Ramanujan Summation, Variations of

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• the_pulp
In summary, the conversation discusses the controversial statement that the sum of all natural numbers is equal to -1/12 and the use of Ramanujan summation and analytic continuation of the Riemann function as proof. It also raises questions about the possibility of obtaining different results by choosing other functions for analytic continuation and the uniqueness of the analytic continuation of the Riemann Zeta function.
the_pulp
Hi, I've seen several videos and documents that state that "the sum of all natural numbers is equal to -1/12". The "proof" in general is using ramanjuan summation and analytic continuation of the riemann function.

In this proof, the election of the riemann function in order to perform the analytic continuation seems just like one of the infinite functions we can choose.

So the questions would be:

1) if we choose other functions in order to perform the analytic continuation, can we obtain other results? (For example 174). Can you give examples?

2a) If that's the case, why in several branch of physics, the "-1/12 result" is used. What characteristic does -1/12 has that physicists tend to use it more than the rest of the numbers?

2b) if it is not the case, where is the demostration that all "analytic continuations proof" of this sum give -1/12?

Thanks in advance for any feedback.

Herer's a simple example. Let ##f(x)=\sum_0^\infty x^n=\frac{1}{1-x}##. ##f'(x)=\sum_0^\infty nx^{n-1}=\frac{1}{(1-x)^2}##. This holds for ##|x|\lt 1##. Let ##x=-1## and get ##\sum_0^\infty (-1)^n n=\frac{1}{4}##.

Sorry, this example is for the sum of n*(-1)^n. I was asking for the sum of n.
I don't know if you haven't read my question properly or i have not understood the connection between your example and my question.

the_pulp said:
Hi, I've seen several videos and documents that state that "the sum of all natural numbers is equal to -1/12". The "proof" in general is using ramanjuan summation and analytic continuation of the riemann function.

In this proof, the election of the riemann function in order to perform the analytic continuation seems just like one of the infinite functions we can choose.

So the questions would be:

1) if we choose other functions in order to perform the analytic continuation, can we obtain other results? (For example 174). Can you give examples?

Perhaps this question is not clear. I am trying to ask about other functions that could be used in order to perform the "analytic continuation trick" for the same serie and that may produce a result different than -1/12.

Just to confirm what you mean by "choice of function"...

We have chosen the function f(x,s)=x^(-s) and then put in the integers n=1,2,3... as x-arguments to that function, which then gives us the series 1 + 2 + 3... as long as we put in s = -1.

We could have chosen some other function g(x,s) that also passes through n when x=n for some special value s0 of s, but differs from f(x,s) in general. Then the sum of the terms g(n,s) for arbitrary complex s would be different, and might have convergent and non-convergent regions -- but the sums might be quite different for s other than -1, including analytically continued values.

Is that what you are saying?

Close! My question would be: is analytic continuation of the sum of g(n;-1) equal to the analytic continuation of the sum of f(n;-1)? (given that g(n;-1)=n)

Thanks!

As far as I know, there is only ONE analytic continuation of the Riemann Zeta function that is complex-differentiable everywhere (but some isolated singularities), and that's the one that provides the (in)famous result ##\zeta(-1)= \frac{-1}{12}##

Also, I suggest you to watch the youtube video by 3Blue1Brown about the Riemann hypothesis. He covers this issue (uniqueness of the analytic continuation) pretty well. He's a top notch explainer.

1. What is Ramanujan Summation?

Ramanujan Summation is a mathematical technique used to assign a numerical value to divergent series, which are series that do not have a finite sum. It was developed by the Indian mathematician Srinivasa Ramanujan and is also known as Ramanujan's method of summation.

2. How does Ramanujan Summation work?

Ramanujan Summation works by assigning a value to a divergent series based on its partial sums. The value is then determined by taking the limit of these partial sums as the number of terms approaches infinity. This technique can be used to assign values to many divergent series that do not have a traditional sum.

3. What are some variations of Ramanujan Summation?

There are several variations of Ramanujan Summation, including Cesàro summation, Abel summation, and Borel summation. These methods use different techniques to assign values to divergent series and can be useful in different situations.

4. What are the applications of Ramanujan Summation?

Ramanujan Summation has applications in various areas of mathematics, including number theory, complex analysis, and quantum field theory. It is also used in physics and engineering to solve problems involving divergent series.

5. What are the limitations of Ramanujan Summation?

Ramanujan Summation is not applicable to all divergent series and may not always provide the most accurate value. It also does not work for series with oscillating terms or series with a slowly converging sum. Additionally, Ramanujan Summation is not a rigorous mathematical method and is based on heuristic reasoning, so its results should be interpreted with caution.

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