# P-adic numbers and the Ramanujan summation

• I
• Spathi
In summary, the Ramanujan summation is a sum that is used in physics for predicting the Casimir effect. It is also used in mathematics, specifically in the p-adic numbers. It is not very easy to obtain using p-adic numbers, and would require some additional assumptions to be made.
Spathi
Gold Member
TL;DR Summary
My question is: can the Ramanujan summation be relatively easily obtained using the p-acid numbers?
In mathematics, there is the Ramanujan summation:

$$1+2+3+4+...=-\frac{1}{12}$$

https://en.wikipedia.org/wiki/1_+_2_+_3_+_4_+_⋯

This sum is used in physics for predicting the Casimir effect:

https://en.wikipedia.org/wiki/Casimir_effect

I have also heard that this sum was used in the string theory (more precisely, in the original bosonic string theory).

Then, in mathematics the p-adic numbers are used:

My question is: can the Ramanujan summation be relatively easily obtained using the p-adic numbers?

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Here is what you need to know to even understand the Ramanujan summation:
https://www.physicsforums.com/insights/the-extended-riemann-hypothesis-and-ramanujans-sum/
and here is explained what p-adic numbers are:

Ramanujan's summation is based on classical analysis, i.e. it depends on the Euclidean metric that induces an Archimedean evaluation. The only connection to p-adic numbers that I am aware of would be Hasse's principle that primarily deals with polynomials, i.e. Diophantic expressions, not series. Since
$$1+2+3+4+\ldots \neq -\dfrac{1}{12}$$
it is not even clear how Hasse's principle could be applied to the continuation of Riemann's zeta-function, let alone being "easier obtained". p-adic numbers have at first glance absolutely nothing to do with Riemann's zeta-function, so the connection between them alone would be quite a challenge.

Spathi said:
My question is: can the Ramanujan summation be relatively easily obtained using the p-acid numbers?

I'm a little confused. What exactly do you think is the relationship between the p-adics and Ramanujan summation?

In any event, ##1+2+3+\ldots## does not converge ##p##-adically for any prime ##p.##

fresh_42
A quick internet search gave me reasonable results about p-adic Dirichlet functions (L-functions), so the question is not completely out of thin air. However, I haven't searched for results in English. The idea is to generalize Euler's formula
$$\zeta(1-n)\;=\;-\dfrac{B_n}{n}$$
and consider ##f(n) =(1-p^{n-1})\zeta(1-n)## (under some technical conditions on the domain) as a unique continuous function ##f\, : \,\mathbb{Z}_p\longrightarrow \mathbb{Q}_p.##

https://www.mathi.uni-heidelberg.de/~mfuetterer/texts/da_fuetterer.pdf

Infrared said:
I'm a little confused. What exactly do you think is the relationship between the p-adics and Ramanujan summation?

In any event, 1+2+3+… does not converge p-adically for any prime
If I am not mistaken, in the video I provided above, there it is shown, how with p-acics the following equation can be obtained:

$$1+x+x^2+x^3+x^4...=\dfrac{1}{1-x}$$

I don't remember how this equation is standardly obtained for x<1, somehow please remind. For x>1, if I am not mistaken, p-adics provide a proper way to get this sum. Maybe the Ramanujan summation and $$1+2+4+8+16...-1$$ are something very similar?
And one more question - maybe we should remove some axioms from "standard" mathematics to get alternative "hyberbolic" mathematics where these sums are fully correct?

One more question: if we find the formula

$$1+x+x^2+x^3+x^4...=\dfrac{1}{1-x}$$

Can we "prove" with a similar formula, that 2+6+18+48+144...=-1? This will mean that (2)2=-1 in ternary system.
If I am not mistaken, (9)9=-1 by definition in pi-adic numbers? I mean that if we sum (9)9 and (0)1, we get 0(0).

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Sorry, but you need to understand what p-adic numbers are and even more what that formula means. ##\sum_{n\in \mathbb{N}_0}x^n=\frac{1}{1-x}## is only valid in ##\mathbb{F}((x)).##

The moment you set ##x=c \in \mathbb{F}## you actually consider a mapping ##f\, : \,\mathbb{F}((x))\longrightarrow \mathbb{F}.## Next, you have to investigate whether and when this function is well-defined, and has finite values. Ramanujan summation is due to a continuation of the zeta-function according to standard analysis. p-adic numbers are not "different" numbers, they only have a different absolute value. This leads to a different meaning of convergence and possibly different results where ##f(c)## converges. However, this isn't obvious, possibly wrong (cp. post #3), and has to be shown - and definitely not by watching a video. After you have done all this, you have to explain what a p-adic zeta-function is, how its continuation is defined, and what it has to do with ##f.## Some of these may exist in exotic papers, like the p-adic L-functions mentioned in the thesis(!) I quoted in post #4, but neither is standard knowledge, so it has either to be proven or referenced.

Bottom line: There is a huge gap between your idea and what has to be done to consider its validity. We cannot close this gap on PF, or in a video on YouTube. I provided some references as first steps to narrow this gap, and maybe there are more papers out there, but it is not on us to find them, and even less to elaborate on the missing theorems. So far we are stuck at the quotient field of the ring of formal power series ##\mathbb{F}((x))## where your formula holds. It is a formal equation with an indeterminate. I have tried to explain how this is connected to Ramanujan's summation in https://www.physicsforums.com/insights/the-extended-riemann-hypothesis-and-ramanujans-sum/
and provided proofs.
$$1+2+3+\ldots=\zeta(-1)=\underbrace{\zeta(1-2)=2(2\pi)^{-2}\Gamma(2)\cos\left(\dfrac{2\pi}{2}\right)\zeta(2)}_{\text{This is the important part!}}=-\dfrac{1}{12}$$
We need something similar in the p-adic case before we can discuss your idea.

Since this is neither trivial nor YouTube a valid reference, I'm afraid we cannot do much about it.

I have a question: how can we prove that in "normal" (non-adic) numbers the expression 1+2+4+9+16...-1 is wrong? Can this be derived from Peano axioms?
Here's a simple reasoning:
When you add a positive number to a positive number, the sum becomes greater than both. The more often new terms are added to the sum, the greater the sum. Since infinity is greater than a finite number, if you add new terms to the sum infinitely, the sum will become even larger. Therefore, such a sum will obviously be greater than a negative number.
Is it possible to translate this reasoning into the language of formal mathematical constructions?

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Spathi said:
I have a question: how can we prove that in "normal" (non-adic) numbers the expression 1+2+4+9+16...-1 is wrong? Can this be derived from Peano axioms?
Yes. If you assume the "normal" Archimedean order, then all squares are positive, and the sum of positive integers is positive again, so it cannot equal a negative number.

Spathi said:
Here's a simple reasoning:
When you add a positive number to a positive number, the sum becomes greater than both. The more often new terms are added to the sum, the greater the sum. Since infinity is greater than a finite number, if you add new terms to the sum infinitely, the sum will become even larger. Therefore, such a sum will obviously be greater than a negative number.
Yes, but any argument that involves the word "infinity" has to be considered as questionable. You cannot treat infinity like you treat numbers. It means "greater than an arbitrarily chosen integer," i.e. it is defined with finite numbers. "Infinity" is a shortcut and common language, it is not suitable as a mathematical term without a precise definition of what it means in a particular context.

Spathi said:
Is it possible to translate this reasoning into the language of formal mathematical constructions?
So a formalization would be along the lines:

Let ##M\in \mathbb{N}.## Then there exists a number ##n\in \mathbb{N},## namely ##n:=\lfloor \log_2 M \rfloor ## such that ##2^n \leq M < 2^{n+1}.## Thus ##2^{n+1} -1 = \sum_{k=0}^n 2^k \geq M>0>-1.##

This shows that your sum becomes greater than any given finite limit ##M,## i.e. infinitely large.

You could refine this argument in many ways if you want to start with Peano. I used the fact that the logarithm is strictly increasing. Without such an analytical argument, you could show the existence of ##n## per induction or dividing ##M## successively by two. Or you could prove the formula for geometric series. If you start with the Peano axioms only, then there is quite a way to go.

DrClaude

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