By 苏剑林 | February 27, 2015
One day, while browsing the Weibo of an editor from Higher Education Press's "i Mathematics," I found a problem written by Kontsevich on a blackboard that he thought was very interesting. The original URL is: http://weibo.com/3271276117/BBrL5foVz.
The problem is as follows:
$$\sum_{n=0}^{\infty} \frac{n! (20n)!}{(4n)!(7n)!(10n)!}x^n\tag{1}$$
The goal is likely to find an expression for the original function.
Asymptotic Series
It should be noted that the series here is not quite "normal." A simple analysis reveals that the coefficients $\frac{n! (20n)!}{(4n)!(7n)!(10n)!}$ grow extremely fast, even exceeding the growth rate of $n!$. As everyone knows, the growth rate of exponential functions is much weaker than factorials. Therefore, according to standard mathematical analysis theory, the convergence region of series $(1)$ is actually just a single point: $x=0$!
However, such a series could be the asymptotic series form of a well-behaved function. A so-called asymptotic series means that the error estimation term has coefficients that grow very quickly. For example, if the Taylor expansion $\sum_{n=0}^{\infty} a_n x^n$ of a function $f(x)$ satisfies
$$(N-1)! \|x\|^N < \left\|f(x)-\sum_{n=0}^{N} a_n x^n\right\| < N! \|x\|^N$$
When taking the first $N$ terms, there is an error estimate. Because of the presence of $\|x\|^N$, when $\|x\|$ is sufficiently small, the error can always be made sufficiently small. But because of the lower bound involving $(N-1)!$, in order to achieve the same precision as $N$ increases, the range of $\|x\|$ must shrink accordingly. When $N \to \infty$, the convergence range for $\|x\|$ is reduced to only 0. This is the concept of an asymptotic series: taking more terms does not necessarily mean more precision. To obtain a reasonable result, one must either restrict the domain (the range of $\|x\|$) or take fewer terms. However, taking fewer terms comes at a price—lack of precision; we might never know the tenth digit after the decimal point.
Borel Resummation
Although the actual convergence region of an asymptotic series is only the origin, the original function of the asymptotic series—that is, the expression before the expansion—is likely to be very well-behaved, potentially even extendable to the complex plane with good properties over most of the area. To restore the original expression of an asymptotic series, we need "resummation" techniques. We can demonstrate this resummation technique by finding the original expression for 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, from the integral expression of the Gamma function, we have $n!=\int_0^{\infty} t^n e^{-t} dt$. Substituting this into the equation above, we get
$$\sum_{n=0}^{\infty} a_n x^n=\sum_{n=0}^{\infty} \frac{a_n}{n!} \left(\int_0^{\infty} t^n e^{-t} dt \right) x^n$$
Swapping the summation and integral signs:
$$\sum_{n=0}^{\infty} a_n x^n=\int_0^{\infty} e^{-t} dt\sum_{n=0}^{\infty} \frac{a_n}{n!} (tx)^n \tag{3}$$
In this way, the series becomes $\sum_{n=0}^{\infty} \frac{a_n}{n!} (tx)^n$. If this is a normal convergent series, we can find the original function and then integrate. In general, this is done by dividing the coefficients by $n!$ to reduce the divergence speed and "force" the series to converge. 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 summation technique is called Borel resummation.
Using Borel resummation, we can quickly transform equation $(1)$ into
$$\int_0^{+\infty}ds\int_0^{+\infty}dt e^{-s-t}\sum_{n=0}^{\infty}\frac{1}{(4n)!(7n)!(10n)!}\left(ts^{20}x\right)^n\tag{5}$$
To facilitate subsequent processing, a further transformation is needed:
$$\int_0^{+\infty}dr\int_0^{+\infty}ds\int_0^{+\infty}dt e^{-r-s-t}\sum_{n=0}^{\infty}\frac{1}{n! (4n)!(7n)!(10n)!}\left(rst^{20}x\right)^n\tag{6}$$
Reciprocal of the Factorial
Now, how do we handle the summation part in $(5)$ or $(6)$? Through Borel resummation, we eliminated the factorial part in the numerator of the coefficients, but what about the denominator? Reflecting on the derivation of Borel resummation, we find its core lies in using the integral form of the factorial—the Gamma function—to replace it. Does the reciprocal of a factorial have a similar form? Yes! However, it requires the help of complex functions.
$$\frac{1}{n!}=\frac{1}{2\pi i}\oint_{\|z\|=1} \frac{e^z}{z^{n+1}}dz\tag{7}$$
The derivation of $(7)$ is very simple; it is merely expressing the coefficients of the series expansion of $e^z$ using the Cauchy integral formula. Interestingly, it is very similar in form to the expression of the Gamma function:
$$\Gamma(x)=\int_0^{+\infty} e^{-t}t^{x-1}dt$$
Considering the relationship between $n!$ and $\frac{1}{n!}$, can $(7)$ be seen as a dual form of the Gamma function?
Through $(7)$, and by mimicking $(4)$, we have
$$\sum_{n=0}^{\infty} a_n x^n=\frac{1}{2\pi i}\oint_{\|z\|=1} e^{z} z^{-1} dz\sum_{n=0}^{\infty} a_n (kn)! (z^{-k} x)^n \tag{8}$$
Using this result, we can transform $(6)$ into
\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}
Where $\left(\int\dots\int\right)$ 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 are complex integrals, and the last three are real integrals. I wonder if this is the desired result?