If the sum of all natural numbers equals [math]-frac{1}{12}[/math], what does the sum of all prime numbers equal?

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

Other urls found in this thread:

en.wikipedia.org/wiki/Divergence_of_the_sum_of_the_reciprocals_of_the_primes
mathworld.wolfram.com/PrimeZetaFunction.html
en.wikipedia.org/wiki/Prime_number_theorem
twitter.com/NSFWRedditImage

why not try doing the math yourself, OP?

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.

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

>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

for how long?

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

Take the limit.

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

otnx

infinity

>If the sum of all natural numbers equals

But it doesn't.

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

relevant article: en.wikipedia.org/wiki/Divergence_of_the_sum_of_the_reciprocals_of_the_primes

Why would a limit exist?

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.

> 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. 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.

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

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

Whats the point of calculating the analytic continuation ?

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

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
en.wikipedia.org/wiki/Prime_number_theorem

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.

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

>I'd probably get in Cambridge
No.
Lrn2probabilly

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

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.

Well meme'd.