The Ramanujan Summation

Preface

We saw in the post Grandi’s Series a very amazing result. We are going to use that result for another and possibly more amazing findings. There is a question to be answered: does it make sense to add all integers to infinity? And what would the result be?

In the YouTube version of the video “ASTOUNDING: 1 + 2 + 3 + 4 + 5 +….= −1/12” (The Ramanujan Summation), Tony Padilla and Ed Copeland, physicists at the University of Nottingham in the UK, appear in and narrate the eight-minute video. Minute by minute, what they explain requires from the viewer no mathematical grounding beyond simple addition and a smattering of the most basic algebra. But in the aggregate, what they argue goes well beyond that simplicity, i.e. that the infinite sum of all integers is equal -1/12. This post resume briefly what they demonstrate in the video. Furthermore we want to go a little be further giving the reader the indication of a more fascinating and rigorous way to achieve the same result.

Problem formulation

Let’s now write what is our scope: we want to see if there is some sense to find the result of the following operation:

$S=\sum _{n=1}^{\mathrm{\infty }}n$     1)

That is

$S=1+2+3+4+5+6+\cdots \cdots \cdots \to \mathrm{\infty }$

Apparently the answer is: it doesn’t make sense since the result is clearly infinite!!

But when we speak about infinite things are not so obvious and clear.

Before proceeding elaborating 1) let’s solve the following intermediate problem

Intermediate problem

In order to deal with expression 1) let’s solve the following problem: what is the result of the following sum?

$T=\sum _{n=1}^{\mathrm{\infty }}{(-1)}^{n+1}n$     2)

That is

T = 1 -2 + 3 -4 + 5 – 6 + …………...   3)

To solve 2) let’s calculate the following

H = T + 0 + T     4)

That is

H = 1 – 2 + 3 - 4 + 5 – 6 + ……………..

 + 0 + 1 – 2 + 3 – 4 +5 -……………….

Summing up term by term from each row lead to the following

H = 1 – 1 + 1 -1 + 1 – 1 +……………….  5)

which is the Grandi’s series G already calculated here

Expression 4) becomes

$H=2T=G=\frac{1}{2}$ and thus

$T=\frac{1}{4}$      6)

The solution

To find the result of 1) let’s develop the following summation

S – T      7)

That means

 1 + 2 + 3 + 4 + 5 + 6 + 7 ………….

-( 1 - 2 + 3 - 4 + 5 + 6 + 7 ………….

The result obviously is:

S – T = 0 + 4 + 0 + 8 + 0 +12 + 0 ………..

and thus

S – T = 4 * (1 + 2 + 3 + 4 +……….) = 4S

That means

3S = -T

recalling that $T=\frac{1}{4}$ we obtain

$S=\sum _{n=1}^{\mathrm{\infty }}n=-\frac{1}{12}$    8)

and it is really surprising.

It is to be noted that this result is used in several fields: as an example the following page

is extracted by one of the most important book related to String Theory (Polchinski - String Theory Volume 1, An Introduction to the Bosonic String, 1998)

In the following we propose a more fascinating approach to the same problem.

Problem reformulation

Let’s recall our problem:

$S=\sum _{n=1}^{\mathrm{\infty }}n$     1)

It is obvious that 1) can be re-written as:

$S=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^{-1}}$

Let’s consider the following function of the real variable x

$S(x)=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^x}$     9)

Also here it is obvious that

S(-1) = S

The study of the function S(x) is not very interesting because it doesn’t add any further information on S. It is a series of function and the convergence interval is x > 1

We can extend our investigation expanding the set of number to be considered. If we move to the complex set we can re-write 9) as

$\zeta (s)=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^s}$     10)

where $s=a+jb$ is a complex variable. Of course it is still valid that

$\zeta (-1)=S$

Expression 10) is the Riemann function. Associated to expression 10) there is the Riemann hypothesis (made by Riemann in mid 19th century) nowadays still to be demonstrated: one of the biggest math problem ever formulated. To be note that linked to the demonstration, there is a big amount of money!! We leave to the interested reader to learn more about that.

At the first view expression 10) does not add anything to 9). It is a series of function and the convergence interval is Re(s) = a > 1; it seems to be the same situation of expression 9).

Instead, the fact to be in a complex field, can allow us to manipulate a little bit further expression 10) as shown below.

A different formulation of $\zeta (s)$

In order to arrive to a different formulation of 10), in the following we will assert some properties of complex functions without however demonstrating them, leaving this task to the interested reader.

In complex analysis a holomorphic function (see Wikipedia) is a complex differentiable function (that means infinitely differentiable) and can be expressed as a Taylor series.

It can be shown that such functions can have analytic continuation: in simple terms, the analytic continuation is identical to the function in its domain of definition, but it is also defined outside it. It can be shown that if analytic continuation exists, it is unique.

Let’s reconsider expression 10): as written above it is defined for a > 1, that means in the half plane with real part greater than 1. Let’s ask the following question:

does an analytic continuation of 10) exist, which extend its domain?

The answer is affirmative and the proof is far from simple: it was done by Riemann in 1859 and of course is omitted here.

But we can use the result, knowing that it is valid and demonstrated.

Riemann demonstrated that an analytic continuation of 10) can be written as:

$\zeta (s)=\frac{\mathrm{\Gamma }(1-s)}{2\pi i}\underset{+\mathrm{\infty }}{\overset{+\mathrm{\infty }}{\int }}\frac{{(-x)}^s}{x({\mathrm{e}}^x-1)}\mathit{dx}$    11)

This means that 11) is identical to 10) in the domain of existence of 10), but it also exists in a wider domain, i.e. in all the complex plane except s = 1.

Expression 11) needs some more explanation.

1. the function Γ (Gamma function) is defined over the Real number set as

   $\mathrm{\Gamma }(x)=\underset{0}{\overset{+\mathrm{\infty }}{\int }}{(-t)}^{x-1}{e}^{-t}\mathit{dt}$ .

   Some proprieties of the Gamma function are:

   $\mathrm{\Gamma }(1)=1$

   $\mathrm{\Gamma }(x)=(x-1)\mathrm{\Gamma }(x-1)$ and if x is a natural number

   $\mathrm{\Gamma }(n)=(n-1)\mathrm{\Gamma }(n-1)=(n-1)(n-2)\mathrm{\Gamma }(n-2)=\dots \dots =(n-1)!$

   This means that Gamma function is the extension of the factorial operation on real number. Gamma function can be extended also to complex number set of course.

2. In expression 11) also the x variable is a complex variable
3. $\zeta (s)$ has a pole for s=1 because Gamma function has a pole in the origin.
4. In expression 11) we note a very strange integral between $+\mathrm{\infty }$ and $+\mathrm{\infty }$ . What is its meaning? Take into account that we are considering functions in the complex set. The integral in expression 11) is to be intended as a curvilinear integral in the Gauss plane. The curve where to integrate the function starts at $+\mathrm{\infty }$ on the real axis, comes towards the origin, makes a circle of radius δ around the origin (clockwise), comes back on the real axis and goes newly to $+\mathrm{\infty }$

Use analytic continuation

Being 11) equal to 10), we can use 11) to calculate $\zeta (s)$ for s = -1, because now the analytic continuation is defined there. Here pure mathematicians have a strong objection: warning, 11) is identical to 10) only for Real(s) > 1, that means in its domain of definition. It doesn’t make sense to calculate it for s = -1, because it is out of its domain of definition. But personally we have another suggestion: the broader vision using complex sets allows us to have a holistic view of the problem: it's like putting on a pair of glasses and discovering a new world. If we remove them we have only a partial view, relative to the set of real numbers. With this vision we can assert that the wider view include and complement the “narrow view” related to the real set. So, let’s continue using expression 11) to calculate expression 10) for s equal to negative natural numebrs

Calculation of $\zeta (s)$ for s equal negative natural numbers

The expression to be calculated is:

$\zeta (-n)=\frac{\mathrm{\Gamma }(1+n)}{2\pi i}\underset{+\mathrm{\infty }}{\overset{+\mathrm{\infty }}{\int }}\frac{{(-x)}^{-n}}{x({\mathrm{e}}^x-1)}\mathit{dx}$    12)

we have already specified the meaning of the integration between $+\mathrm{\infty }$ and $+\mathrm{\infty }$ : let’s now apply that meaning splitting the integral in 12) in three parts:

$\zeta (-n)=\frac{\mathrm{\Gamma }(1+n)}{2\pi i}\underset{+\mathrm{\infty }}{\overset{+\mathrm{\infty }}{\int }}\frac{{(-x)}^{-n}}{x({\mathrm{e}}^x-1)}\mathit{dx}=\frac{\mathrm{\Gamma }(1+n)}{2\pi i}[\underset{+\mathrm{\infty }}{\overset{\delta }{\int }}\frac{{(-x)}^{-n}}{x({\mathrm{e}}^x-1)}\mathit{dx}+\underset{\left|x\right|=\delta }{\int }\frac{{(-x)}^{-n}}{x({\mathrm{e}}^x-1)}\mathit{dx}+\underset{\delta }{\overset{\mathrm{\infty }}{\int }}\frac{{(-x)}^{-n}}{x({\mathrm{e}}^x-1)}\mathit{dx}]$ 13)

It can be shown that the first and third ingratal in 13) are both zero. Expression 13) becomes simply:

$\zeta (-n)=\frac{\mathrm{\Gamma }(1+n)}{2\pi i}\underset{\left|x\right|=\delta }{\int }\frac{{(-x)}^{-n}}{x({\mathrm{e}}^x-1)}\mathit{dx}$   14)

Expression 14) is not very difficult to calculate taking into account that

1. $\mathrm{\Gamma }(1+n)=n!$
2. $\frac{x}{{\mathrm{e}}^x-1}$ can be expressed in Taylor series as $\frac{\sum _0^{\mathrm{\infty }}{B}_n{x}^n}{n!}$ where $B_n$ are Bernoulli numbers

After some elaboration (not shown here) we achieve the following result:

$\zeta (-n)={(-1)}^n\frac{B_{n+1}}{n+1}$   15)

It is useful now recall the Bernoulli number values (or at least few of them) in the following table

 

n	0	1	2	3	4	5	6	7	8
Bn	1	$-\frac{1}{2}$	$\frac{1}{6}$	0	$-\frac{1}{30}$	0	$\frac{1}{42}$	$-\frac{1}{30}$	0

From 15) we can now calculate $\zeta (-1)$

$\zeta (-1)={(-1)}^1\frac{B_2}{2}=-\frac{1}{12}$   16)

and here we have reached the same result as in 8). Unbelievable!!!

This is really an amazing result!!!!!