[Physics FAQ] - [Copyright]

By Don Koks, 2016.

Can All the Natural Numbers be Summed?
(or, does $1 + 2 + 3 + \dots$ equal $-1/12$?)

No, of course the natural numbers can't be summed.  $1 + 2 + 3 + \dots$ has no sum; or we might just as well say that it sums to infinity.  The real question is: why do some people write $1 + 2 + 3 + \ldots = -1/12$?  The answer involves some maths, some physics, and some analysis of common misunderstandings about what mathematicians and physicists are saying.  Some physicists mistakenly believe that mathematicians have summed the series to give $-1/12$.  And some mathematicians mistakenly believe that physicists have summed the series experimentally to give $-1/12$.  Neither are right, but so much finger pointing of each to the other's discipline has occurred that many laymen now believe that maths and physics have both proved that the sum is $-1/12$.  The subject is an old one, but gained a new lease of life in 2014 with the appearance of a notorious youtube clip presented by an academic well outside his zone of expertise, who succeeded only in breaking the rules of elementary maths.

The Relevant Maths

Of course, it's physically impossible to use a calculator, abacus, or pen and paper to actually sum a series that is infinitely long, so mathematicians long ago realised that such an expression must be carefully defined for it to have any useful meaning.  They define it in a way that matches everyone's expectation of what such an expression should mean: begin the addition term by term in the order written, and keep an eye on the running sum (also known as the "sequence of partial sums") as each term is added.  If this running sum gets ever closer to some number, then that number will be unique and is called the sum of the series.  If the running sum doesn't behave in that way, then we say the series has no sum.  If you start with 1, then add 2 (running sum is 3), then add 3 (running sum is 6), then add 4 (running sum is 10), those partial sums 1, 3, 6, 10, get bigger and bigger and don't get arbitrarily close to any number at all.  So the series $1 + 2 + 3 + \dots$ has no sum.  But you knew that anyway.

But didn't the mathematician Ramanujan sum the series to give $-1/12$?  Ramanujan's letter of almost a century ago to the mathematician Hardy, in which he wrote the sum, dates from a different time.  Ramanujan certainly had his feet on the ground enough to know that putting one orange into a big pit, followed by 2 more oranges, then 3 more oranges, and so on forever, is not going to result in there being $-1/12$ oranges in the pit.  He was a trailblazer of another time, and went very far by experimenting with the fewer boundaries that existed back then.  Since that time, more boundaries have been drawn, and modern mathematics knows perfectly well where these lie.  Those boundaries were established by setting axiomatic properties of numbers, to keep mathematics from running off the rails and all hell breaking loose.  Early mathematicians were allowed to do what they did, but modern mathematicians and physicists no longer work under the paradigm that was current back then.

The reason why some modern physicists think that mathematics has summed the natural numbers actually has nothing to do with simple algebraic manipulations of the series.  So let's take it in stages, and begin with some much simpler ideas.

For example, what do we mean by the repeating decimal $0.3333\dots$?  This represents the infinite series $3/10 + 3/100 + 3/1000 + \dots$.  It can be shown that the ordered sequence $0.3, 0.33, 0.333, 0.3333, \dots$ converges to 1/3 (or "has a limit of 1/3", or "tends toward 1/3"), and so we define $0.3333\dots$ to equal 1/3.  (Technically, that means we can always find an element in that sequence of $0.3, 0.33, 0.333, 0.3333, \dots$ which is as close to 1/3 as we wish, and such that all successive elements lie even closer to 1/3.)  By the same token, the repeating decimal $0.9999\dots$ equals 1, because the ordered sequence $0.9, 0.99, 0.999, 0.9999, \dots$ converges to 1.  In contrast, the ordered sequence $1,\; 1 + 2,\; 1 + 2 + 3, \dots$ has no limit at all that you can find on a number line.  By convention we then say that the sum tends to infinity; although you can't find infinity on a number line, mathematicians do supplement the number system with a "number" called infinity, and so it can be said that $1 + 2 + 3 + \dots$ equals infinity.

How about the series $1 + x + x^2 + x^3 + \dots$, where $x$ is a real number?  This series converges only when $|x| < 1$.  When it does converge, it sums to $1/(1-x)$, but you must always remember that the procedure that yields this sum is valid only when $|x| < 1$.  It is certainly nonsensical to set $x$ equal to 5 and conclude that $1 + 5 + 5^2 + 5^3 + \dots$ equals $-1/4$.  But it turns out that this is essentially what those are doing who say that $1 + 2 + 3 + \dots$ equals $-1/12$.

If we were to assume that $1 + 5 + 5^2 + 5^3 + \dots$ equals some number that we can point to on the real-number line, then it would be easy to find that number.  Call it $S$: \begin{equation}\label{eqn-for-S} S = 1 + 5 + 5^2 + 5^3 + \dots \,. \end{equation} Since we are assuming that $S$ is a number, it must be amenable to the usual rules of algebra.  So multiply equation \eqref{eqn-for-S} by 5, giving \begin{equation}\label{eqn-for-5S} 5S = 5 + 5^2 + 5^3 + 5^4 + \dots \,, \end{equation} and then subtract \eqref{eqn-for-S} from \eqref{eqn-for-5S} to conclude that \begin{equation} 4S = -1 \,. \end{equation} We find that if $S$ is assumed to exist, then it must equal $-1/4$.  But does $S$ exist?  The sum \eqref{eqn-for-S} deals only with whole and positive quantities $1, 5, 5^2, 5^3, \dots$, and so it should apply to whole objects such as eggs or electrons; but it's clear that adding $1\text{ egg} + 5\text{ eggs} + 5^2\text{ eggs} + 5^3\text{ eggs} + \dots$ can never give $-1/4$ egg (what is $-1/4$ of an egg, anyway?); nor can it give $-1/4$ of an electron (is that $1/4$ of a positron?)  So we've proved by contradiction that our initial idea to call the series $S$ was wrong, and that means that all the mathematical manipulations we did after introducing $S$ were invalid.  It follows that $1 + 5 + 5^2 + 5^3 + \dots$ cannot equal any number at all that you can point to on the real-number line.

This idea is routinely analysed using partial sums, and then the reason for why it doesn't work becomes very obvious.  Denote the $n^\text{th}$ partial sum by $S_n$, which is certainly possible because these are just normal everyday sums: \begin{equation} S_n = 1 + 5 + 5^2 + 5^3 + \dots + 5^n . \end{equation} Then \begin{equation} 5S_n = 5 + 5^2 + 5^3 + 5^4 + \dots 5^{n+1} . \end{equation} Subtracting the first line from the second gives \begin{equation} 4S_n = 5^{n+1} - 1 \,, \end{equation} and $S_n = (5^{n+1} - 1)/4$.  This is all quite correct, but notice that as $n$ gets larger, $S_n = (5^{n+1} - 1)/4$ grows without limit.  So the sequence of partial sums $S_0, S_1, S_2, \dots$ doesn't converge to any number that we can locate on the real-number line.  That's a formal mathematical proof of the obvious statement that $1 + 5 + 5^2 + 5^3 + \dots$ doesn't converge to any real number.

Analytic Continuation

Now let's picture a scenario, one that more usually occurs in the realm of complex numbers, but we can also explore it in the realm of real numbers.  The example here is simple enough to highlight the main points, but don't be misled by its simplicity into believing that more complicated examples are as transparent as this one.

I have a function $f(x) = 1 + x + x^2 + x^3 + \dots$ that I have worked out is defined for $|x| < 1$.  Suppose I don't know what the sum converges to, but I just know that it converges for, and only for, all $|x| < 1$.

But as I dabble with various expressions, I happen to notice that the expression $1/(1-x)$ equals the sum of my series $f(x)$ for all $|x| < 1$.  I also know that $1/(1-x)$ is defined for all other values of $x$ except $x = 1$.  It turns out that of all "sufficiently smooth" functions, $1/(1-x)$ is the unique function with these properties.  For suppose, on the contrary, that I was able to find another function $g(x)$ that also matched my series $f(x)$ for all $|x| < 1$.  That would mean $g(x) - 1/(1-x)$ was zero everywhere inside $|x| < 1$.  But if $g(x) - 1/(1-x)$ is "sufficiently smooth", it turns out that it will have to equal zero everywhere.  But that means $g(x) = 1/(1-x)$, and so I haven't really found another function—I've only found $1/(1-x)$ again.  So $1/(1-x)$ is a unique extension of my function $f(x)$ to the wider world of all $x$ (except $x = 1$).  This $1/(1-x)$ is called the analytic continuation of $f(x)$.  We have started with a series $f(x)$ valid in a limited interval $|x| < 1$, and managed to come up with a unique function that agrees with $f(x)$ in that interval, but is also valid on a larger interval.

The crucial point here is that $1/(1-x)$ is not the sum of $1 + x + x^2 + x^3 + \dots$ outside the interval $|x| < 1$.  It's simply a function that equals $f(x)$ within $|x| < 1$ but is also defined outside that interval, and is unique if we require our functions to be sufficiently smooth.  What I could do is define a function $f(x)$ as follows (where "$\equiv$" means "is defined to be"): \begin{equation} f(x) \equiv \begin{cases} 1 + x + x^2 + x^3 + \dots & \text{for } |x| < 1 \,,\\[2ex] 1/(1-x) & \text{for all other $x$ except } x = 1 \,. \end{cases} \end{equation} I can then correctly write $f(5) = 1/(1-5) = -1/4$.  But I'm certainly not saying that $1 + 5 + 5^2 + 5^3 + \dots = -1/4$.  I have simply enlarged my original definition of $f(x)$ outside its original realm of applicability by using a different expression: $1/(1-x)$ in place of $1 + x + x^2 + x^3 +\dots$.  Within the original realm of applicability, $1/(1-x)$ equals $1 + x + x^2 + x^3 + \dots$ of course.  But that's neither here nor there, because 5 is not within the original realm of applicability.

The same ideas apply more generally to functions defined over complex numbers.  A "sufficiently smooth" function is called analytic.  The series $f(z) = 1 + z + z^2 + z^3 + \dots$ certainly exists for all complex numbers $z$ in the disk $|z| < 1$, and again a unique analytic continuation exists, which is the function $1/(1-z)$ that is defined for all other $z$ except $z = 1$.  And, as usual, we only say that $1 + z + z^2 + z^3 + \dots = 1/(1-z)$ for $|z| < 1$, but we can redefine $f(z)$ outside that disk by setting it equal to $1/(1-z)$ there.

This series $1 + z + z^2 + z^3 + \dots$ was a simple example.  Long ago, mathematicians became interested in the following series: \begin{equation} 1 + 1/2^z + 1/3^z + \dots \,. \end{equation} This series converges only when the real part of $z$ is greater than 1, and is called the zeta function, $\zeta(z)$, in this region of the complex numbers: \begin{equation} \label{zeta-defn-real-z-greater-than-1} \zeta(z) \equiv 1 + 1/2^z + 1/3^z + \dots \,,\text{ provided real}(z) > 1 \,. \end{equation} This series doesn't converge when $\text{real}(z) \leqslant 1$, and so we cannot write $\zeta(z) = 1 + 1/2^z + 1/3^z + \dots$ when $\text{real}(z) \leqslant 1$.  And we don't.  But because an analytic continuation of the zeta function has been found for all $z$ not equalling 1, the zeta function has been redefined to include this wider definition.  But note that this has nothing to do with the series in \eqref{zeta-defn-real-z-greater-than-1}!

It turns out that the zeta function can first be analytically continued to the region $\text{real}(z) > 0$ with $z$ not equal to 1, by using the following series: \begin{equation} \zeta(z) \equiv {1\over 1-2^{1-z}} \left(1 - {1\over 2^z} + {1\over 3^z} - {1\over 4^z} + \dots\right), \text{ provided } \text{real}(z) > 0 \text{ and } z \neq 1 \,. \end{equation} A second analytic continuation now extends the definition of the zeta function to all $z\neq 1$.  This turns out to be possible with the following definition: \begin{equation} \label{zeta-defn-real-z-less-than-1} \zeta(z) \equiv 2(2\pi)^{z-1} \sin(\pi z/2)\; \Pi(-z)\; \zeta(1-z) \,, \text{ provided } \text{real}(z) < 1 \,, \end{equation} where $\Pi(z)$ is the "Pi" or factorial function as defined over all complex numbers, and can also be written as "$z\textit{!}\,$".  [There are actually several ways that the factorial can be defined over all complex numbers, but \eqref{zeta-defn-real-z-less-than-1} uses the most common one.  Note also that $\Pi(z)$ is often written as the "Gamma function" $\Gamma(z+1)$, that has a confusing yet bizarrely fashionable shift of 1 whose origins presumably trace back to a mathematician from a bygone century having had a little too much to drink, or something.  Use $\Gamma$ only if you want to confuse people.]

It turns out that when $\zeta(z)$ is analytically continued in the way of the above, that $\zeta(-1) = -1/12$.  Of course, this has nothing to do with the original series \eqref{zeta-defn-real-z-greater-than-1} that started $\zeta(z)$ off, which was only defined when the real part of $z$ is greater than 1.  But picture a mathematicians' party where the joke goes around that if we set $z = -1$ in the series $1 + 1/2^z + 1/3^z + \dots$, and then say that the result equals $\zeta(-1)$ (which it most certainly does not), then we'll end up by saying that $1 + 2 + 3 + \dots$ equals $-1/12$.  It's a bit of humour that probably works well at that party precisely because the mathematicians understand what's really going on, that it is just a joke.  But a problem arises when this joke leaks out into the wider world and gets taken seriously by those who aren't in on the joke.  And that is precisely what has happened, and why so many people think that mathematicians say that $1 + 2 + 3 + \dots$ equals $-1/12$.  Consider it to be just a joke that got out of hand.

Of course, mathematicians are well within their remit to search for any connection that might exist between two functions when one is the analytic continuation of the other.  That's a very interesting technical question, and one that currently has no answer.  In other words, how is the most general definition of $\zeta(z)$ related to the restricted definition in \eqref{zeta-defn-real-z-greater-than-1}, $\zeta(z)=1 + 1/2^z + 1/3^z + \dots$, that holds only for $\text{real}(z) > 1$?  Perhaps one day someone will figure that out.  But currently that is a task specific to pure mathematics.  Although a few isolated values of the zeta function turn up in physics, the function itself (as a whole entity) has not yet found any application in physics.

In the meantime, I don't think anyone ever sets $z$ to 0 in the series $1 + 1/2^z + 1/3^z + \dots$ and equates the answer to $\zeta(0) = -1/2$, to say that $1 + 1 + 1 + \dots$ equals $-1/2$.  And if they do, how do they find any consistency with the fact that they could equally well have done the same thing for the series $1 + z + z^2 + z^3 + \dots$ by substituting $z = 1$ into the expression $1/(1-z)$ to arrive at infinity?  Nor does anyone set $z$ to $-2$ in the series $1 + 1/2^z + 1/3^z + \dots$ and equate the answer to $\zeta(-2) = 0$, to say that $1^2 + 2^2 + 3^2 + \dots$ equals 0.  And yet, the mathematicians' joke that $1 + 2 + 3 + \dots$ equals $-1/12$ clings stubbornly on in the imaginations of many non mathematicians—and even of some mathematicians.

Other ideas of summing infinite series exist.  One that you will find mentioned in connection with summing the natural numbers is "Cesaro summation", in which we write a sequence of arithmetic means of the partial sums of the infinite series.  If that sequence converges to some limit, then that limit is called the Cesaro sum of the infinite series.  Cesaro sums can sometimes be better behaved than the usual type.  Also, any series that sums in the usual sense of partial sums will be Cesaro summable to the same number.  But some series that are not summable in the usual sense are Cesaro summable—although the series $1 + 2 + 3 + \dots$ is not one of those.  In principle there is an infinite number of different "flavours" of Cesaro sum that one could define for any one series, purely because there is an infinite number of different ways to define an average: the arithmetic, geometric, and harmonic means are only three examples of an infinite number of different means.  Cesaro summing uses the usual arithmetic mean, but there is no reason why we shouldn't use any other flavour of mean instead.

Here's an example of a series that is not summable in the usual sense, but is Cesaro summable: \begin{equation} \label{summing-1-minus-1} 1 - 1 + 1 - 1 + 1 - 1 + \dots \,. \end{equation} The sequence of running sums is $1, 0, 1, 0, 1, 0, \dots$. The average of these tends to 1/2 as the sequence gets longer, and so the Cesaro sum of \eqref{summing-1-minus-1} is 1/2, even though the usual sum doesn't exist.  Does the Cesaro sum have any physical use here?  Note that if you give a man 1 dollar in the first second, then 1/2 dollar in the next second, then 1/4 dollar, then 1/8 dollar and so on, he will eventually be able to spend a sum of money arbitrarily close to \$2 (but not more than that) if he waits long enough.  But if you give him 1 dollar, then take it away, then give it back, then take it away, repeating these actions indefinitely, he'll never be able to spend anything at all, unless he spends \$1 very quickly and then sells what he bought for \$1 so that he can give it back to you.  He certainly won't be able to spend 1/2 a dollar and keep what he buys.  So whereas the usual way of summing says the series \eqref{summing-1-minus-1} has no sum—which corresponds to something about the real world here—the Cesaro sum does not correspond to anything about the real world.

Cesaro summing is really one version of a more general idea in which we define a function, say $f$, that takes the elements of the infinite series in order and returns a number that we find interesting.  Those who wish to say that the sum of the natural numbers is $\zeta(-1)$ are really defining such a function via "$f(1,2,3,\dots) = \zeta(-1)$".  That can certainly be done, but is it useful?  There are any number of infinite series in $z$ that we could write down that reduce to $1 + 2 + 3 + \dots$ when some $z$ is set to some number, and the result won't necessarily be $-1/12$.  So now we'll have to introduce another function, call it $g$, such that $g(1,2,3,\dots)$ equals whatever number is returned by that new series of interest.  An infinite number of functions will need to be invented to cover all series.  The subject of infinite summation will then be running off the rails, and departing from the very basic idea that a sum is all about piling objects on top of each other and seeing what, if anything, is taking shape.

It's sometimes argued that if the series $1 + 2 + 3 + \dots$ has no sum in the usual sense, that we are free to define it to equal something of use.  The analogy is used that the square root of 2 doesn't exist in the rational numbers, so we extend those numbers to become the real numbers, in which case the square root of 2 does then exist.  Likewise the square root of $-1$ doesn't exist in the real numbers, so we extend the real numbers to become the complex numbers, and then the square root of $-1$ does exist.  So, some argue, we should be allowed to define $1 + 2 + 3 + \dots$ to be $-1/12$.  But this line of reasoning has no logical content.  The only way that the square root of 2 and the square root of $-1$ were given meaning was precisely by extending the realm of known numbers to a larger set that contained new entities.  For example, we can't just say the square root of 2 equals 4/5; we have to invent a new type of number, an irrational number.  Likewise we can't just say the square root of $-1$ equals $-5$ or 349.  We need to extend the realm of numbers to give the square root of $-1$ meaning; we do that by inventing a new number, $i$, for the square root of $-1$.  Likewise, we cannot set $1 + 2 + 3 + \dots$ equal to a real number, because we already know that $1 + 2 + 3 + \dots$ doesn't converge to any such number.  So can we extend the complex numbers by defining a new object that equals $1 + 2 + 3 + \dots$?  Certainly, and this was done long ago: that new object is "infinity".  Defining it hasn't proved as fruitful as defining irrational numbers and complex numbers, but it has been done.  The important point is that infinity does not equal $-1/12$.  Maintaining that an infinite sum is just too inconvenient and must be replaced by some real number—as some do—is about as daft as demanding that Earth's radius should be redefined to be 10 metres to make it quicker to travel to another country.

The Relevant Physics

Some mathematicians and physicists think that physicists have measured $1 + 2 + 3 + \dots$ to equal $-1/12$ in the laboratory.  But no such measurement has ever been made; nor will it ever be made.  This measurement is described in the FAQ entry What is the Casimir Effect?.  It concerns a prediction made by the Dutch physicist Hendrick Casimir in 1948.  Casimir analysed two parallel plates using the language of quantum mechanics.  He considered a field that might fill the universe, and in particular the space between the plates.  This field was purely a theoretical construct until Casimir came along.

The usual line of reasoning says that the value of this field in some region is to be interpreted in a quantum mechanical sense as being related to the probability of finding a related particle in that region.  Since presumably no particles can exist at the plates themselves, the field is required to vanish at the plates.  If the field is constrained that way and Fourier-analysed into a sum of "modes", then only a countably infinite number of these modes can exist between the plates.

Casimir calculated the quantum mechanical energy of the mode-restricted field between the plates.  As expected, it was infinite, because quantum mechanics allocates a unit of energy to each mode, and there are an infinite number of modes between the plates.  His calculation arrived at the sum "$1 + 2 + 3 + \dots$", which expressed this infinite energy.

The idea of infinite energy is something of a problem in practice.  What Casimir did next got him through that problematic infinity.  First, he made the series somewhat like a geometrical one by inserting a "cutoff" factor that could go to 1 in an appropriate limit.  This new series was well defined and able to be summed exactly to some number $S_L$, where $L$ is the plates' separation.  Next, Casimir calculated the difference between this expression and the analogous expression for the absence of plates: that is, when the the plate separation $L$ went to infinity.  So he calculated $S_L - S_\infty$.  The divergent part of each series was contained in the cutoff, but this part cancelled in the subtraction—thus no "limit as cutoff goes to 1" ever had to be considered.  What was left included a factor of $-1/12$.  I'll call this the "Energy Difference".

This difference in energies was treated like a difference in potential energies.  Now, potential energy is only defined up to an additive constant, so there is no unique potential energy for, say, a mass in a gravitational field (unless we stipulate the additive constant, which we certainly do in practice).  But the difference between two potential energies certainly is well defined and related to forces.  The difference in field energies with and without the plates was then related to the existence of a tiny force between the plates, and the existence of this force is known as the Casimir Effect.  In fact, a force is a spatial gradient of potential energy, but spatial gradients actually have nothing to do with the difference in field energies with and without plates.  So this very standard explanation of the Casimir Effect doesn't quite work.  But it has become standard in the field.

Historically, that's where the Casimir Effect stands, but the above procedure has been re-interpreted by many practitioners in the following way.  They note that the original divergent sum (i.e., with plates present) will give what I called the Energy Difference if they simply replace its "$1 + 2 + 3 + \dots$" part with $-1/12$.  I think that's of no great consequence: the mathematics of the sums in the Casimir Effect is not complicated, so it should not be surprising to find that the Energy Difference has a lot in common with the non-divergent overall factor in the original divergent sum.

But notice what this interpretation has done: it has taken a divergent quantity and observed that replacing its $1 + 2 + 3 + \dots$ part by $-1/12$ gives the difference between two related sums ($S_L$ and $S_\infty$, that each have a cutoff).  Then, because $S_L - S_\infty$ is in some sense measured in the lab to involve a factor of $-1/12$, this interpretation concludes that $1 + 2 + 3 + \dots$ equals $-1/12$.  There is no real logic here, so don't look too hard for it.  Just remember the bottom line: neither mathematics nor physics says that $1 + 2 + 3 + \dots$ equals anything other than infinity.  And because the meaning of "$+$" was already extended long ago to apply to infinite series in the particular way described above, it cannot now be given some new, arbitrary meaning to suit the tastes of those who think that they have a licence to change the rules of elementary mathematics.