Starting from a Series Problem Kontsevich Wrote on a Blackboard

从Knotsevich在黑板上写的级数题目谈起 苏剑林 · 2015-02-27 · Mathematics series integral

The other day, while browsing the Weibo account run by the "i-Math" editors at Higher Education Press, I came across a problem that Kontsevich apparently found interesting enough to write on a blackboard. The original post can be found here: http://weibo.com/3271276117/BBrL5foVz.

The series problem Kontsevich wrote on the blackboard

The problem goes like this:

$$\sum_{n=0}^{\infty} \frac{n! (20n)!}{(4n)!(7n)!(10n)!}x^n\tag{1}$$

The general goal, it seems, is to find an expression for the original function. More below the fold.

Asymptotic series

It's worth noting that this series isn't particularly "well-behaved" — a bit of simple analysis shows that the coefficients $\frac{n! (20n)!}{(4n)!(7n)!(10n)!}$ grow extremely fast, in fact faster than $n!$. As we know, exponential growth is far slower than factorial growth, so by ordinary standards of mathematical analysis, the region of convergence of the series $(1)$ actually consists of a single point: $x=0$!

That said, the expression above may still be the asymptotic series of some perfectly nice function. An asymptotic series is one in which the error term has coefficients that grow extremely fast. For instance, if the Taylor expansion $\sum_{n=0}^{\infty} a_n x^n$ of a function $f(x)$ satisfies

$$(N-1)! |x|^N < \left|f(x)-\sum_{n=0}^{N} a_n x^n\right| < N! |x|^N$$

then, having fixed the first $N$ terms, we get an error estimate. Because of the $|x|^N$ factor, the error can always be made arbitrarily small as long as $|x|$ is small enough. But since there's also a lower bound involving $(N-1)!$, achieving the same precision as $N$ increases forces the admissible range of $|x|$ to shrink. When $N\to\infty$, the range over which $|x|$ converges shrinks down to just the origin. This is exactly the notion of an asymptotic series: taking more terms doesn't necessarily make things more accurate. To get a sensible result, you either need to shrink the domain — i.e., restrict the range of $|x|$ — or take fewer terms. But taking fewer terms comes at a cost: reduced precision. We may never know, say, the tenth digit after the decimal point.

Borel resummation

Even though the actual radius of convergence of an asymptotic series may be just a single point, the underlying function — the one whose expansion gave rise to the asymptotic series in the first place — is often quite well-behaved, and may even extend nicely (over most of the domain) to the complex plane. To recover this original expression from the asymptotic series, we need some "resummation" techniques. Let's illustrate this resummation approach by finding the closed form of the series

$$\sum_{n=0}^{\infty} (n!)x^n \tag{2}$$

First, we have

$$\sum_{n=0}^{\infty} a_n x^n=\sum_{n=0}^{\infty} \frac{a_n}{n!} (n!) x^n$$

Then, using the expression for the Gamma function, we have $n!=\int_0^{\infty} t^n e^{-t} dt$. Substituting this into the expression above gives

$$\sum_{n=0}^{\infty} a_n x^n=\sum_{n=0}^{\infty} \frac{a_n}{n!} \left(\int_0^{\infty} t^n e^{-t} dt \right) x^n$$

Now swap the order of summation and integration:

$$\sum_{n=0}^{\infty} a_n x^n=\int_0^{\infty} e^{-t} dt\sum_{n=0}^{\infty} \frac{a_n}{n!} (tx)^n \tag{3}$$

The series has now become $\sum_{n=0}^{\infty} \frac{a_n}{n!} (tx)^n $. If this is a genuinely convergent series, we can find the original function and then integrate. In essence, the trick is to divide the coefficients by $n!$ to slow down their rate of divergence, thereby "coaxing" the series into converging. More generally, we have

$$\sum_{n=0}^{\infty} a_n x^n=\int_0^{\infty} e^{-t} dt\sum_{n=0}^{\infty} \frac{a_n}{(kn)!} (t^k x)^n \tag{4}$$

This resummation technique is called Borel resummation.

Using Borel resummation, we can quickly rewrite expression $(1)$ as

$$\int_0^{+\infty}ds\int_0^{+\infty}dt e^{-s-t}\sum_{n=0}^{\infty}\frac{1}{(4n)!(7n)!(10n)!}\left(ts^{20}x\right)^n\tag{5}$$

and, to make the subsequent manipulations more convenient, we further rewrite this as

$$\int_0^{+\infty}dr\int_0^{+\infty}ds\int_0^{+\infty}dt e^{-r-s-t}\sum_{n=0}^{\infty}\frac{1}{n! (4n)!(7n)!(10n)!}\left(rst^{20}x\right)^n\tag{6}$$

The reciprocal of the factorial

Now, how do we handle the summation part in expressions $(5)$ and $(6)$? Through Borel summation, we managed to eliminate the factorial in the numerator of the coefficients — but what about the one in the denominator? Reflecting on the derivation of Borel summation, we see that its key idea was to replace the factorial with its integral representation, i.e., the Gamma function. So, is there an analogous integral representation for the reciprocal of the factorial? Indeed there is — but we need to bring in complex analysis to get it.

$$\frac{1}{n!}=\frac{1}{2\pi i}\oint_{|z|=1} \frac{e^z}{z^{n+1}}dz\tag{7}$$

The derivation of $(7)$ is quite simple: it's just a matter of expressing the coefficients of the series expansion of $e^z$ using Cauchy's integral formula for complex integrals. What's interesting is how strikingly similar it looks, in form, to the expression for the Gamma function:

$$\Gamma(x)=\int_0^{+\infty} e^{-t}t^{x-1}dt$$

Given the relationship between $n!$ and $\frac{1}{n!}$, might we regard $(7)$ as a kind of dual to the Gamma function?

Using expression $(7)$, and following the same pattern as expression $(4)$, we obtain

$$\sum_{n=0}^{\infty} a_n x^n=\frac{1}{2\pi i}\oint_{|z|=1} e^{z} z^{-1} dz\sum_{n=0}^{\infty} a_n (kn)! (z^{-k} x)^n \tag{8}$$

With this result, we can now rewrite expression $(6)$ as

$$\begin{aligned}&\left(\frac{1}{2\pi i}\right)^3\left(\int\dots\int\right) e^{-r-s-t+u+v+w}(uvw)^{-1}\times\sum_{n=0}^{\infty}\frac{1}{n!}\left(rst^{20}u^{-4}v^{-7}w^{-10} x\right)^n\\ =&\left(\frac{1}{2\pi i}\right)^3\left(\int\dots\int\right) e^{-r-s-t+u+v+w+rst^{20}u^{-4}v^{-7}w^{-10} x}(uvw)^{-1} \end{aligned}$$

Here, $\left(\int\dots\int\right)$ refers to

$$\oint_{|u|=1}du\oint_{|v|=1}dv\oint_{|w|=1}dw\int_0^{+\infty}dr\int_0^{+\infty}ds\int_0^{+\infty}dt$$

The first three integrals are complex integrals, and the last three are real integrals. I'm not entirely sure whether this is the kind of result that was being sought after.