thanks if not sarcastic.

Here is a great video where this sum is mentioned.
youtube.com/watch?v=sD0NjbwqlYw
You have to see it in the context of the Riemann Zeta function, otherwise it is incorrect.

More simply, you CANNOT change the order of terms in a divergeant sum.

yall bitches read this

smoke tarry tao ery day

*<
sorry :/

at least post the real judy.

Without changing the order. See

By inducyion on the number of adds!
Base case (1 add): 0 + 1 = 1 > 0 is positive
Now suppose the sum up to K is positove. We add K+1 which is >= 2 and therefore positove. So the sum cannot be negative assuming omega consistency

t. CS first year

so what's anorexia gonna do when she hits 30?

...

this was amazing actually

m.youtube.com/watch?v=sD0NjbwqlYw
best math channel

you actually have no idea what you are talking about mate

Anzu is a pure princess. she is very healthy!

>Do you guys believe this,if not demonstrate why not

It's not really a sum...

It's the y intercept of a slope that represents the infinite series.

en.wikipedia.org/wiki/1_+_2_+_3_+_4_+_⋯

– the sum isn't uncountable
–math doesn't have a purpose fuck u
– although the sum doesn't equal \frac{-1}{12} it behaves like it algebraically so that's what physicists do to wave some hands and make some progress in their nonsensical (imo) field

ty for being the voice of reason, user

>dat cute little ass
*heavy breathing*

The conclusion isn't irrational

I was sincere

>math doesn't have a purpose

Yes it does. There's a reason we construct math so that 1 doesn't equal 2 -- we want math to coorespond to reality. We could construct it any way we want, but we choose the method we do because it works best as a tool. There's nothing else to it.

Hey that's my math lecturer!