# Ramanujan Summation, Variations of

• A
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.

mathman
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 dont know if you havent read my question properly or i have not understood the connection between your example and my question.

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. Im 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.