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 mathematicians Euler and 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.  Euler's interest was similar to that of Ramanujan: he wanted to see where the rules of mathematics could take him, so he assumed that the sum existed and performed some mathematical gymnastics to arrive at $-1/12$.  Euler and Ramanujan certainly had their 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.  They were trailblazers of other times, and they went very far by experimenting with the fewer boundaries that existed back then.  Since that time, proper boundaries have been drawn, and modern mathematics knows perfectly well where these lie: those boundaries were established by setting axiomatic properties of numbers: these properties keep mathematics from running off the rails and all hell breaking loose.  Euler's early work belongs to his time and is part of mathematical history.  He was allowed to do what he did, but modern mathematicians and physicists no longer work under the paradigm that was current in Euler's time.  They now work under established rules that weren't available to Euler.

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 $S$: $$\label{eqn-for-S} S = 1 + 5 + 5^2 + 5^3 + \dots \,.$$ 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 $$\label{eqn-for-5S} 5S = 5 + 5^2 + 5^3 + 5^4 + \dots \,,$$ and then subtract \eqref{eqn-for-S} from \eqref{eqn-for-5S} to conclude that $$4S = -1 \,.$$ 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: $$S_n = 1 + 5 + 5^2 + 5^3 + \dots + 5^n .$$ Then $$5S_n = 5 + 5^2 + 5^3 + 5^4 + \dots 5^{n+1} .$$ Subtracting the first line from the second gives $$4S_n = 5^{n+1} - 1 \,,$$ 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"): $$f(x) \equiv \begin{cases} 1 + x + x^2 + x^3 + \dots & \text{for } |x| < 1 \,,\\[2ex] 1/(1-x) & \text{for all other x except } x = 1 \,. \end{cases}$$ 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: $$1 + 1/2^z + 1/3^z + \dots \,.$$ 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: $$\label{zeta-defn-real-z-greater-than-1} \zeta(z) \equiv 1 + 1/2^z + 1/3^z + \dots \,,\text{ provided real}(z) > 1 \,.$$ 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: $$\zeta(z) \equiv {1\over 1-2^{1-z}} \left(1 - {1\over 2^z} + {1\over 3^z} - {1\over 4^z} + \dots\right), \text{ provided } \text{real}(z) > 0 \text{ and } z \neq 1 \,.$$ 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: $$\label{zeta-defn-real-z-less-than-1} \zeta(z) \equiv 2(2\pi)^{z-1} \sin(\pi z/2)\; \Pi(-z)\; \zeta(1-z) \,, \text{ provided } \text{real}(z) < 1 \,,$$ 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.