Starting from a Series Problem Written by Knotsevich on a Blackboard

Asymptotic Series

It should be noted that the series here is not quite "normal." A simple analysis reveals that the coefficient \frac{n! (20n)!}{(4n)!(7n)!(10n)!} grows extremely fast, even exceeding the growth rate of n!. As is well known, the growth rate of exponential functions is far weaker than that of factorials. Therefore, according to standard mathematical analysis theory, the convergence region of series (1) is actually just a single point: x=0!

However, the aforementioned function could also be the asymptotic series form of some well-behaved function. A so-called asymptotic series refers to a series where the error estimation terms have coefficients that grow very rapidly. 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. Due to the presence of |x|^N, the error can always be made sufficiently small when |x| is small enough. However, because of the existence of the lower bound involving (N-1)!, the range of |x| must shrink as N increases to maintain the same level of precision. As N \to \infty, the convergence range of |x| becomes only 0. This is the concept of an asymptotic series: taking more terms does not necessarily lead to higher accuracy. To obtain a reasonable result, one must either restrict the domain (the range of |x|) or take fewer terms. But taking fewer terms comes at a cost—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 (the expression before expansion) may actually be very well-behaved, potentially even extending to the complex plane with good properties over most regions. To recover the original expression of an asymptotic series, we need "resummation" techniques. We demonstrate this 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 expression of the Gamma function, we have n! = \int_0^{\infty} t^n e^{-t} dt. Substituting this into the above equation 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 Exchanging the summation and integration 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 normally convergent series, we can find the original function and then integrate. In general, this involves dividing the coefficients by n! to reduce their 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.

Through 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} For convenience in 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 Factorial

How do we handle the summation part in equation (5) or (6)? Through Borel summation, we eliminated the factorial part in the numerator of the coefficients, but what about the denominator? Reflecting on the derivation of the Borel summation method, we find that its core lies in using the integral form of the factorial—the Gamma function—to replace it. Does the reciprocal of the 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 equation (7) is simple; it is merely expressing the coefficients of the series expansion of e^z using Cauchy’s integral formula for complex integrals. 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 equation (7) be regarded as the dual form of the Gamma function?

Using equation (7) and following the logic of equation (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 equation (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} 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 latter three are real integrals. I wonder if this is the desired result?

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