Sum Divergent Series, II « The Everything Seminar

Sum Divergent Series, II

By mnoonan

Last time we saw hints of how the geometric series was trying to tell us that there was some “truth” behind the equation

$1 + 2 + 4 + 8 + 16 + \dots = -1$

So how do we get at the truth behind such an audacious claim? One fantastically powerful way to sum divergent series such as this is through regularization: finding an analytic function of $z$ through which we can relate the divergent sum to some limit in the complex plane. In particular, we can find reasonable values for surprisingly unreasonable sums like

$1 + 1 + 1 + 1 + \dots$

We know that the canonical limiting procedure (at least since Cauchy got his hands on analysis) does not give a finite value to the sum $\sum 2^n$ . However, as Tom Leinster pointed out in the last post, there are other analytic methods that we could use to assign a value to this sum. Perhaps the simplest is the Abel sum: we take a small number $z$ and multiply the $n^{th}$ term by $z^n$ . If the terms of our sum are not growing too quickly, the resulting power series

$S(z) = 1 + 2z + 4z^2 + 8z^3 + \dots$

will have a nonzero radius of convergence in the complex plane. On the disc where this function converges, $S$ agrees with the function $1/(1 - 2z)$ . As a result, if we analytically continue $S$ to the rest of $\mathbb{C}$ we find that

$S(z) = \frac{1}{1 - 2z}$

Now we can let our regularizing parameter $z$ get larger and larger to define the Abel sum

$1 + 2 + 4 + 8 + \dots = \lim_{z \rightarrow 1} \frac{1}{1 - 2z} = -1$

The exact same regularization method lets us sum the geometric series for all values of $x$ except for $x = 1$ , and the answer is exactly what we would expect.

It is not clear that any of this has left the realm of “cute but unimportant” yet, so let me present the divergent sum which first made me a believer. I learned about this result while reading John Baez’s lecture notes on generating functions. Let

$T(z) = 1 + z + 2 z^2 + 5 z^3 + 14 z^4 + 42 z^5 + \dots$

be the generating function for the Catalan numbers. Explicitly, the coefficient of $x^n$ is the number of binary rooted trees with $n + 1$ leaves. By saying exactly what a tree is in the language of generating functions, you can easily compute that

$T(z) = \frac{1 - \sqrt{1 - 4z}}{2z}$

(or now that you know the answer, you can verify by direct computation that the power series given above is the Taylor series for $T$ around zero). Happily, there is no pole at one so the series is summable with this regularization method! And what sum do we get?

$1 + 1 + 2 + 5 + 14 + 42 + \dots = \frac{1 - \sqrt{1 - 4}}{2} = \frac{1}{2} - i\frac{\sqrt{3}}{2}$

Well, if we were happy thinking we could sum positive integers to get a negative, it doesn’t seem like such a huge leap to sum positive integers and get a complex number…

Now here is the beautiful thing that made my Platonic heart flutter: we now know that

$(1 + 1 + 2 + 5 + \dots)^7 = 1 + 1 + 2 + 5 + \dots$

which in turn means that it is possible for an isomorphism to exist between 1-tuples and 7-tuples of trees (though not for any nontrivial number smaller than 7!) And, in fact, such an isomorphism exists! This was apparently discovered by Lawvere and written up in the fantastic paper Seven Trees in One by Andreas Blass. The amazing thing is that nobody would have expected such an isomorphism to exist if it were not for this seemingly phony calculation using divergent series. So not only can we give these series finite values, but these values can lead us towards new and surprising facts about other standard mathematical objects that we know and love. These regularized divergent sums are not off in some far-flung corner of the mathematical universe — they are intimately connected with math as concrete as combinatorics, and give us honestly new insight into these areas.

Before we get too carried away with this regularization scheme, let us look at some of the things that we still can’t sum. The sum $1 + 1 + 1 + 1 + \dots$ regularizes to the function $1 / (1 - x)$ which has a singularity at $x = 1$ , so even the regularized sum appears to diverge. On the other hand, we can do the sum

$f(x) = 1 - x + x^2 - x^3 + \dots = \frac{1}{1 + x}$

to prove that $1 - 1 + 1 - 1 + 1 - 1 + \dots = 1/2$ . Now, let us do a little bit of arithmetic: the reader can verify that

$1 + x + x^2 + x^3 + \dots$

$= f(x) + 2xf(x^2) + 4x^3f(x^4) + 8x^7f(x^8) + \dots$

$= \sum 2^k x^{2^k - 1} f(x^{2^k})$

The right side of this equation allows us to plug in $x = 1$ and, using the previously-computed $f(1) = 1/2$ , we find that

$1 + 1 + 1 + 1 + \dots = \sum 2^k \cdot \frac{1}{2}$

$= \frac{1}{2} (1 + 2 + 4 + 8 + \dots) = -1/2$

though it is entirely unclear what universe this spacey calculation happened in.

Another simple sum that troubles us is $1 + 2 + 3 + 4 + 5 + \dots$ . Using our regularization method, this gives

$N(z) = 1 + 2z + 3z^2 + 4z^3 + \dots = \frac{d}{dz} \frac{1}{1 - z} = \frac{1}{(1 - z)^2}$

which still has an honest singularity at $z = 1$ . However, we can do another spacey calculation to compute this value. Let us begin with the regularized sum

$f(z) = 1 - z + z^2 - z^3 + \dots = \frac{1}{1 + z}$

Then we can honestly compute

$f(z)^2 = 1 - 2z + 3z^2 - 4z^3 + \dots = N(z) - 4z N(z^2)$

which is a true statement about the meromorphic functions $f$ and $N$ . But now we can let $z$ approach 1 to get

$(\frac{1}{2})^2 = N(1) - 4 N(1)$

proving(?) the infamous formula

$N(1) = 1 + 2 + 3 + 4 + 5 + \dots = -1/12$

Coming up next: zeta regularization and the promised puzzle! (apologies for dragging it across three posts, but the post length was getting a bit too divergent…) In the meantime, please check out John Baez’s fantastic lectures on generating functions, structure types, and summing divergent series for fun and profit here and the recent discussion on divergent sums and Tom Leinster’s work on generalized Euler characteristics (one of the best motivations for paying attention to divergent sums) at the n-Category Cafe. If you’ve got a lot of time on your hands, there is an entire fascinating book by G. H. Hardy called Divergent Series available at the Internet Archive. The calculation of $1 + 2 + 3 + 4 + 5 + \dots = -1/12$ above comes from the first few pages of chapter 1.

Finally, something that has been puzzling me for quite some time. When I showed Josh Bowman the generating function

$T(z) = 1 + z + 2 z^2 + 5 z^3 + \dots = \frac{1 - \sqrt{1 - 4z}}{2z}$

he asked me “is there a reason that $T$ is so close to the inverse of the Koebe function?” To be precise, $z = \frac{T - 1}{T^2}$ and making the Euclidean substitution $w = 1 - T$ yields the Koebe function $z(w) = w / (1 - w)^2$ . Now: what on earth do binary trees have to do with conformal mappings?

This entry was posted on July 30, 2007 at 8:59 pm and is filed under Basic Grad Student, Matt, Undergraduate. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

17 Responses to “Sum Divergent Series, II”

Isabel Says:
July 30, 2007 at 10:56 pm | Reply
The Motzkin numbers (which count unary-binary trees, i. e. trees where each node has zero, one, or two children) have the generating function M(z) = (1-z-sqrt(1-2z-3z^2))/(2z); thus M(1) = -i. Perhaps there is some bijection between 4-tuples of “Motzkin trees” and Motzkin trees?

(I don’t think there’s anything special about Motzkin trees, or even if that’s what they’re usually called; I just flipped through a book until I found a generating function that looked like it might give a nice result.)
mnoonan Says:
July 31, 2007 at 12:25 pm | Reply
(edited “4″s to “5″s -m)

Great find, Isabel! Sure enough, there is an isomorphism from the Motzkin trees to 5-tuples of Motzkin trees! The best one I could find involves twelve applications of the defining isomorphism M = 1 + zM + zM^2 (this gives a slightly different generating function, counting Motzkin trees with n edges in the nth degree term). The solution is here, but everybody should try to work it out first. These calculations are really fun!

edited to add: Despite reading John Baez’s sci.math threads with some degree of religiousness, I somehow missed this computation of the Motzkin tree cardinality almost exactly a year ago: http://tinyurl.com/2fs4to though it is unclear that the isomorphism M^5 = M has been constructed before. Can we find some non-tree examples of good generating functions? Is there a reason that different sorts of trees seem to like being roots of unity?
Kea Says:
July 31, 2007 at 8:27 pm | Reply
I’m probably being stupid, but don’t you mean 5-tuples of Motzkin trees?
mnoonan Says:
July 31, 2007 at 11:30 pm | Reply
No, no, you are right. And in the solution I wrote up, you of course end up with M^5 = M. Somehow I translated that into a 4 in the comment, though..

I guess that makes it clear that the “isomorphism” M^4 = M has never been constructed before!

Actually, I think there is a second mistake: what I actually computed seems to be the generating function for the Schroder numbers. The two functions are very similar: the Schroder-number generating function satisfies $S(z) = 1 + zS(z) + zS(z)^2$ while the Motzkin-number generating function satisfies $M(z) = 1 + z M(z) + z^2 M(z)^2$ (which gives the equation that Isabel posted). However, they both have the same value $-i$ at 1 and result in the same substitution law $X = 1 + X + X^2$ , so they both satisfy the equation $X^5 = X$ . The Schroder numbers $S_n$ count the number of Motzkin trees with $n$ non-leaf vertices, so from this perspective it is less surprising that the equality

$S(1) = 1 + 2 + 6 + 22 + 90 + \dots = 1+ 2 + 4 + 9 + 21 + 51 + \dots = M(1)$

holds. But this really is remarkable: these two divergent series, “counting” the number of Motzkin trees in two very different ways, agree that the answer should be $-i$ !
Tom Leinster Says:
August 2, 2007 at 2:13 pm | Reply
I like very much your calculation of 1 + 1 + … = 1/2. In section
4.1 of Cartier’s excellent paper “Mathemagics” -

http://www.mat.univie.ac.at/~slc/wpapers/s44cartier1.html

- there’s a different elementary derivation of this “identity”. But I
think the advantage of yours is that it’s clearer what rules are being
used. If I wanted to figure out exactly what universe this identity
is valid in – and I do! – I’d use your proof.

While we’re talking about Motzkin numbers/trees, there’s a phenomenon so weird that I can’t resist describing it.

When you want to study a polynomial equation (e.g. x^n + y^n = z^n) a
standard procedure is to study the free ring on generators satisfying
the polynomial (e.g. Z[x, y, z]/(x^n + y^n = z^n)). We’re interested
in the polynomial equation

M = 1 + M + M^2,

which in some sense defines the collection M of Motzkin trees
(described above by Isabel). Here it’s more appropriate to use rigs
(semirigs) than rings, so to study Motzkin trees, we should study the
rig

N[x]/(x = 1 + x + x^2)

- that is, the rig freely generated by one generator x satisfying this
equation. (N denotes the natural numbers.)

If we wanted to turn this rig into a ring, we could do so by formally
adjoining an additive inverse to every element. The result is the
ring Z[i] = Z[x]/(x = 1 + x + x^2) of Gaussian integers. (This
follows from the general categorical fact that adjunctions can be
composed.) But what’s weird is that our rig also *contains* the
Gaussian integers as a sort-of-subrig!

I’ll make that precise. An element of our rig can be regarded
as an equivalence class of polynomials. It’s fairly easy to see that
a constant polynomial is never equivalent to a non-constant
polynomial, so we can unambiguously define

S = {equivalence classes of non-constant polynomials}.

Writing R = N[x]/(x = 1 + x + x^2), it’s clear that S is a subset of R
closed under addition and multiplication. It’s equally clear that S
isn’t a subrig, since 0 and 1 aren’t in S. But three surprising
things happen:

(i) S *is* a rig, but with different additive and multiplicative units
than R
(ii) S is in fact a *ring*
(iii) the ring S is nothing but Z[i].

So we have a partition of the rig R into two pieces:

– the rig N of natural numbers (corresponding to the constant
polynomials, which are all inequivalent), and
– the ring Z[i] of Gaussian integers (corresponding to the equivalence
classes of non-constant polynomials).

We’d like to say that R is some kind of “sum” of N and Z[i], but it’s
a “sum” of an unusual kind: Z[i] isn’t a subrig, and it’s a
set-theoretic disjoint union. So from the standard algebraic point of
view it’s a mess, but it’s a true fact!
James Says:
August 2, 2007 at 6:45 pm | Reply
Hi Tom. Can you say what your exotic ring structure on S is? (After all, saying a set admits a ring structure isomorphic to Z[i] is saying nothing more than it’s countably infinite.)
Tom Leinster Says:
August 3, 2007 at 2:24 am | Reply
Hi James. The addition and multiplication of S are the same as those of R. (I should have made that more explicit.) The additive unit is 1 + x^4 and the multiplicative unit is x + x^5. To see get an idea of why that’s plausible, remember that x is behaves something like i.

That was the reason for the phrase “sort-of-subrig”: it’s got the same + and ., but different 0 and 1.
Tom Leinster Says:
August 3, 2007 at 2:40 am | Reply
Sorry, talking rubbish. The zero is 1 + x^2 (or rather, its equivalence class), the multiplicative unit is 2 + x^2, and while I’m at it, negatives are given by multiplication by x^2. But the addition and multiplication really are the same as in R!
James Says:
August 4, 2007 at 11:34 pm | Reply
Thanks. And now of course I see that you did say + and x were the same.
David Corfield Says:
August 23, 2007 at 8:31 am | Reply
The Motzkin observation goes back to sci.math.research in Jan 2003. Tom Leinster and Marcelo Fiore prove the isomorphism here.
Alucard Says:
October 13, 2008 at 1:46 pm | Reply
A sum of real numbers, whether finite or infinite, cannot equal a complex number. If you apply complex conjugation to each side you can see the contradiction. Another example is plugging x=-2 into the Mercator series for log(1+x); the series is x-xx/2+xxx/3-… and the regularized sum does not yield i pi, but in fact yields 0. One way to think about this is that it yields i pi when x is just above the real line, and -i pi when x is just below the real line, hence the series yields the average of the two values surrounding the discontinuity. This series is only suitable as an inverse exponential function in the slit plane and not for negative numbers.
Alucard Says:
October 13, 2008 at 2:16 pm | Reply
BTW your series 1+2z+5zz+14zzz+… does yield the complex number you claimed, or its complex conjugate, if z is very close to 1 but with an infinitesimal imaginary part.
Lubos and divergent series « The Gauge Connection Says:
February 23, 2009 at 6:44 am | Reply
[...] http://cornellmath.wordpress.com/2007/07/30/sum-divergent-series-ii/ [...]
Yasiru Says:
February 27, 2009 at 9:27 am | Reply
See my blog at,

http://mathrants.blogspot.com
Paul Jackson Says:
April 18, 2009 at 6:38 am | Reply
Hi
I found your page by chance as I was looking for info on divergent series, and found it most interesting as I have made some observations, which I will be surprised that no one has thought of them before, but here goes.

If it holds that we can use f(x) = 1/(1+x) to sum 1-1+1-1+ …… , and
(f(x))^2 = (1/(1+x))^2,to sum 1-2+3- 4+ ….. does it follow that we can sum 1- 3+ 6 – 10 + ….. the triangles, and then the tetrahedral numbers and so forth each having sum (½)^n? Thus we are able to sum the alternating members of what we can look at as being the diagonals of Pascal’s Triangle!
But as the first and second diagonals of Pascal’s Triangle are 1,1,1,1,….. and 1,2,3,,4,….., and have sums
zeta(0), and zeta(-1), can we then deduce that the sums of all the diagonals with members positive are related to zeta(-n) ? We know that zeta(-2) = 1^2 + 2^2 + 3^2+ ….., and the sum of two consecutive triangles is a square, and while playing with with zD(1/(1-z)) (where D means d/dz), from ‘Euler’s Proof that 1+2+3+ …… = -1/2, by John Baez, I was able to find that the squares, cubes and so on could each be expressed as the sum of, 2, 3, … n, consecutive, triangles, tetrahedrals, and so forth, respectively, with coefficients that could be found iteratively. So for example each positive cube can be expressed in the form 1xT(n) + 4xT(n+1) + 1xT(n+2), where the T are consecutive tetrahedrals, i.e
3^3 = 1+4×4+10
4^3 = 4 + 4×10 + 20, so is it possible that the diagonals with all terms positive could also have finite sums?

Also the sums of the reciprocals of the members of the diagonals all have sums, (i.e (n-1)/(n-2) for n>2), except for the second, the harmonic series.
Trivially the rows of Pascal’s Triangle each have sum 2^n, and if what I have said above about the sums of the diagonals makes any sense, there seems to be some kind of relationship between the rows and the diagonals, with the harmonic series being fancifully seen as a pivot around infinity.

Is there something going on here, some kind of global symmetry, or is this just some kind of trivial consequence of the fact that we are messing around with Binomial coefficients, and generating functions, got from the geometric series?
cheers Paul
The Catalan numbers, regular languages, and orthogonal polynomials « Annoying Precision Says:
June 7, 2009 at 4:02 am | Reply
[...] here involving what I’ll refer to as “generalized cardinality”; see, for example, The Everything Seminar regarding the [...]
Carl Says:
November 30, 2009 at 5:11 pm | Reply
Hi

For quite some time i’m thrilled by divergent series and consistent ways to assign a number to their ’sum’.

Has anyone got a pointer what that sum would be for the sum of all primes?
N = 2 + 3 + 5 + 7 + 11 + … ?

Thanks
Carl

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Sum Divergent Series, II

17 Responses to “Sum Divergent Series, II”

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