Interpretation of Ramanujan summation of infinite divergent series

Question

I am not mathematician by any means so this question might be rather stupid. I came across this Wikipedia article on Ramanujan's summation and found this bewildering formula,

$$1 + 2 + 3 + \dots = - \frac1{12}$$

The article also says that "Ramanujan summation of a divergent series is not a sum in the traditional sense". I am wondering then what this summation actually implies?

Is there any interpretation of this confounding result in the physical world that someone not so math-savvy can understand?

Answer 1

There are answers in -as I think- in mathoverflow for "what are physical interpretations of zeta at negative arguments" (or the like, I don't have the actual words in mind). Just browse a bit through MO or MSE using that terms as filter.

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0) Need of a new concept - and a basic requirement: Make the general term explicite

For the interpretation and assignment of a meaningful value in your series you must define, how the general term, say $a_k$, shall be constructed, i.e. how the value of $a_k$ is dependend on its index $k$. This is a requirement which shall be demanded in all discussions of divergent series (and should not been forgot when such series, written only in their "obvious" numerical forms, are considered at all).

In this example the value of the general term, say $a_k$, reduces simply to the index $k$ if taken from $k=1 \ldots \infty $ and we have actually the formal statement $S = \sum_{k=1}^\infty k$. But because the summing procedure is obviously divergent we must introduce some new concept if we want assign something meaningful to it.

1) Geometric series

A useful idea here was (for instance already by L. Euler) that we can cofactor a variable $x$ at the terms $a_k$ such that for some $x$ the series becomes convergent and evaluatable to some finite value and see what happens with the sequence of resulting sum-values, when the expressions of $x$ approximate $1$ . For instance, we could redefine your sum as $$ S(x) = 1 + 2x + 3x^2 + 4x^3 + ... \\ S = \lim_{x \to 1^-} S(x) $$ because there are some $x$ (actually it is important, that there exists a continuous interval of $x$) where this is convergent, for instance for $x=0 \ldots \frac12$. We can then even observe that with that definition the sum $S(x)$ has a closed form $$S(x) = {d\over dx}{1\over 1-x } = {1\over (1-x)^2 }$$ This is already a very nice representation, because it allows now even negative $x$ which result in finite values for the alternating sum, even if divergent, because it is accepted to assign that fraction's value also to its formal power series-expression even if the latter is divergent - except, well, except if $x=1$.
But what we want here is actually just the latter, so we have not yet a satisfying answer.

2) Dirichlet series

Another idea is to apply a function of $x$ to the exponent of the terms $a_k$ and then let $x$ go to $1$ . Thus we consider the notation $$ S(x) = 1^x + 2^x + 3^x +4^x + ... $$ which is convergent for a continuous interval of $x$, for instance of $x = -\infty \ldots -2$ (and even a bit more). But still, for $x=1$ we get no obvious answer.
But interestingly, for the whole interval of convergence we have also the functional relation $$ \begin{matrix} \text{let } &A(x) &=& 1^x - 2^x + 3^x - 4^x + \ldots \\ \text{then } &S(x) &=& A(x) + 2 \cdot 2^x \cdot S(x) \\ \text{and} & S(x) &=& A(x)/(1-2 \cdot 2^x) \end{matrix} $$ And now we can assign a meaningful value to $S = \lim_{x \to 1^- }$. Either by evaluating $A(1)$ as conditionally converging series in the given form, or by the above geometric-series interpretation and its derivative at $x=-1$ where both ways of evaluation give the same rational expression $A(1)=\frac 14$.
After that nothing more divergent is there and we get $$ S = \lim_{x \to 1^-} S(x) = \frac 14 / (1-2\cdot 2^x) = \frac 14 / (1-4)= - \frac 1{12} $$ having now a proposal for a meaningful assignment of a finite value for the infinite divergent series.

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3) Caveat

Of course that value must conform with all and any cases where such series occur in mathematics, and one example of a contradiction with known results taken from the mathematic without divergent summation would invalidate that procedure!
Moreover, we would hope that even in the physical world, where we model some observations with that series, such an assignement of values to a divergent series would hold. Interestingly such observations exist in the real world and it seems that the whole process and also the final value meets the modeling of that observations. (Examples are given to the according questions either in MSE here or in mathoverflow (I've just found (1)(2)), you can do a search for the important words "zeta" and "at negative arguments" or similar)

Answer 2

This result may be counterintuitive and makes sense at a high level of complex analysis dealing with Zeta function's and analytic number theory. Btw this has not only to do with math, this is even used in Physic but again at a very high level when we treat string theory or the Casimir effect (and other phenomena that can be solved using Zeta regularization).

If you're asking yourself why those kind of results make sense that's why in math we usually like to generalize things: when we have something that doesn't make sense in the "usual way" we give it a new sense with the only condition that the other results are preserved. In this case Ramanujan summation (along with other formulas like Cesaro or Abel summation) gives the usual results with convergent series and new results with divergent ones, that can be used in the same way as the others.

Answer 3

Is there any interpretation of this confounding result in the physical world that someone not so math-savvy can understand?

We all learn about long division in primary school. And long division is one particular way to generate an infinite series with. Using long division you can derive the formula for the expansion of the function $\frac{1}{1-x}$ in powers of $x$ of

$$\frac{1}{1-x} = 1 + x + x^2 + \ldots \tag{1} $$

But thsi doesn't converge for $|x|\geq 1$. But how are we then to interpret this divergent series for the case $|x|\geq 1$? My opinion diverges a bit from most other people, I insist that we go back to what the math actually is telling us when we derive the series (1).

Does the math tell us to sum an infinite number of terms of (1)? No, it doesn't! When you derive (1) using algebra pof long division, you are not using any concepts of analysis like taking the limit of the partial sums. And addition is only defined from the fundamental axioms for a finite number of terms anyway. So, the math used to generate the series cannot possibly tell you that you need to sum an infinite number of terms of (1).

What the math does tell you is that the expression you are calculating is given by a finite number of terms of the series you are generating plus a reminder term. We all learn about the remainder in primary school when learning about long division! You can choose where you truncate the series, and there is then a remainder term corresponding to the point where you decide to truncate the series.

So, what the math itself is telling you is that (1) should be interpreted as:

$$\frac{1}{1-x} = \sum_{k=0}^{n}+ R(n) \tag{2} $$

where $R(n)$ is the remainder term corresponding to truncating the infinite series at the nth term. And this is then valid for both divergent and convergent series.

When the series converges, the limit of $n$ to infinity of the remainder term will be zero (strictly speaking it can tend to any value, but a nonzero value implies that the expanded function not analytic). So, in case of a convergent series, the limit of the partial sums will yield the correct value of the series. In case of a divergent series, the value of the series can be perfectly finite, despite the limit of the partial sums not existing. This is not a contradiction because the limit of $R(n)$ for $n$ to infinity doesn't exist in this case, so we don't get rid of the unknown $R(n)$ by taking this limit.

As I've pointed out in detail here, Ramanujan summation can be interpreted as a formula for the remainder term. And oen can on that basis also derive that the value of an infinite series for which the partial sum of the first $n$ terms is $S(n)$, is given by:

$$\operatorname*{con}_{N}\int_{N-1}^N S(x) dx$$

where $\displaystyle \operatorname*{con}_{N}$ denotes the constant term in the large $N$ expansion of the expression. So, in case of the sum of the positive integers, we have $S(x) = \frac{1}{2} x (x+1)$, and:

$$\int_{N-1}^N S(x) dx = \frac{N^2}{2} - \frac{1}{12}$$

The constant term is then $-\frac{1}{12}$. The interpretation of this is then that whenever the series $\sum_{k=0}^{\infty} k$ arises as a result of computing some well-defined quantity using a method that then generates this infinite series, that the value of that quantity is $-\frac{1}{12}$ when assuming that the quantity as a function of the expansion parameter doesn't have any additional nonanalytic behavior than implied by the series.