Is the sum 1+2+3+4+... really equal to -1/12?

[Shaun Culver]

How can

1+2+3+4+... = -1/12

?

Apparently, this series has been used in quantum physics - giving it physical significance! True/False?

 

[ice109]

yea that's definitely wrong, that series diverges

 

[Noone]

it has signifcance to others due to the perception of what they came up with there thought. it seems like you don't want to grant it the same significance the others have.

you asked (how can) so find how meany ways if can't be then find how meany ways it can't be, then compair to what others have said. mainly to bring your self more understanding of what makes it and why its true or false -.- this way you can teach your self the same concepts they were teached by others. anyways... its true in a few ways and false in more... just ask what would make it false or true and why, then you would know how it can and how it can't without being told by others. So i won't spoil your learning of the workings of this world, by just telling you. but i will say a way how you can learn for your self :D

 

[sutupidmath]

first signt it is wrong: the sum of any real positive numbers cannot be negative.

second: even if the result you posted was positive, 1+2+3+4+... its terms are of an arithmetic sequence and therefore it diverges. so it cannot equal any real number.

 

[Xislaben2]

How can

1+2+3+4+... = -1/12

?

Apparently, this series has been used in quantum physics - giving it physical significance! True/False?

It's a trick, a trick of notation.

This equation is of historic importance, and comes from Srinivasa Ramanujan (1887 - 1920).

He was basically a poor Indian who didn't have any formal mathematical schooling, so he made up his own syntax. What that series really represents is:

1/(1^-1) + 1/(2^-1) + 1/(3^-1) + 1/(4^-1) + ... + 1/(n^-1) = -1/12

It was a pretty important discovery for analyzing the Riemann zeta landscape, and Ramanujan did the whole thing on his own apparently, which greatly impressed Brittan's Hardy and Littlewood.

As for the quantum physics usage, I believe the distribution in the Riemann zeta looks similar to something with electron distribution.

 

[Shaun Culver]

It's a trick, a trick of notation.

This equation is of historic importance, and comes from Srinivasa Ramanujan (1887 - 1920).

He was basically a poor Indian who didn't have any formal mathematical schooling, so he made up his own syntax. What that series really represents is:

1/(1^-1) + 1/(2^-1) + 1/(3^-1) + 1/(4^-1) + ... + 1/(n^-1) = -1/12

It was a pretty important discovery for analyzing the Riemann zeta landscape, and Ramanujan did the whole thing on his own apparently, which greatly impressed Brittan's Hardy and Littlewood.

As for the quantum physics usage, I believe the distribution in the Riemann zeta looks similar to something with electron distribution.

Thank you. This is elegant. This is profound!

 

[Shaun Culver]

How would you set this series up in mathematica code to give a sensible answer?

 

[Big-T]

This is because Mathematica uses summation by partial sums. When using this method of summation; the series diverges.

Have a look at:
http://en.wikipedia.org/wiki/Ramanujan_summation .

 

[Xislaben2]

How would you set this series up in mathematica code to give a sensible answer?
http://files.liveadmaker.com/F/11767471/Image.gif
This equation produces an error.

Couldn't tell you, I'm much better at reading about mathematics than actually putting it into practice :(


du Satoy, Marcus . The Music Of Primes: Searching to Solve The Greatest Mystery in Mathematics. New York: Harpercollins, 2004. (p137)

 

[Shaun Culver]

Thank you for the link (Big-T) & the reference (Xislaben). I have read (on John Baez's website) that this equality is important in string theory.

 

[Xislaben2]

The Stability of Electron Orbital Shells based on a Model of the Riemann-Zeta Function

http://www.ptep-online.com/index_files/2008/PP-12-01.PDF

 

[stabu]

Sorry, I don't get it

\sum^n_{r = 1} r = \sum^n_{r = 1} \frac{1}{r^{-1}}

Says the same thing to me, unless we define the $-$ sign differently.

I don't see the trick in the notation.

 

[Don Carnage]

Sorry, I don't get it

\sum^n_{r = 1} r = \sum^n_{r = 1} \frac{1}{r^{-1}}

Says the same thing to me, unless we define the $-$ sign differently.

I don't see the trick in the notation.

The problem is that you compare the Ramanujan's sum with a normal sum..

 

[CRGreathouse]

Ramanujan summation is a formal way to sum series that would otherwise diverge. I can give only a small example of 'how it works'; I can't even prove the 1 + 2 + 3 + ... example yet.

Let a_n = (-1)^n. Then the series is
S = 1 - 1 + 1 - 1 + 1 - ...
-1+S = -1 + 1 - 1 + ...

Adding the two,
2S - 1 = 0 + 0 + 0 + ...
2S - 1 = 0
S = 1/2

 

[Santa1]

This was discussed here =D

https://www.physicsforums.com/showthread.php?t=149279

 

[alxm]

The Stability of Electron Orbital Shells based on a Model of the Riemann-Zeta Function

http://www.ptep-online.com/index_files/2008/PP-12-01.PDF

Ridiculous numerology coming from a journal I've never heard of.

Just a cursory glance turns up several factual errors, no real physical justifications, and essentially, the whole paper amounts to the ridiculously shallow observation that the Z of atoms with a single electron in their valence shell (a set of 17 numbers) corresponds to some of the Riemann zeta primes.

The reality is that orbital filling is perfectly well-understood as following from the Pauli principle and allowed combinations of quantum numbers. That sequence is trivially derivable and looks nothing like the Riemann zeta function.

Stupidest thing I've seen all day.

 

[ThirstyDog]

I was under the impression that it was more sum
S_{p} = \sum_{n=1}^{\infty} \frac{1}{n^{p}}
was only convergent for p greater than one. But if you take its analytic continuation of the function then you obtain the result
S_{-1} = \frac{-1}{12}. [/itex]<br /> So I think it reduces to a question of analytic continuation. <br /> <br /> This is what I remember from a course in string theory but at the same time I felt a little strange about it as well.