Why is it that divergent series make sense?

Specifically, by basic calculus a sum such as $1 - 1 + 1 ...$ describes a divergent series (where divergent := non-convergent sequence of partial sums) but, as described in these videos, one can use Euler, Borel or generic summation to arrive at a value of $\tfrac{1}{2}$ for this sum.

The first apparent indication that this makes any sense is the claim that the summation machine 'really' works in the complex plane, so that, for a sum like $1 + 2 + 4 + 8 + ... = -1$ there is some process like this:

enter image description here

going on on a unit circle in the complex plane, where those lines will go all the way back to $-1$ explaining why it gives that sum.

The claim seems to be that when we have a divergent series it is a non-convergent way of representing a function, thus we just need to change the way we express it, e.g. in the way one analytically continues the Gamma function to a larger domain. This claim doesn't make any sense out of the picture above, & I don't see (or can't find or think of) any justification for believing it.

Futhermore there is this notion of Cesaro summation one uses in Fourier theory. For some reason one can construct these Cesaro sums to get convergence when you have a divergent Fourier series & prove some form of convergence, where in the world does such a notion come from? It just seems as though you are defining something to work when it doesn't, obviously I'm missing something.

I've really tried to find some answers to these questions but I can't. Typical of the explanations is this summary of Hardy's divergent series book, just plowing right ahead without explaining or justifying the concepts.

It would be so great if somebody could just say something that links all these threads together.

  • 3
    $\begingroup$ As I ask in my answer, can you please clarify how your diagram is constructed from the series? I don't understand it! $\endgroup$ Jan 10, 2014 at 17:12
  • 2
    $\begingroup$ A possibly interesting fact is that $\sum_{i=0}^{\infty} 2^i$ is a convergent summation in the $2$-adic numbers. $\endgroup$
    – user14972
    Jan 11, 2014 at 3:48
  • $\begingroup$ mathoverflow.net/questions/115743/an-algebra-of-integrals/… $\endgroup$
    – Anixx
    Nov 16, 2019 at 18:16

2 Answers 2


The logic is as follows:

  • As yet, there is no meaning to the sequence of symbols $\sum_{n=1}^\infty a_n$, given the values of $a_n$, if the sum does not converge in the classical sense.
  • We want to give this expression some meaning, perhaps because computing the answer to some problem in some naïve way gives such a divergent sum as an answer, which is obviously nonsensical, but yet which is connected to a useful answer in a way we want to determine. (Frequently the case in physics.)
  • The only way we could ever hope to gain anything useful from our definition is if
    • it agrees with the convergent case where both are defined; and
    • it shares key properties with the convergent case.
  • Let's try to choose some potential properties first. Write $S_n[a_n] = \sum_{n=1}^\infty a_n$ Here are three ideas:
    1. $S_n[a_n] = a_1 + S_n[a_{n+1}]$ (stability), i.e. $$a_1 + a_2 + \cdots = a_1 + (a_2 + \cdots)$$
    2. $S_n[a_n + b_n] = S_n[a_n] + S_n[b_n]$ (linearity part 1)
    3. $S_n[k a_n] = k S_n[a_n]$ (linearity part 2)
  • These are compatible with the normal rules, and therefore we can also require that:
    1. for convergent series, $S_n$ is indeed the usual summation (regularity).
  • They are also consistent, so we shouldn't be introducing any contradictions using this definition.
  • Would these rules suffice to compute actual results for divergent sums? Yes! For example, stability and linearity mean that $1 + 2 + 4 + \cdots$ can be computed by observing $S_n[a_n] = 1 + 2 S_n[a_n]$ and so $S_n[a_n] = -1$.

So now one question we can answer is why do loads of different summation techniques give you this same answer? If the technique is constructed to obey the above rules too! If these rules are sufficient to compute the sum, then regardless of what else you specify about $S_n$, you will have to get the above answer - provided it is defined. (On the other hand, Cesàro summation does not give a value to the above sum, for instance.)

However, whilst Nørlund means (including Cesàro sums) and Abelian means are regular, linear and stable, some other techniques are not. (Note that these are not equivalent: this means some sums cannot be computed using just regularity, linearity and stability even though these schemes do define them whilst obeying these properties.) Notably, stability is quite often dropped or weakened.

An example? Consider the series $S=1+1+1+\cdots$. By zeta function regularization, this has an answer of $-\frac 1 2$. However, if we ascribed it a value in a stable scheme, we would have $S = 1+S$, which is a contradiction because $1 \neq 0$. Fine, this just means that no stable scheme will define this quantity. Therefore zeta function regularization is not stable.

Weird, huh? Regularity and linearity can also be dropped, but then we are getting beyond the realm of most physics applications, I think.

Apparently (Wikipedia says) Borel summation is not stable, yet it still often gives the same answers as other summation methods. I'm not quite sure why, but I suspect Borel summation will turn out to be stable 'in certain situations'.

The analogy with the Gamma function is good. If one chooses some integral to define $\Gamma(z)$ which only works for some (let's say open) domain $z\in D$, then one has the problem of trying to decide what $\Gamma$ should be outside $D$. But noticing that within $D$ the $\Gamma$ function has the property that it is analytic (meromorphic perhaps) one can just impose two requirements on $\Gamma(z)$:

  • it agrees with the previous definition on $D$; and
  • it is meromorphic on $\mathbb C$.

These are exactly analogous to the two requirements made of $S_n$ above - and because they once more turn out to uniquely specify $\Gamma$ everywhere, they will agree with any other method of expressing $\Gamma$, so long as one keeps the meromorphic property.

As far as Cesàro summation and Fourier series go, this is slightly different. Here, one takes a Fourier series which does convergence pointwise. The basic point is that Fourier series can oscillate rapidly causing uniform convergence issues (there is always a big error somewhere, but not in the same place as earlier), but if one smooths out this oscillation one can make the convergence uniform.

This is not a very mysterious application of Cesàro summation - it just decreases the effect of adding a term which will eventually be cancelled by later terms.

Edit: In answer to your concerns that this is all "nonsense" - your problem is that you are thinking about summation techniques as being used for improving numerical convergence, whereas in general there are lots more properties of summations that you are interested in. Here is one.

Suppose you have a physical system which is described by some property $x$, which might be a coupling constant for instance. Now suppose that you are interested in some property $F(x)$ which is a measurable output from the system. Further, suppose you reckon that $F(x)$ is an analytic function of $x$.

In fact, if you assume $x$ is really small, you are very happy to find $F(x) = b_0 + b_1 x + b_2 x^2 + \cdots$, where perhaps $b_n = (-2)^n$. For nice small numbers, you can work out the answer by taking a few terms.

Now imagine you're interested in measuring at $x=37.4$. Suppose further that you didn't really understand about series only working when they're small, so you just plugged in $x$.

You get $S = \sum (37.4)^n (-2)^n = \sum (-74.8)^n = 1 - 74.8 + 74.8^2 - \cdots$ and think "~%*£%!". But now you laugh cheekily, and apply the rules from above. You find $S = 1 - 74.8 S$ and so $S = 1/75.8 = 0.0132\ldots$.

That's rubbish, right?

Wrong. Notice that for small $x$, $$F(x) = 1 - 2x + 4x^2 - \cdots = \frac1{1+2x}$$and the unique analytic continuation of this results in $$F(37.4) = \frac{1}{1+2\times 37.4} = \frac 1 {75.8} = 0.0132\ldots$$

Magic? Nope. Just the fact that our definition plays nice with analyticity. Whenever you sum up analytic functions outside their radius of convergence, you should get the right answer.

Now the real places where these tricks are used (and not always really understood) in physics are things like the critical dimension of a string, or in the quick computation of the magnitude of the Casimir effect. The deal is similar to the above, except that one doesn't have an $x$ so one doesn't work out the dependence on a parameter; one simply computes a single quantity in an expansion, then hopes and prays that the physics in the background is really nice (like the analyticity above) so that it plays ball with your summation - and then you can use one of these summation tricks and it should give you the right answer, even if the series you wrote down diverged!

Having thought about the pictorial aide, I think all Carl Bender is trying to say is that - working on the Riemann sphere to make things nice and clear, where once again complex infinity is a unique point at the northern hemisphere - there is a definite sense in which lines can go to infinity and come back from the 'other side'. http://upload.wikimedia.org/wikipedia/commons/7/7c/Riemann_sphere1.jpg

It's not clear to me that this is anything more than a vague hint though, most notably because if one plots the partial sums on the sphere (and treat it as a normal sphere) it is clear that they accumulate to the point of complex infinity, and can never reach it or pass it. I'm curious about this though, and whether it can be shown to have any rigorous value.

Perhaps a better way to look at this is to say that if one does the sum by a technique equivalent to analytic continuation, then one can obtain any value for the function. If one calculated $1+2+4+\cdots$ by writing $1+2x+4x^2+\cdots = 1/(1-2x)$ then analytically continued in the complex plane from near 0 to 1 to avoid the singularity at $\frac 1 2$, one can get negative results because one is wandering around the complex plane; specifically if one makes a really tight circle around $\frac 1 2$ then the value for the sum goes round a 'small' circle near complex infinity. Drawing this on the sphere links up with the intuition above.

  • $\begingroup$ Thanks for the answer, I got the picture idea from this video youtu.be/LMw0NZDM5B4?t=4m50s I understand the summation machine axioms, but I don't understand why the fact that we got to infinity 'linearly' somehow means that the infinity we got is not actually infinity, I just can't shake off the feeling that we're just placing definitions on absolute nonsense because we want it to be so. I need to get rid of that feeling, because my motivation here is perturbation series, which assigns physical values to what I think of as nonsense, but I need to see why it's not nonsense, if you get me $\endgroup$
    – bolbteppa
    Jan 10, 2014 at 17:31
  • $\begingroup$ Also, I guess Cesaro summation is similar to a Shanks transformation, or Richardson extrapolation. For these two methods, I can see their use - you get the approximate value of the sum of a series after a few calculations. However, what is the point in applying tricks to get uniform convergence? The original series may not converge uniformly, what is the use/reason for creating a new series that does converge uniformly, if you know one off hand? Thanks very much $\endgroup$
    – bolbteppa
    Jan 10, 2014 at 17:35
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    $\begingroup$ (1) You never 'get infinity'. It is not the case that the series "$=\infty$". You just have some numbers, and you want a rule to give another number to them if possible. I'll add something to my answer to hint at why you might find this definition useful. But the one thing it is not is "nonsense". It is a 100% perfectly consistent set of definitions for something that wasn't defined before. You are not computing $1 + 2 = 3, 3 + 4 = 7, 7 + 8 = 15, ...$ and then "finding the answer". You are deciding what is a sensible thing to do with $1+2+4+8+\cdots$. $\endgroup$ Jan 10, 2014 at 17:42
  • $\begingroup$ (2) I've seen it used as a mathematical tool more than as a practical aid for rapid summation. But nonetheless, imagine you have the data for the Fourier series of a compactly supported signal, and want to reconstruct it without having wild oscillations near the ends of the interval. Then using Cesàro summation would probably give you just that. $\endgroup$ Jan 10, 2014 at 17:44
  • $\begingroup$ @bolbteppa I've added a simple exploration of this in action. $\endgroup$ Jan 10, 2014 at 18:01

We can compute finite value of the sum of a divergent series as general limit of their partial sum: https://m4t3m4t1k4.wordpress.com/2015/11/25/general-method-for-summing-divergent-series-using-mathematica-and-a-comparison-to-other-summation-methods/

We can use that general limit to compute finite value of divergent integrals, here is 3 examples: https://m4t3m4t1k4.wordpress.com/2015/02/14/general-method-for-summing-divergent-series-determination-of-limits-of-divergent-sequences-and-functions-in-singular-points-v2/


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