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13 / 3 / 13 / ?

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  No. 236
>>237 >>239 >>242 >>247 >>837
-1/12

contradiction or too complex for mere mortals to understand?
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  No. 237
>>236
there's probably a dividing by 0 going on somewhere
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  No. 239
>>236
It's genuinely a false statement, it should read infinity. Negative one over 12 comes from the Riemann Zeta function, which can be evaluated using the above formula for values where it converges. To define it for values where it doesn't converge, they used different equations that agree for those values but have a larger domain. These larger domain equations are where the negative one over 12 comes from.
https://en.wikipedia.org/wiki/Analytic_continuation
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  No. 242
>>236
https://en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_⋯
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  No. 247
>>236
Ramanujan's brain surpasses mere mortals. I wouldn't be surprised if it were true, as I haven't seen the proof for this. Nevertheless this is just one of the things in mathematics that contradict logic
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  No. 253 >>278

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yes.
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  No. 278
>>253
For a group that believes themselves to be "rational", mathematicians sure love creating literal cults around people.
Humans are disgusting.
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  No. 283
Doesn't work per the common definition of an infinite sum
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  No. 286
abuse of notation
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  No. 304

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This physicist's treatment of the -1/12 issue from Zee's QFT book is the best treatment of the question I've seen.
>Appallingly, in an apparent attempt to make the subject appear even more mysterious than it is, some treatments simply assert that the sum is by some mathematical sleight-of-hand equal to -1/12. Even though it would have allowed us to wormhole from (1) to (3) instantly, this assertion is manifestly absurd. What we did here, however, makes physical sense.
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  No. 305
Behold, the Barnett-Tooker-Wildberger conjecture.

Let

ζ^b(s)=∑k=1∞^−bn−s\qquad \hat\zeta_b(s) = \sum_{k=1}^{\hat\infty - b} n^{-s}


then

ζ^e+π+2(−1)=eiπ11+0.999…\qquad \hat\zeta_{e + \pi + \sqrt 2}(-1) = \frac{e^{i\pi}}{11 + 0.999\dots}
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  No. 315
I still don't get it.
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  No. 811
Contraddiction, because in Ramanujan's proof there are mistakes i.e. the term equals zero changes the number of step in the serie and also it's true that
2>Σn=0+∞12n 2 > \Sigma_{n=0}^{+\infty} \frac{1}{2}^n
>>
  No. 837
>>236
it diverges up, right?
and the limit is the first value something dosent reach
therefore, the limit of the series must be down

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