Does 1+1+1+... = 1/2 or -1/2?

depends on whether the riemann hypothesis is true, false, or undecidable in ZFC

(it's equivalent to the wave/particle duality)

>>8722695
>I paid for a 4chan pass to shitpost more efficiently

Jesus Fucking Christ anon.

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>>8722710
>willingly supporting the google/NSA captcha database
>not willingly supporting the last bastion of online free speech
Jesus Fucking Christ anon.

>>8722715
>giving up your anonymity

>>8722717
>thinking you're anonymous to 4chan

>>8722686
neither

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>>8722717
who am i?

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>>8722715
>4chan
>bastion of free speech

>>8722715
>willingly funding gmookt's bitcoin mining

>>8722686
>1+1+1+...
Not a valid notation.

>>8723351
But you still understood it.

>>8723351
>Not a valid notation.
elaborate, why not?

1+1+1...=1+1+1...
a=1+1+1+...
0.999a=0.999+0.999+0.999...
a=0.999+0.999+0.999.../0.999
1+1+1+...=0.999(1+1+1+...)/0.999(1+1-1+1-...)
1+1+1+...=0.999/0.999
1+1+1+...=1
i crack the jee with this looool

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>>8723423
>0.999+0.999+0.999.../0.999 = 0.999(1+1+1+...)/0.999(1+1-1+1-...)

I guess /sci/ is just brainlets.

>>8722686
yes

>>8722686
>1+1+1+... =
>...
what do you mean by ...?

>>8723973
I think you can guess.

>>8722686

Being a geometric series whose R is equal to or greater than one, it diverges and has no finite sum

Meme degreers will disagree

>>8723973
it means "etc"

>>8723360
unless he misunderstood it. hard to sayyyyyy

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>>8723351
>>8723973
Brainlet pedants btfo.
https://terrytao.wordpress.com/2010/04/10/the-euler-maclaurin-formula-bernoulli-numbers-the-zeta-function-and-real-variable-analytic-continuation/

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[math] \sum_{n=1}^\infty n^m z^n + (-1)^m\dfrac{m!}{\log(z)^{m+1}} = -\dfrac{1}{m+1} B_{m+1} + {\mathcal O}((z-1)) [/math]

So

[math] \lim_{z\to 1} \left( \sum_{n=1}^\infty n^m z^n + (-1)^m\dfrac{m!}{\log(z)^{m+1}} \right) = -\dfrac{1}{m+1} B_{m+1} [/math]

and e.g.

[math] \lim_{z\to 1} \left( \sum_{n=1}^\infty n z^n - \dfrac{1} {\log(z)^{2}} \right) = -\dfrac{1}{2} \dfrac{1}{6} [/math]

and

[math] \lim_{z\to 1} \left( \sum_{n=1}^\infty z^n + \dfrac{1}{\log(z)} \right) = - \dfrac{1}{2} [/math]

If you think it's [math] + \dfrac{1}{2} [/math], you may have been looking at a source which uses the alternative Bernoulli numbers (pic related), but those aren't the ones matching up with the zeta function, e.g. in formulas such as

[math] \zeta(2n) = \dfrac{(2\pi)^{2n}}{(2n)!} (-1)^{n+1} \dfrac{1}{2} B_{2n} [/math]

>>8725062
>those aren't the ones matching up with the zeta function
How do you know?

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>>8725071
Depends on what you expect as an answer. You can take Mathematica.
I can also rant about
https://en.wikipedia.org/wiki/Euler%E2%80%93Maclaurin_formula#The_formula
where the standard BernoulliB's are used

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>>8725094
ad.:
I remembered I also at one point tried to write down a derivation that only assumes a Taylor expansion, pic related
Although of course this avoids the Bernoullis

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ad 2.:
okay I played around with this series regulation a bit and I find that OPs sum is indeed the one where several expressions that have 1+1+1+1+ as limit all are regularized by the same term (the -1/log(z)) and that one it gives you any result.

Specifically limit z->1 via

1 + z + z^2 + z^3 + z^4 ---> 1+1+1+1+

gives you +1/2 and limit z->1 via

z + z^2 + z^3 + z^4 ---> 1+1+1+1+

gives you -1/2.

>>8725132
ty

>>8723351
>series aren't valid

>the power of american education

[eqn]\sum_{n=0}^\infty z = \frac{1}{1-z} [/eqn]

now take the limit as z->1

[eqn]1+1+1... = \lim_{z \rightarrow 1} \frac{1}{1-z} = -\frac{1}{2}[/eqn]

viola!

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>>8722695
Are you saying that the value of this series is someone related to wave/particle duality? If so why can't we just do experiments in quantum mechanics and verify if this sequence is correct, and therefore find the truth about the Riemann hypothesis?

>>8725291
he was trolling

>>8725156
nice limit of 1/(1-z)