If the sum of all natural numbers equals [math]-\frac{1}{12}[/math],

>>8929804
why not try doing the math yourself, OP?

>>8929804
You'd have to figure out a way to generate all the prime numbers first. As far as I know, that's really hard to do.

>>8929804
[eqn]\frac {-1}{7}[/eqn]

>>8930003
>Take largest prime you know
>Add 2
>Check if it's prime by simply dividing by all numbers smaller than it
>Continue adding 2 and checking
Wew that was hard

>>8930028
for how long?

>>8930028
>largest prime i know is 2
Checked and maited, mathematicist.

>>8930031
Take the limit.

>>8930033
Sorry I'm not putting different cases that would only apply to brainlets anyway.

>>8930042
otnx

>>8929804
infinity

>>8929804
>If the sum of all natural numbers equals

But it doesn't.

>what does the sum of all prime numbers equal?
It diverges

>>8929804
relevant article: https://en.wikipedia.org/wiki/Divergence_of_the_sum_of_the_reciprocals_of_the_primes

>>8930042
Why would a limit exist?

http://mathworld.wolfram.com/PrimeZetaFunction.html

I don't know how the analytic continuation goes, but apparently you can't find "the prime sum" the same way you can sum natural numbers.
While the source only says that you can't go past the imaginary axis and not that there is nothing there, I would think that the sum does not exist even in the loose sense used here.

>>8929804
> If the sum of all natural numbers equals -1/12

No, the series diverges, so the sum is undefined.

What you're thinking about is called the "analytic continuation" of the series.

For a good example of the difference between the two, look at the difference between the function 1/(1-x) and the geometric series (which is the sum of x^k for k=0..∞). Just because those two functions yield the same value when |x|<1 doesn't mean they're the same function. For example, if x = -1, then 1/(1-x) yields 1/2, but the geometric series diverges (see: Grandi's series) -- which proves that they're not the same function.

What 1/(1-x) gives you is the "analytic continuation" of the geometric series for cases where |x|>=1. But having an analytic continuation function like 1/(1-x) IN NO WAY implies that it's the same function as the original series -- and they can definitely yield different results on some values.

For your series, the appropriate function is called "zeta(-1)", which is undefined -- but using analytic continuation techniques on it will yield -1/12.

>>8930028
>Wew that was hard
Doing it for all of eternity would be.

>>8929804
[math]-\frac{1}{12}[/math]

>>8931160
Whats the point of calculating the analytic continuation ?

>>8931399
To make the sum equal to [math]-1/12[/math].

>>8931399
To get a everywhere complex differential function that coincidences with a function for which you know the expansion at some places.
For uses of the zeta function in particular, consider
https://en.wikipedia.org/wiki/Prime_number_theorem

>>8929804
S = 2 + 3 + 5 + 7 + 11 + ... ;
S = (1 + 1) + (1+1+1) + (1+1+1+1+1) + (1+1+1+1+1+1+1) + (1+1+1+1+1+1+1+1) + ...

It is clear we can keep doing this indefinitely, and because addition is associative:

S = 1 + (1+1) + (1+1+1) + (1+1+1+1) + ...
Thus.
S = 1+2+3+4+5+...
Which we know to be -1/12.
Thus the sum of all primes is -1/12.

>>8931433
I'd probably get in Cambridge if I wrote this and I was a 3rd world shitskin.

If the sum of all naturals is 1/12 then you know the sum of primes must be slightly less, like 1/7 for crude approximation

>>8931434
>I'd probably get in Cambridge
No.
Lrn2probabilly

>>8931433
lmao
>the sum of all prime numbers is -1/12

>>8930028
brainlet detected.
Even for a quick" off the top of your head" algorithm (watered down so brainlets can understand it) this is absolutely horrible.

>Take largest prime you know
2. 2 is therefore the only prime number.

>Add 2
Fair enough.

>Check if it's prime by simply dividing by all numbers smaller than it

What's the point of adding 2 (instead of 1), when you're going to include even numbers in the checking process anyway. That triggered me while being triggered, nice assassination attempt, but I'm mentally robust.

>>8931433
Well meme'd.