Padé Approximations and Irrationality Measures on Values of Confluent Hypergeometric Functions

Abstract

Padé approximations are approximations of holomorphic functions by rational functions. The application of Padé approximations to Diophantine approximations has a long history dating back to Hermite. In this paper, we use the Maier–Chudnovsky construction of Padé-type approximation to study irrationality properties about values of functions with the form $f\left(x\right)={\displaystyle \sum _{k=0}^{\infty }}{\displaystyle \frac{x^k}{k!(bk+s)(bk+s+1)\cdots (bk+t)}},$ where $b,t,s$ are positive integers and obtain upper bounds for irrationality measures of their values at nonzero rational points. Important examples includes exponential integral, Gauss error function and Kummer’s confluent hypergeometric functions.

1. Introduction

Ever since Hippasus of Metapontum made the astonishing discovery of the existence of irrational numbers, the irrationality or transcendence of special values has attracted much attention. From the simplistic proof for the irrationality of $\sqrt{2}$ and e, mathematicians have developed a variety of different measures to prove the irrationality of special values, as well as describe its irrationality in terms of rational approximations.

The French mathematician Charles Hermite first came up with the idea of using rational functions to approximate and prove the irrationality or transcendence of values of exponential functions at rational points. This method of approximation by rational functions is called Padé approximation. This method motivated many mathematicians to study Diophantine approximations, for example, Maier [1], Siegel [2] and Chudnovsky [3].

One of the most memorable and significant yet surprising findings in the irrationality of values is Apéry’s proof of $\zeta \left(3\right)$’s irrationality in 1978. The original sketch of proof published by Apéry involves two novel series $A_n$ and $B_n$ named Apéry numbers and uses their quotients to approach the value of $\zeta \left(3\right)$ (See [4]), which was once considered miraculous and unexpected.

However, such series $A_n$ and $B_n$ for many other Diophantine approximation problems naturally appear in Padé approximations or Padé-type approximations, especially in the study of generalized hypergeometric functions. A number of significant works have been conducted by Maier [1], followed by Chudnovsky [3], who sketched a general form of Padé approximants to generalized hypergeometric functions.

In this paper, we will follow the idea of using Padé or partial Padé approximants to study the irrationality of values of special functions. We will first recall the result of irrationality proof of the exponential values at rational points, and then study the irrationality property of some special types of generalized hypergeometric functions, construct the Maier–Chudnovsky type approximations and derive upper bounds for the irrationality measures. More specifically, we will derive the following main theorem

Theorem 1.

For the following special formal power series,

$$f\left(x\right)=\sum _{k=0}^{\infty }{\displaystyle \frac{x^k}{k!(bk+s)(bk+s+1)\cdots (bk+t)}},$$

where $b\ge 2,s,t$ are positive integers such that $s\le t$ and there exists an integer in $[s,t]$ that is co-prime with b. Then, there is a nonzero polynomial $g\left(x\right)\in \mathbb{Q}\left[x\right]$ depending on $b,s,t$; if $x_0\in \mathbb{Q}-\left\{0\right\}$ is not a root of $g\left(x\right)$, then $f\left(x_0\right)$ is irrational and the irrationality measure is bounded by:

$$2\le \mu \left(f\left(x_0\right)\right)\le b+2.$$

See Proposition 5 for the explicit form of $g\left(x\right)$.

A detailed definition of irrationality measure can be found in Definition 1. This concept is also called the Liouville–Roth constant or irrationality exponent. It is used to measure how well an irrational real number can be approximated by rational numbers. Finding the values or bounds of irrationality measure is a very important topic in Diophantine approximation. For example, the irrationality measure $\mu \left(x\right)=2$ for irrational algebraic numbers is the Fields Medal-winning work of Roth. The main contribution in this paper is the bounds for irrationality measure for values of the function $f\left(x\right)$ at nonzero rational numbers x. Special cases of the functions are summarized below.

Example 1.

The form of formal power series in Theorem 1 includes the following important functions upon choosing different parameters:

1. 

When $b=0,\hspace{0.17em}s=t=1,\hspace{0.17em}f\left(x\right)=e^x$ irrational at nonzero irrational points was studied by Hermite [5]. In this degenerate case, the irrationality measure conclusion still holds; in other words, we have $\mu \left(e^x\right)=2$ for nonzero rational number x.

2. 

When $b=1,\hspace{0.17em}s=t=0,\hspace{0.17em}f\left(x\right)={\sum }_{k=1}^{\infty }{\displaystyle \frac{x^k}{k·k!}}$ is related to the exponential integral by

$$E_1\left(z\right)={\int }_z^{\infty }{\displaystyle \frac{e^{-t}}{t}}dt=-\gamma -lnz-f(-z)$$

with γ being the Euler–Mascheroni constant. The irrationality of $f\left(x\right)$ was studied by Maier [1].

3. 

When $b=2,\hspace{0.17em}s=t=1,\hspace{0.17em}f\left(x\right)={\sum }_{k=0}^{\infty }{\displaystyle \frac{x^k}{(2k+1)·k!}}$ is related to the Gauss error function $\mathrm{erf}\left(x\right)$ via the following

$$\mathrm{erf}\left(z\right)={\displaystyle \frac{2}{\sqrt{\pi }}}{\int }_0^z{e}^{-t^2/2}dt={\displaystyle \frac{2z}{\sqrt{\pi }}}f(-z^2)$$

This error function is widely used in probability, statistics and PDE.

4. 

When $b=1$ or $s=t$, the function represents a large class of Kummer’s confluent hypergeometric functions (also known as confluent hypergeometric function of the first kind), which is the solution of Kummer’s differential equations.

The main ingredients in the proof Theorem 1 are the following three steps:

• The first step is to use a combinatorial cancellation lemma (Lemma 2) to construct partial Padé approximations to $f\left(x\right)$. The key idea here is to sacrifice the order of remainder functions to obtain control of denominators of coefficients in the polynomials.
• The second step is to use a mod-p method to prove the non-vanishing of the remainder terms and thus implies the irrationality of those values.
• The last step is to apply Tannery’s theorem to estimate the polynomials and remainder terms (Lemmas 5 and 6), and obtain the upper bound for irrationality measure based on a folklore lemma (Lemma 1).

The paper is organized as follows. In Section 2, we review some basic background including Padé approximations and irrationality measures, and the folklore lemma (Lemma 1) on bounds of irrationality measures. In Section 3, we introduce the combinatorial cancellation lemma (Lemma 2) and reconstruct the Padé approximation of exponential functions. This also leads to precise estimates of denominators and remainders, and hence the exact value of the irrationality measure of $e^x$ at nonzero rational x. In Section 4, we construct explicit partial Padé-type approximants to general $f\left(x\right)$. In Section 5, we prove the non-vanishing nature of the remainder via a modulo-p argument similar to Maier [1]. In Section 6, we obtain estimates of denominators and remainders, hence the proof of the main theorem. In Section 8, we point out some rational values for parameters excluded from the theorem and study the differential systems satisfied by the function. In Section 9, we discuss the recent important results of Fischler and Rivoal about irrationality measures on values of Type-E functions.

2. Background

2.1. Padé Approximation

Taylor expansion can be viewed as the approximation of complex analytic functions $f\left(x\right)$ by polynomials. Padé approximation is introduced to provide a better approximation of meromorphic functions with larger domains of convergence using rational functions by Henri Padé in [6].

Let $f\left(x\right)$ be a complex analytic function defined on a neighborhood of $a\in \mathbb{C}$. For a pair of positive integers $m,n$, Padé approximation is usually defined by:

$$f_{[n,m]}\left(x\right)={\displaystyle \frac{B_{[n,m]}\left(x\right)}{A_{[n,m]}\left(x\right)}},x\to a$$

where $B_{[n,m]}\left(x\right)$ is a polynomial of degree m and $A_{[n,m]}\left(x\right)$ is a monic polynomial of degree n, and

$$f\left(a\right)=f_{[n,m]}\left(x\right)$$

$$f^{\prime }\left(a\right)=f_{[n,m]}^{\prime }\left(x\right)$$

$$f^{″}\left(a\right)=f_{[n,m]}^{″}\left(x\right)$$

$$...$$

$$f^{(m+n)}\left(a\right)=f_{[n,m]}^{(m+n)}\left(x\right)$$

or alternatively, Padé approximants of f could be defined as:

$$f\left(x\right)-{\displaystyle \frac{B_{[n,m]}\left(x\right)}{A_{[n,m]}\left(x\right)}}=O\left({(x-a)}^{n+m+1}\right),x\to a$$

or:

$$A_{[n,m]}\left(x\right)f\left(x\right)-B_{[n,m]}\left(x\right)=O\left({(x-a)}^{n+m+1}\right),x\to a$$

if we linearize this condition.

2.2. Irrationality Proof by Padé Approximants

Using Padé approximation, the proof of the irrationality for specific values of a function can be directly derived with a non-vanishing remainder term $R\left(x\right)$ that approaches 0 fast enough as $m,n$ approaches infinity. This will directly lead to the condition:

• For $\alpha =f\left(x\right)$ with x fixed, there exists infinitely many pairs of integers $(p_n,q_n)$ such that

  $$0<|q_n\alpha -p_n|\to 0.$$

  If such pairs exist, then we can obtain the irrationality of $\alpha$.

Based on the previous idea, Maier [1] investigated the irrationality of some special generalized hypergeometric functions using partial Padé approximation, which scarifies accuracy of the approximation order $O\left({(x-a)}^{n+m}\right)$, but still gives an irrationality proof.

$$A_{[n,m]}\left(x\right)f\left(x\right)-B_{[n,m]}\left(x\right)=O\left({(x-a)}^{r(m,n)}\right),x\to a$$

(1)

where $r(m,n)$ is an integer-valued function of $(m,n)$.

Inspired by Maier’s work, Siegel studied a more general class of functions: Type-E functions. He proved the irrationality of Type-E functions satisfying the “normal condition”, which was removed by Shidlovskii in his later works. This result has been generally cited as the “Siegel–Shidlovskii Theorem”, and was later refined by F. Beukers (See [7]). This theorem relates the $\overline{\mathbb{Q}}\left(x\right)$-algebraic independence of some Type-E functions to the $\overline{\mathbb{Q}}$-algebraic independence of their values at $\overline{\mathbb{Q}}$-points excluding finitely many points. Those functions form solutions to a linear differential system with $\overline{\mathbb{Q}}\left(x\right)$ coefficients and the points excluded are singular points of the system together with zero. The $\overline{\mathbb{Q}}\left(x\right)$-algebraic independence or explicit transcendence degree is usually obtained case by case for different types of functions. The values of one of the functions in the system can still be rational; for example, see the discussion in Section 8. Furthermore, the results of explicit Padé-type approximation can give more information, for example, bounds for irrationality measures.

2.3. Irrationality Measure

First, we recall the definition of irrationality measure (or irrationality exponent).

Definition 1.

Let α be a real number. The irrationality measure $\mu \left(\alpha \right)$ is the smallest possible value for μ such that

$$|x-{\displaystyle \frac{p}{q}}|>{\displaystyle \frac{1}{q^{\mu +ϵ}}}$$

is satisfied for any fixed $ϵ>0$ and integer pairs $(p,q)$ with q large enough.

The irrationality measure for the rational number is 1. By Dirichlet’s approximation theorem, the irrationality measure for the irrational real number is greater than or equal to 2. The Thue–Siegel–Roth theorem implies that actually this is the best bound for irrational algebraic real numbers. Roth was awarded the Fields Medal for this important result in Diophantine approximation. Beside the effort in proving the irrationality of unknown functions, the irrationality measure (or irrationality exponent, in some works) of known irrational function values has also attracted much attention. Many have joined the competition to obtain better bounds or accurate values for irrationality measures of important numbers. One of the most recent examples is the result about bounds of irrationality measure of $\pi$ given by Zeilberger and Zudilin [8], $\mu \left(\pi \right)<7.10320534$, improving the previous upper bound $7.606308$ by Salikhov [9].

In this paper, we study a type of function with a similar form as confluent hypergeometric functions with rational parameters and generalize Maier’s results following his methods, including some interesting examples such as the error function of the normal distribution. By explicit construction of Padé or partial Padé approximants to the functions, we also derive an upper bound for the irrationality measures of the rational values of these functions by the following lemma.

Lemma 1

(Upper bound of irrationality measure). Let $\theta >0$ be a constant and $\sigma :\mathbb{N}\to \mathbb{R}>0$ be a increasing function satisfying ${lim}_{n\to \infty }\sigma \left(n\right)=+\infty$, ${lim}_{n\to \infty }{\displaystyle \frac{\sigma (n+1)}{\sigma \left(n\right)}}=1$. Let $\alpha \in \mathbb{R}$ and $q_n,p_n\in \mathbb{Z}$, satisfying

1. 

${lim}_{n\to \infty }{\displaystyle \frac{log|q_n|}{\sigma \left(n\right)}}=\gamma >0$;

2. 

$0\ne r_n=q_n\alpha -p_n$ and ${lim\; sup}_{n\to \infty }{\displaystyle \frac{log|r_n|}{\sigma \left(n\right)}}\le -\theta$.

Then $\alpha \notin \mathbb{Q};\mu \left(\alpha \right)\le 1+{\displaystyle \frac{\gamma }{\theta }}$.

Later in our application of this lemma, we take $\sigma \left(n\right)=nlogn$. This lemma is well known and has appeared in people’s work frequently. However, we found that recent works that use this lemma usually cite the version in [10] (Excercise 3, p. 387), which actually requires more to obtain the upper bound of the irrationality measure. The above version we cited was originally in Hata’s paper (see [11]) and the proof is a line-by-line argument in [11] (Lemma 2.1), communicated to us by Li Lai.

3. Combinatorial Cancellation Lemma and Padé Approximation

From Equation (1), we need enough vanishing for coefficients in the expansion of $A\left(x\right)f\left(x\right)$ to obtain explicit partial Padé expansions. An important ingredient in those constructions for hypergeometric functions is the following combinatorial cancellation lemma of binomial numbers.

Lemma 2

(Cancellation lemma). Let $S\left(x\right)\in \mathbb{C}\left[x\right]$ be a polynomial of degree strictly less than n, then

$$\sum _{i=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{i}}\right){(-1)}^{i}S\left(i\right)=0$$

Proof. 

We first prove the case of $S\left(i\right)=i^k$ with $k<n$. Define the logarithmic differential operator $\delta =x{\displaystyle \frac{\sigma }{\sigma x}}$. Notice that

$${\left(\delta \right)}^k{x}^i={\left(i\right)}^k{x}^i.$$

$$\begin{array}{ccc}\hfill \sum _{i=0}^{\infty }\left({\displaystyle \genfrac{}{}{0pt}{}{n}{i}}\right){(-1)}^i{i}^k& =& \sum _{i=0}^{\infty }\left({\displaystyle \genfrac{}{}{0pt}{}{n}{i}}\right){(-1)}^i{\left(\delta \right)}^k{x}^i{|}_{x=1}\hfill \\ & =& {\left(\delta \right)}^k{(1-x)}^n{|}_{x=1}\hfill \\ & =& 0.\hfill \end{array}$$

By the linearity of S in the left-hand side of the equality, we conclude the proof of the cancellation lemma. □

The lemma was used by Maier [1] and later Chudnovsky [3] to construct explicit forms of partial Padé approximations.

It is worth mentioning that there are also interesting developments using the combinatorical properties of binomial coefficients to approximate certain operators, see [12,13,14,15].

3.1. $[n,m]$-Padé Approximants of $e^x$

In order to show the power of the cancellation lemma, we find $[n,m]$-Padé approximants of $e^x$ by this lemma, and use Padé approximants to find the exact value of the irrationality measure of $e^x$ at nonzero rational number x. This can be viewed as a degenerate case of our main Theorem 1 with $b=0$. More specifically, we will show the following result by Lemmas 1 and 2.

Proposition 1.

The irrationality measure of $e^x$ with nonzero rational number x is 2.

Hermite first proved the irrationality of $e^x$ and Siegel [2] later interpreted the proof using Padé approximation of exponential function. The proof by Siegel uses the appropriate differential operator acting on both sides of equation (1). Here, we approach the construction by Lemma 2.

Proposition 2

(Padé approximation of $e^x$). Let $n\le m+1$ be a pair of positive integers. Define polynomials $A_{[n,m]}\left(x\right)$ of degree n and $B_{[n,m]}\left(x\right)$ of degree m by

$$A_{[n,m]}\left(x\right)=\sum _{i=0}^n{(-1)}^i\left({\displaystyle \genfrac{}{}{0pt}{}{n}{i}}\right){\displaystyle \frac{(m+i)!}{m!}}{x}^{n-i}$$

$$B_{[n,m]}\left(x\right)=\sum _{k=0}^m\sum _{i=max(0,n-k)}^n{(-1)}^i\left({\displaystyle \genfrac{}{}{0pt}{}{n}{i}}\right){\displaystyle \frac{(m+i)!}{m!(k-n+i)!}}{x}^k$$

Then $A_{[n,m]}\left(x\right)e^x-B_{[n,m]}\left(x\right)=O\left(x^{m+n+1}\right)$.

Proof. 

Using the Taylor expansion of $e^x$, we have

$$\begin{array}{ccc}\hfill A_{[n,m]}\left(x\right)e^x& =& \sum _{k=0}^{\infty }\sum _{i=0}^n{(-1)}^i\left({\displaystyle \genfrac{}{}{0pt}{}{n}{i}}\right){\displaystyle \frac{1}{m!}}{\displaystyle \frac{(m+i)!}{k!}}{x}^{k+n-i}\hfill \\ & =& \sum _{k=0}^{\infty }\sum _{i=max(0,n-k)}^n{(-1)}^i\left({\displaystyle \genfrac{}{}{0pt}{}{n}{i}}\right){\displaystyle \frac{1}{m!}}{\displaystyle \frac{(m+i)!}{(k-n+i)!}}{x}^k\hfill \end{array}$$

The summation above for $0\le k\le m$ forms $B_{[n,m]}\left(x\right)$. So we only need to prove that

$$\sum _{k=m+1}^{n+m}\sum _{i=0}^n{(-1)}^i\left({\displaystyle \genfrac{}{}{0pt}{}{n}{i}}\right){\displaystyle \frac{1}{m!}}{\displaystyle \frac{(m+i)!}{(k-n+i)!}}{x}^k=0$$

For k such that $m+1\le k\le m+n$, we have

$${\displaystyle \frac{(m+i)!}{(k-n+i)!}}=(i+m)(i+m-1)\cdots (i+k-n+1).$$

This expression is a polynomial of i with degree $m-k+n$, which is strictly less than n. So by Lemma 2, we have

$$\sum _{i=0}^n{(-1)}^i\left({\displaystyle \genfrac{}{}{0pt}{}{n}{i}}\right){\displaystyle \frac{1}{m!}}{\displaystyle \frac{(m+i)!}{(k-n+i)!}}=0.$$

Hence we have the conclusion. □

This proposition also gives us the explicit form of the remainder term $R_{[n,m]}\left(x\right)=A_{[n,m]}\left(x\right)e^x-B_{[n,m]}\left(x\right)$ as follows

$$\begin{array}{ccc}\hfill R_{[n,m]}\left(x\right)& =& \sum _{k=m+n+1}^{\infty }\sum _{i=0}^n{(-1)}^i\left({\displaystyle \genfrac{}{}{0pt}{}{n}{i}}\right){\displaystyle \frac{1}{m!(m+1+i)\cdots (k-n+i)}}{x}^k\hfill \\ & =& {\displaystyle \frac{x^{m+n+1}}{m!}}\sum _{k=0}^{\infty }\sum _{i=0}^n{(-1)}^i\left({\displaystyle \genfrac{}{}{0pt}{}{n}{i}}\right){\displaystyle \frac{1}{(m+i+1)\cdots (k+m+i+1)}}{x}^k\hfill \\ & =& {\displaystyle \frac{x^{m+n+1}}{m!}}\sum _{i=0}^n{(-1)}^i\left({\displaystyle \genfrac{}{}{0pt}{}{n}{i}}\right){\displaystyle \frac{1}{m+i+1}}{}_{1}F_1\left[{\displaystyle \genfrac{}{}{0.0pt}{}{1}{m+i+2}};x\right]\hfill \end{array}$$

Here, ${}_{1}F_1\left[{\displaystyle \genfrac{}{}{0.0pt}{}{a}{b}};x\right]$ is the confluent hypergeometric function and it has the integration form

$${}_{1}F_1\left[{\displaystyle \genfrac{}{}{0.0pt}{}{a}{b}};x\right]={\displaystyle \frac{\Gamma \left(b\right)}{\Gamma \left(a\right)\Gamma (b-a)}}{\int }_0^1{e}^{xu}{u}^{a-1}{(1-u)}^{b-a-1}du.$$

Plugging this into the remainder term, we obtain

$$\begin{array}{ccc}\hfill R_{[n,m]}\left(x\right)& =& {\displaystyle \frac{x^{m+n+1}}{m!}}{\int }_0^1{e}^{xu}\sum _{i=0}^n{(-1)}^i\left({\displaystyle \genfrac{}{}{0pt}{}{n}{i}}\right){(1-u)}^{m+i}du\hfill \end{array}$$

(2)

$$\begin{array}{ccc}& =& {\displaystyle \frac{x^{m+n+1}}{m!}}{\int }_0^1{e}^{xu}{u}^m{(1-u)}^{n}du\hfill \end{array}$$

(3)

3.2. Irrationality Measure of $e^x$

In this section, we calculate the irrationality measure of $e^x$ at nonzero rational x.

Proposition 3.

The irrationality measure of $e^x$ for $x\in \mathbb{Q}-\left\{0\right\}$ is equal to 2.

Proof. 

For positive integers $n\le m$, we have the following integral representation for $A_{[m,n]}\left(x\right)$ from the expansion formula above:

$$\begin{array}{ccc}\hfill A_{[n,m]}\left(x\right)& =& \sum _{i=0}^n{(-1)}^i\left({\displaystyle \genfrac{}{}{0pt}{}{n}{i}}\right){\displaystyle \frac{\Gamma (m+i+1)}{m!}}{x}^{n-i}\hfill \\ & =& {\displaystyle \frac{x^{n-i}}{m!}}\sum _{i=0}^n{(-1)}^i\left({\displaystyle \genfrac{}{}{0pt}{}{n}{i}}\right){\int }_0^{\infty }{t}^{m+i}{e}^{-t}dt\hfill \\ & =& {\displaystyle \frac{1}{m!}}{\int }_0^{\infty }{t}^m{(x-t)}^n{e}^{-t}dt\hfill \\ & =& {\displaystyle \frac{1}{m!}}{\int }_0^x{t}^m{(x-t)}^n{e}^{-t}dt+{\displaystyle \frac{1}{m!}}{\int }_x^{\infty }{t}^m{(x-t)}^n{e}^{-t}dt\hfill \end{array}$$

When n is large enough, we will always have

$${\int }_x^{\infty }{(t-x)}^{n+m}{e}^{-t}dt<{\int }_x^{\infty }{t}^m{(x-t)}^n{e}^{-t}dt<{\int }_x^{\infty }{t}^{m+n}{e}^{-t}dt\sim \Gamma (m+n+1)$$

We take $m=n$, the growth rate of $A_{[n,n]}\left(x\right)$ is given by ${\displaystyle \frac{\left(2n\right)!}{n!}}$.

Now we estimate the remainder term:

$$R_{[n,m]}\left(x\right)={\displaystyle \frac{x^{2n+1}}{n!}}({\int }_0^1{(u-u^2)}^n{e}^{ux}du)\ne 0$$

The factor $e^{ux}$ could be neglected as it gives a factor not related to n.

Using trigonometric substitution $u=si{n}^{2}s$:

$$\begin{array}{ccc}\hfill {\int }_0^1{(u-u^2)}^{n}du& =& {\int }_0^{{\displaystyle \frac{\pi }{2}}}{\left({sin}^{2}s{cos}^{2}s\right)}^{n}dsinscoss\hfill \\ & =& {\displaystyle \frac{\left(2n\right)!!}{(2n+1)!!2^{2n}}}\hfill \end{array}$$

So $R_{[n,n]}\left(x\right)\sim x^{2n+1}·{\displaystyle \frac{n!}{(2n+1)!}}$. Hence, $A_{[n,n]}\left(x\right)R_{[n,n]}\left(x\right)\sim O({\displaystyle \frac{x^{2n+1}}{2n+1}})$. Notice that the coefficients in $A_{[n,n]}\left(x\right)$ and $B_{[n,n]}\left(x\right)$ are integers. When $x={\displaystyle \frac{p}{q}}$ with p and q nonzero integers, we need to multiply $q^n$ on both sides of $A_{[n,n]}\left(x\right)e^x-B_{[n,n]}\left(x\right)=R\left(x\right)$ and we still have the following inequality for n large enough

$$\left(q^n{R}_{[n,n]}\left(x\right)\right)<{\displaystyle \frac{1}{{\left(q^n{A}_{[n,n]}\left(x\right)\right)}^{1-ϵ}}}$$

with any $ϵ>0$ fixed. So Lemma 1 implies that the irrationality measure of $\mu \left(e^x\right)=2$. □

More generally, the precise estimates above helped Davis to obtain the following.

Theorem 2

(Davis [16]). For any $ϵ>0$, there exists an infinite sequence of rational numbers ${\displaystyle \frac{p}{q}}$ such that

$$\left|e-{\displaystyle \frac{p}{q}}\right|<\left({\displaystyle \frac{1}{2}}+\epsilon \right){\displaystyle \frac{lnlnq}{q^{2}lnq}}$$

The constant ${\displaystyle \frac{1}{2}}$ is not improvable.

4. Partial Padé Approximation

In this section, we study the Partial Padé approximation to a special function defined by:

$$f\left(x\right)=\sum _{k=0}^{\infty }{\displaystyle \frac{x^k}{k!(bk+s)(bk+s+1)\cdots (bk+t)}}$$

where $b,s,t$ are positive integers, $gcd(b,t)=1$. We use a similar method of explicitly constructing the Padé approximants, and we generalize the “mod-p” proof for the non-vanishing nature of the remainder terms as Maier [1], to complete the proof of irrationality for this type of function. Two special choices of parameters $b,s,t$ give two different types of confluent hypergeometric functions. Notice that when $b=1$,

$$f\left(x\right)={\displaystyle \frac{1}{s(s+1)\cdots t}}{}_{1}F_1\left[{\displaystyle \genfrac{}{}{0.0pt}{}{s}{t+1}};x\right];$$

When $t=s$,

$$f\left(x\right)={\displaystyle \frac{1}{b}}{}_{1}F_1\left[{\displaystyle \genfrac{}{}{0.0pt}{}{\frac{t}{b}}{\frac{t}{b}+1}};x\right]-{\displaystyle \frac{1}{b}}+{\displaystyle \frac{1}{t}}.$$

We will first briefly introduce similar work by Maier. Historically, in Maier’s original paper [1], he provided the proof for the irrationality of a similar function $\zeta \left(q\right)$:

$$\zeta \left(q\right)=\sum _{k=1}^{\infty }{\displaystyle \frac{q^k}{k!k}}$$

His construction of Padé approximants was complicated as it involves differential operators of two variables (see [1]). With Lemma 2, however, we can directly construct an explicit form of Padé approximants for this kind of function.

We will first construct a general form of Padé approximants to $f\left(x\right)$:

$$A\left(x\right)f\left(x\right)-B\left(x\right)=R\left(x\right)$$

Unlike the case for exponential functions, the coefficients in the exact Padé approximation are no longer integers, and the least common multiples of denominators may grow rapidly. The vanishing order of the remainder terms is not enough to beat the growth rate to apply Lemma 1 to bound the irrationality measure. So the basic idea is to sacrifice the vanishing order of $R\left(x\right)$ to compensate the integrality of the coefficients in $A\left(x\right)$ and $B\left(x\right)$.

Proposition 4.

Let $m,n$ be two positive integers such that $n\ge (b+1)m-1$, and

$$d=\mathrm{lcm}\{{\Pi }_{i=s}^{t}i,{\Pi }_{i=s}^t(b+i),{\Pi }_{i=s}^t(2b+i),\cdots ,{\Pi }_{i=s}^t(nb+i)\}$$

the least common multiple of those numbers inside. We define polynomials $A_{[n,m]}\left(x\right),B_{[n,m]}\left(x\right)\in \mathbb{Z}\left[x\right]$ by

$$A_{[n,m]}\left(x\right)=d\sum _{i=0}^n{(-1)}^i\left({\displaystyle \genfrac{}{}{0pt}{}{n}{i}}\right){\displaystyle \frac{(m+i)!}{m!}}\left({\displaystyle \genfrac{}{}{0pt}{}{b(m+i)+t}{bi+s}}\right)x^{n-i}$$

$$\begin{array}{ccc}\hfill B_{[n,m]}\left(x\right)& =& \sum _{k=0}^n\sum _{i=max\{n-k,0\}}^n{(-1)}^i\left({\displaystyle \genfrac{}{}{0pt}{}{n}{i}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{m+i}{i}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{b(m+i)+t}{bm+t-s}}\right)\hfill \\ & & \times {\displaystyle \frac{i!}{(k-n+i)!}}{\displaystyle \frac{d}{\left(b\right(k-n+i)+s)\cdots \left(b\right(k-n+i)+t)}}{x}^k\hfill \end{array}$$

Then $R_{[n,m]}\left(x\right)=A_{[n,m]}\left(x\right)f\left(x\right)-B_{[n,m]}\left(x\right)$ is $O\left(x^{m+n+1}\right)$.

Proof. 

Similar to the exponential function, we have

$$\begin{array}{ccc}\hfill A_{[n,m]}\left(x\right)f\left(x\right)& =& d\sum _{k=0}^{\infty }\sum _{i=0}^n{(-1)}^i\left({\displaystyle \genfrac{}{}{0pt}{}{n}{i}}\right){\displaystyle \frac{(m+i)!}{m!}}\left({\displaystyle \genfrac{}{}{0pt}{}{b(m+i)+t}{bi+s}}\right)\hfill \\ & & \times {\displaystyle \frac{1}{k!(bk+s)(bk+s+1)\cdots (bk+t)}}{x}^{n-i+k}\hfill \\ & =& d\sum _{k=0}^{\infty }a(n,k)x^k,\hfill \end{array}$$

where

$$\begin{array}{ccc}\hfill a(n,k)& =& \sum _{i=max\{n-k,0\}}^n{(-1)}^i\left({\displaystyle \genfrac{}{}{0pt}{}{n}{i}}\right){\displaystyle \frac{(m+i)!}{m!(k-n+i)!}}\left({\displaystyle \genfrac{}{}{0pt}{}{bi+bm+t}{bi+s}}\right)\hfill \\ & & ·{\displaystyle \frac{1}{\left(b\right(k-n+i)+s)\left(b\right(k-n+i)+s+1)\cdots \left(b\right(k-n+i)+t)}}.\hfill \end{array}$$

When $n+1\le k\le m+n$, we have

$$\begin{array}{ccc}\hfill a(n,k)& =& {\displaystyle \frac{1}{m!(bm+t-s)!}}\sum _{i=0}^n{(-1)}^i\left({\displaystyle \genfrac{}{}{0pt}{}{n}{i}}\right){\displaystyle \frac{(m+i)!}{(k-n+i)!}}\hfill \\ & & ·{\displaystyle \frac{(bi+bm+t)(bi+bm+t-1)\cdots (bi+s+1)}{\left(b\right(k-n+i)+s)\left(b\right(k-n+i)+s+1)\cdots \left(b\right(k-n+i)+t)}}.\hfill \end{array}$$

The term ${\displaystyle \frac{(m+i)!}{(k-n+i)!}}$ is a degree-$(m+n-k)$ polynomial of i and

$${\displaystyle \frac{(bi+bm+t)(bi+bm+t-1)\cdots (bi+s+1)}{\left(b\right(k-n+i)+s)\left(b\right(k-n+i)+s+1)\cdots \left(b\right(k-n+i)+t)}}$$

is a degree-$(bm-1)$ polynomial of i. So the total degree of i is $(b+1)m-1+n-k$, which is less than n. So lemma 2 implies $a(n,k)=0$ for all $n+1\le k\le m+n$. Let

$$B_{[n,m]}\left(x\right)=d\sum _{k=0}^{n}a(n,k)x^k.$$

Next we check $B_{[n,m]}\left(x\right)\in \mathbb{Z}\left[x\right]$. The coefficients of $B_{[n,m]}\left(x\right)$ are

$$\begin{array}{ccc}\hfill da(n,k)& =& \sum _{i=n-k}^n{(-1)}^i\left({\displaystyle \genfrac{}{}{0pt}{}{n}{i}}\right){\displaystyle \frac{(m+i)!}{m!(k-n+i)!}}\left({\displaystyle \genfrac{}{}{0pt}{}{bi+bm+t}{bi+s}}\right)\hfill \\ & & ·{\displaystyle \frac{d}{\left(b\right(k-n+i)+s)\left(b\right(k-n+i)+s+1)\cdots \left(b\right(k-n+i)+t)}}.\hfill \end{array}$$

Since $k\le n$, we have

$${\displaystyle \frac{(m+i)!}{m!(k-n+i)!}}\in \mathbb{Z}.$$

From the choice of d, we have

$${\displaystyle \frac{d}{\left(b\right(k-n+i)+s)\left(b\right(k-n+i)+s+1)\cdots \left(b\right(k-n+i)+t)}}\in \mathbb{Z}.$$

So $da(n,k)\in \mathbb{Z}$ for $0\le k\le n$.

In conclusion, we have $A_{[n,m]}\left(x\right),B_{[n,m]}\left(x\right)\ne 0\in \mathbb{Z}\left[x\right]$ and the remainder term

$$R_{[n,m]}\left(x\right)=d\sum _{k=m+n+1}a(n,k)x^k$$

is $O\left(x^{m+n+1}\right)$. □

More explicitly, the remainder term has the following form

$$\begin{array}{c}\hfill R_{[n,m]}\left(x\right)={\displaystyle \frac{x^{m+n+1}}{m!}}\sum _{k=0}^{\infty }\sum _{i=0}^n{(-1)}^i\left({\displaystyle \genfrac{}{}{0pt}{}{n}{i}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{b(m+i)+t}{bm+t-s}}\right)\end{array}$$

(4)

$$\begin{array}{c}\hfill {\displaystyle \frac{(m+i)!}{(k+m+i)!}}{\displaystyle \frac{d}{\left(b\right(k+m+i)+s)\cdots \left(b\right(k+m+i)+t)}}{x}^k\end{array}$$

(5)

5. Nonvanishing of Remainder Term

In the irrationality proof of $e^x$ for $x\in \mathbb{Q}-\left\{0\right\}$, the remainder term is nonzero by the integral formula (3). In our construction of partial Padé approximants, the remainder term does not have a convenient integration formula to show the non-vanishing nature of the remainder term. In this section, we apply a mod-p argument for infinitely many primes p similar to Maier [1].

In the following discussion, we assume $b\ge 2$. Let $w_0$ be the minimal integer such that $s\le w_0\le t$ and $gcd(w_0,b)=1$. In addition, we let

• $w=w_0$, if $w_0=s$;
• $w=w_0-b$, if $w_0>s$.

From the choice of $w_0$ and w, we have $w\le s\le w_0$. Since $gcd(b,w)=1$, there are infinitely many prime numbers p such that ${\displaystyle \frac{p-w}{b}}$ is an integer, by the Dirichlet prime number theorem. We choose such prime numbers p and let

$$m={\displaystyle \frac{p-w}{b}}-a,n=p+{\displaystyle \frac{p-w}{b}}.$$

(6)

where a is a fixed positive integer such that $ba>t-2w$.

Let

$$v_p:{\mathbb{Q}}^{\times }\to \mathbb{Z}$$

be the p-adic valuation of nonzero rational numbers at prime number p. We use the convention that $v_p\left(0\right)=+\infty$ for convenience. We have the following results for the p-adic valuation of d.

Lemma 3.

Under the choice of $m,n,p$ in (6), when p is large enough, the p-adic valuation of integer d is $v_p\left(d\right)=1$.

Proof. 

From the choice of p, we have $b(m+a)+w=p$. For a large enough p, there exists $k=m+a$ or $m+a-1$ such that $0\le k\le n$. We have $v_p(bk+w_0)=1$ and $bk+w_0\mid d$. So $v_p\left(d\right)\ge 1$. Since $bn+t<p^2$, so $v_p\left(d\right)<2$ and $v_p\left(d\right)=1$. □

We need Lucas’s theorem about the results of the binomial coefficient modulo p.

Theorem 3

(Lucas [17]). Let p be a prime number. Assume $y={\sum }_{i=0}^k{y}_i{p}^i$ and $z={\sum }_{i=0}^k{z}_i{p}^i$ are base p expansions of non-negative integers y and z, respectively. Then

$$\left({\displaystyle \genfrac{}{}{0pt}{}{y}{z}}\right)=\prod _{i=0}^k\left({\displaystyle \genfrac{}{}{0pt}{}{y_i}{z_i}}\right)modp$$

with the convention that $\left({\displaystyle \genfrac{}{}{0pt}{}{y_i}{z_i}}\right)=0$ if $y_i<z_i$.

Definition 2.

We use the (rising) Pochhammer symbol defined by

$${\left(q\right)}_k=q(q+1)\cdots (q+n-1).$$

We have the following proposition on the non-vanishing of $B_{[n,m]}\left(x_0\right)$ modulo p for nonzero rational numbers $x_0$.

Proposition 5.

Assume $b\ge 2$. Let $c=\left[{\displaystyle \frac{t-w}{b}}\right]$ and $g\left(x\right)={\sum }_{K=0}^c{g}_{c-K}{x}^K\in \mathbb{Q}\left[x\right]$ be a nonzero degree-c rational polynomial defined by

$$g_K={(-1)}^{K+\delta }{({\displaystyle \frac{w}{b}})}_{K+\delta }·{\displaystyle \frac{1}{(s-w_0-bK)(s+1-w_0-bK)\cdots (-1)·1\cdots (t-w-bK)}},$$

where $\delta =0$ if $w_0=w$ and $\delta =1$ if $w_0\ne w$. Let $x_0$ be a nonzero rational number that $g\left(x_0\right)\ne 0$. Then, for a large enough p such that $b\mid p-w$ and $m,n$ are chosen in (6), the rational number $B_{[n,m]}\left(x_0\right)\ne 0$ modulo p, or in other words, $B\left(x_0\right)\ne 0$ and $v_p\left(B\left(x_0\right)\right)=0$.

Proof. 

We choose p large enough so that $v_p\left(x_0\right)=0$. We show that most of the coefficients of $B\left(x\right)$ are multiples of p. The coefficient $da(n,k)$ is the product of the following integer factors

$$\begin{array}{c}\hfill {(-1)}^i\left({\displaystyle \genfrac{}{}{0pt}{}{n}{i}}\right),\left({\displaystyle \genfrac{}{}{0pt}{}{m+i}{m}}\right),{\displaystyle \frac{i!}{(k-n+i)!}},\left({\displaystyle \genfrac{}{}{0pt}{}{bi+bm+t}{bm+t-s}}\right),\\ \hfill {\displaystyle \frac{d}{\left(b\right(k-n+i)+s)\left(b\right(k-n+i)+s+1)\cdots \left(b\right(k-n+i)+t)}}\end{array}$$

Now we assume integer $da(n,k)\ne 0$ modulo p, in other words, each of the factors is not zero modulo p, and look for all the possible pairs $k,i$. If the last factor is not zero modulo p, then we have

$$b(k-n+i)+f=lp$$

for positive integers $k,i,f,l$ such that $k\le n,n-k\le i\le n$ and $s\le f\le t$. Since

$$k-n+i\le n\le (1+{\displaystyle \frac{1}{b}})p.$$

So $l\le b+1$. We consider the following two cases:

• If $1\le l\le b$, then $b(k-n+i)+f\le bp$, so $k-n+i<p$. Since the third factor

  $${\displaystyle \frac{i!}{(k-n+i)!}}\ne 0$$

  modulo p, we have $i<p$. So, the first factor $\left({\displaystyle \genfrac{}{}{0pt}{}{n}{i}}\right)\ne 0$ modulo p implies $n-i\ge p$. So $i\le {\displaystyle \frac{p-w}{b}}$. In this case, $k-n+i\le {\displaystyle \frac{p-w}{b}}$ and $b(k-n+i)+f\le p-w+f$. So $l=1$ and $i\ge {\displaystyle \frac{p-t}{b}}$. The second factor $\left({\displaystyle \genfrac{}{}{0pt}{}{m+i}{m}}\right)\ne 0$ modulo p because $m+i<p$. Consider the fourth factor $\left({\displaystyle \genfrac{}{}{0pt}{}{bi+bm+t}{bm+t-s}}\right)$. Since $bm+t-s<p$ from the choice of a, and $bi+bm+t>p$, we have $bi+bm+t-p\ge bm+t-s$ from Theorem 3. So $bi+s\ge p$, in other words, $i\ge {\displaystyle \frac{p-s}{b}}$. When $s=w_0$, then ${\displaystyle \frac{p-i}{s}}$ is an integer and $w=s$ and $i={\displaystyle \frac{p-s}{b}}$. When $s<w_0$, then ${\displaystyle \frac{p-s}{b}}$ is not an integer for large p, and ${\displaystyle \frac{p-w_0}{b}}<{\displaystyle \frac{p-s}{b}}\le i\le {\displaystyle \frac{p-w}{b}}$. So in either case, we have $i={\displaystyle \frac{p-w}{b}}$. Plugging this into $b(k-n+i)+f=p$, we have

  $$(k,f)=(n-w_0+w,w_0),(n-w_0+w-1,w_0+b),\cdots ,(n-c-w_0+w,w_0+bc),$$

  where $c=\left[{\displaystyle \frac{t-w_0}{b}}\right]$.
• If $l=b+1$, then we have $b(k-n+i)+f=(b+1)p$. So $k-n+i\ge p$ and the third factor is nonzero modulo p. Since $p\le i\le 2p$, we have $\left({\displaystyle \genfrac{}{}{0pt}{}{n}{i}}\right)\equiv \left({\displaystyle \genfrac{}{}{0pt}{}{n}{n-i}}\right)\equiv \left({\displaystyle \genfrac{}{}{0pt}{}{n-p}{n-i}}\right)\equiv \left({\displaystyle \genfrac{}{}{0pt}{}{n}{i-p}}\right)modp$. So the first factor being nonzero modulo p implies that $n-(i-p)\ge 0$. Notice that if we replace i by $i-p$, then all the binomial factors are the same modulo p. So the argument in the first case implies that $i=p+{\displaystyle \frac{p-w}{b}}$ and

  $$(k,f)=(n-w_0+w,w_0),(n-w_0+w-1,w_0+b),\cdots ,(n-c-w_0+w,w_0+bc),$$

  where $c=\left[{\displaystyle \frac{t-w_0}{b}}\right]$.

Denote by $u={\displaystyle \frac{p-w}{b}}$. In the two cases above, we have

$$\left({\displaystyle \genfrac{}{}{0pt}{}{n}{i}}\right)\equiv 1,\left({\displaystyle \genfrac{}{}{0pt}{}{m+i}{i}}\right)\equiv \left({\displaystyle \genfrac{}{}{0pt}{}{2u-a}{u}}\right),\left({\displaystyle \genfrac{}{}{0pt}{}{bi+bm+t}{bm+t-s}}\right)\equiv \left({\displaystyle \genfrac{}{}{0pt}{}{p+t-ba-2w}{p+t-ba-s-w}}\right)modp,$$

and the product of the three terms is independent of k. We denote the product by M and it is nonzero modulo p. Let $K={\displaystyle \frac{f-w_0}{b}}$, then $n-k=K+\delta$, and third factor is

$${\displaystyle \frac{i!}{(k-n+i)!}}\equiv (-{\displaystyle \frac{w}{b}})(-{\displaystyle \frac{w}{b}}-1)\cdots (-{\displaystyle \frac{w}{b}}+k-n+1)={(-1)}^{K+\delta }{({\displaystyle \frac{w}{b}})}_{K+\delta }modp$$

The right-hand side only depends on $K=n-k$ and we denote it by $C\left(K\right)$. The last factor is congruent to

$${\displaystyle \frac{d}{lp}}·{\displaystyle \frac{1}{(s-f)(s+1-f)\cdots \widehat{(f-f)}\cdots (t-f)}}$$

We denote the factor $D\left(K\right)={\displaystyle \frac{1}{(s-f)(s+1-f)\cdots \widehat{(f-f)}\cdots (t-f)}}$ since it also depends on $K=n-k$. In conclusion, we have.

$$\begin{array}{c}\hfill B\left(x_0\right)\equiv {(-1)}^u{\displaystyle \frac{d}{p}}(1-{\displaystyle \frac{1}{b+1}})M{x}_0^{n-\delta }\sum _{K=0}^{c}C\left(K\right)D\left(K\right)x_0^{-K}modp\end{array}$$

(7)

The summation on the right-hand side is equal to

$$\sum _{K=0}^{c}C\left(K\right)D\left(K\right)x_0^{-K}=x_0^{-c}g\left(x_0\right).$$

Since $g\left(x_0\right)\ne 0$, this term is not equal to zero modulo p when p is large enough. So we have $B_{[n,m]}\left(x_0\right)\ne 0$ modulo large prime number p. □

In conclusion, when $b\ge 2$, we have infinitely prime numbers p such that

$$v_p\left(A_{[m,n]}\left(x_0\right)\right)\ge 1\mathrm{and}{v}_p\left(B_{[m,n]}\left(x_0\right)\right)=0,$$

as long as $x_0\in \mathbb{Q}-\left\{0\right\}$ is not a root of polynomial $g\left(x\right)$. If $f\left(x_0\right)$ is a rational number, this implies that $R_{[m,n]}\left(x_0\right)\ne 0$. Next we show that $R_{[m,n]}\left(x_0\right)\to 0$ as $p\to +\infty$. Then, it contradicts $f\left(x_0\right)$ being rational. In fact, we show more about the growth rate of $A_{[m,n]}\left(x_0\right)$ and $R_{[m,n]}\left(x_0\right)$ to obtain bounds of the irrationality measure of $f\left(x_0\right)$.

6. Estimates of Growth Rate

In this section, we carry out the estimate of $A_{[m,n]}\left(x\right)$ and $R_{[m,n]}\left(x\right)$. First we fix $a_0$ a positive integer. Let $n=(b+1)m+a_0$ and m go to infinity.

Lemma 4

(Estimate of d). There is a positive number C such that $d\le C^n$ for $n=(b+1)m+a_0$ large enough.

Proof. 

Let $ϵ>0$ be any positive real number. From the prime number theorem, it is well known that $\mathrm{lcm}\{1,2,\cdots ,k\}\le e^{k(1+ϵ)}$ for k large enough; for example, see [4] (Footnote 5). From the definition of d in Proposition 4, we have

$$d\mid {\left(\mathrm{lcm}\{1,\cdots ,bn+t\}\right)}^{t-s+1}.$$

So $d\le e^{(t-s+1)(bn+t)(1+ϵ)}$. In conclusion, there is $C>0$ such that $d\le C^n$ for large n. □

Lemma 5

(Estimate of $R_{[m,n]}\left(x_0\right)$). For any $ϵ>0$ and $x_0\in \mathbb{Q}$, the following inequality

$$|R_{[m,n]}\left(x_0\right)|<{\left({\displaystyle \frac{1}{m!}}\right)}^{(1-ϵ)},$$

holds for $n=(b+1)m+a_0$ large enough.

Proof. 

Recall the expression of $R_{[m,n]}\left(x\right)$,

$$\begin{array}{c}\hfill R_{[n,m]}\left(x\right)={\displaystyle \frac{x^{m+n+1}}{m!}}\sum _{k=0}^{\infty }\sum _{i=0}^n{(-1)}^i\left({\displaystyle \genfrac{}{}{0pt}{}{n}{i}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{b(m+i)+t}{bm+t-s}}\right)\\ \hfill {\displaystyle \frac{(m+i)!}{(k+m+i)!}}{\displaystyle \frac{d}{\left(b\right(k+m+i)+s)\cdots \left(b\right(k+m+i)+t)}}{x}^k.\end{array}$$

Notice that the binomial numbers have bounds

$$\left({\displaystyle \genfrac{}{}{0pt}{}{n}{i}}\right)\le 2^n$$

and

$${\displaystyle \frac{(m+i)!}{(k+m+i)!}}\le {\displaystyle \frac{1}{k!}}.$$

So

$$R_{[n,m]}\left(x_0\right)\le {\displaystyle \frac{C^m}{m!}}$$

for some fixed positive number C. So the lemma is proved. □

Lemma 6

(Estimate of $A_{[m,n]}\left(x_0\right)$). For any $ϵ>0$, the following inequality

$${\left({\displaystyle \frac{(n+m)!}{m!}}\right)}^{1-ϵ}<\left|A_{[m,n]}\left(x\right)\right|<{\left({\displaystyle \frac{(n+m)!}{m!}}\right)}^{1+ϵ}$$

holds for $n=(b+1)m+a_0$ large enough.

Proof. 

We rewrite $|A_{[m,n]}\left(x\right)|$ as

$$|A_{[m,n]}\left(x\right)|=\left|d\sum _{i=0}^n{(-1)}^i\left({\displaystyle \genfrac{}{}{0pt}{}{n}{i}}\right){\displaystyle \frac{(m+i)!}{m!}}\left({\displaystyle \genfrac{}{}{0pt}{}{b(m+i)+t}{bm+t-s}}\right)x^{n-i}\right|$$

$$=d\left|{\displaystyle \frac{(m+n)!}{m!}}\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{(m+n-k)!}{(m+n)!}}\left({\displaystyle \genfrac{}{}{0pt}{}{bm+bn-bk+t}{bm+t-s}}\right){(-x)}^k\right|$$

For a fixed non-negative integer k, we have

$$\begin{array}{ccc}& & \underset{m\to \infty }{lim}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{(m+n-k)!}{(m+n)!}}\hfill \\ & =& {\displaystyle \frac{1}{k!}}\underset{m\to \infty }{lim}{\displaystyle \frac{n}{m+n}}{\displaystyle \frac{n-1}{m+n-1}}\cdots {\displaystyle \frac{n-k+1}{m+n-k+1}}\hfill \\ & =& {\displaystyle \frac{1}{k!}}{({\displaystyle \frac{b+1}{b+2}})}^k\hfill \end{array}$$

Next we study the behaviour of

$$\left({\displaystyle \genfrac{}{}{0pt}{}{bm+bn-bk+t}{bm+t-s}}\right)$$

as $m\to +\infty$ for each fixed k. We rewrite it as

$$\left({\displaystyle \genfrac{}{}{0pt}{}{b_{1}m+a_1}{b_{2}m+a_2}}\right)={\displaystyle \frac{(b_{1}m+a_1)!}{\left((b_1-b_2)m+a_1-a_2\right)!(b_{2}m+a_2)!}}$$

where $b_1=b(b+2)$, $b_2=b$, $a_1=-bk+a_3$ and $a_2,a_3$ are constants independent of k and m. By Stirling’s formula, we have

$$\begin{array}{ccc}& & ln\left({\displaystyle \genfrac{}{}{0pt}{}{b_{1}m+a_1}{b_{2}m+a_2}}\right)\hfill \\ & =& (b_{1}m+a_1)ln(b_{1}m+a_1)-\left((b_1-b_2)m+a_1\right)ln\left((b_1-b_2)m+a_1\right)\hfill \\ & & -(b_{2}m+a_2)ln(b_{2}m+a_2)+{\displaystyle \frac{1}{2}}ln{\displaystyle \frac{(b_{1}m+a_1)}{2\pi \left((b_1-b_2)m+a_1-a_2\right)(b_{2}m+a_2)}}+o\left(1\right)\hfill \end{array}$$

Apply the expansion of $ln(um+v)$

$$ln(um+v)=lnm+lnu+{\displaystyle \frac{v}{u}}{\displaystyle \frac{1}{m}}-{\displaystyle \frac{v^2}{u^2}}{\displaystyle \frac{1}{m^2}}+o({\displaystyle \frac{1}{m^2}}).$$

Then we obtain

$$ln\left({\displaystyle \genfrac{}{}{0pt}{}{b_{1}m+a_1}{b_{2}m+a_2}}\right)=c_{1}m+c_{2}lnm+c_3+o\left(1\right)$$

where constants $c_1,c_2,c_3$ do not depend on $m,k$.

In other words, there are constants $d_1,d_2,d_3$ such that

$$\underset{m\to \infty }{lim}\left({\displaystyle \genfrac{}{}{0pt}{}{bm+bn-bk+t}{bm+t-s}}\right)/d_1{e}^{d_{2}m}{m}^{d_3}=1$$

Applying Tannery’s theorem to the summation formula of $A_{[m,n]}\left(x\right)/(d{d}_1{\displaystyle \frac{(m+n)!}{m!}}{e}^{d_{2}m}{m}^{d_3})$, we have

$$\underset{m\to \infty }{lim}{A}_{[m,n]}\left(x\right)/(d{d}_1{\displaystyle \frac{(m+n)!}{m!}}{e}^{d_{2}m}{m}^{d_3})=e^{-x}.$$

Applying Stirling’s formula to the denominator, we obtain the growth rate estimate of $A_{[m,n]}\left(x_0\right)$. □

Proof of Theorem 1. 

Firstly, from Lemma 5, we know that $R_{[m,n]}\left(x_0\right)\to 0$ as $m\to +\infty$. So by the non-vanishing result of $R_{[m,n]}\left(x_0\right)$, we obtain the irrationality of $f\left(x_0\right)$.

By Lemmas 5 and 6, we have that for any positive number $ϵ$,

$$|R_{(m,n)}\left(x\right)|<{\displaystyle \frac{C}{{|A_{(m,n)}\left(x\right)|}^{\frac{1-ϵ}{b+1}}}}$$

for $n=(b+1)m+a_0$ large enough. So Lemma 1 implies that

$$\mu \left(f\left(x_0\right)\right)\le b+2.$$

□

7. Special Case $\mathit{s}=\mathit{t}$

When $s=t$ and $b\ge 2$, then the polynomial $g\left(x\right)$ in Proposition 5 is a nonzero constant. This implies that $f\left(x\right)$ is irrational for all nonzero rational numbers x. Recall that under this assumption, function $f\left(x\right)$ has the form

$$f\left(x\right)={\displaystyle \frac{1}{b}}{}_{1}F_1\left[{\displaystyle \genfrac{}{}{0.0pt}{}{\frac{t}{b}}{\frac{t}{b}+1}};x\right]-{\displaystyle \frac{1}{b}}+{\displaystyle \frac{1}{t}}.$$

This applies to irrational values for confluent hypergeometric functions of the following form

$${}_{1}F_1\left[{\displaystyle \genfrac{}{}{0.0pt}{}{\alpha }{\alpha +1}};x\right]$$

for $\alpha \in {\mathbb{Q}}_{>0}-\mathbb{Z}$ and $x\in \mathbb{Q}-\left\{0\right\}$. We use Humber’s symbol to denote this function $\Phi (\alpha ,\alpha +1;x)$. Then, the derivative of this function is

$${\displaystyle \frac{d}{dx}}\Phi (\alpha ,\alpha +1;x)={\displaystyle \frac{\alpha }{\alpha +1}}\Phi (\alpha +1,\alpha +2;x)$$

and still has the same form. So we can conclude the irrationality of $f\left(x\right)$ and all its derivatives at nonzero rational x.

8. Rational Values and Differential Systems

In this section, we discuss the values of the function f for special choices of $b,s,t$. In particular, we show that the theorem for $b=1$ fails. We further discuss the differential systems for general $f\left(x\right)$.

Let $b=1,\hspace{0.17em}s=2,\hspace{0.17em}t=2$. Then, the corresponding function

$$f\left(x\right)={\displaystyle \frac{1}{2}}{}_{1}F_1\left[{\displaystyle \genfrac{}{}{0.0pt}{}{2}{3}};x\right]={\displaystyle \frac{e^x(x-1)+1}{x^2}}.$$

So it takes a rational value for $x=1$. On the other hand, this value is not the root of $g\left(x\right)$ in Proposition 5. More generally, we obtain the following formula for $f\left(x\right)$ when $b=1,$ $s=2,\hspace{0.17em}t\ge 2$

$$f\left(x\right)={\displaystyle \frac{1}{s\cdots t}}{}_{1}F_1\left[{\displaystyle \genfrac{}{}{0.0pt}{}{s}{t+1}};x\right]={\displaystyle \frac{1}{x^t}}(e^x(x-t+1)-P\left(x\right))$$

where $P\left(x\right)$ is the Taylor expansion of $e^x(x-t+1)$ up to the term $x^{t-1}$. So $f\left(x_0\right)$ is rational when $x_0=t-1$. So in general, the irrationality result fails for general nonzero rational x when $b=1$. Next we will see that this is not the singular points of the differential system appearing in Siegel–Shidlovskii’s theorem.

It is well known that $y={}_{1}F_1\left[{\displaystyle \genfrac{}{}{0.0pt}{}{\alpha }{\beta }};x\right]$ satisfies a second-order differential equation

$$x{\displaystyle \frac{d^2}{d{x}^2}}y+(\beta -x){\displaystyle \frac{d}{dx}}y-\alpha y=0.$$

For example, see [18] (Equation (2), Chapter 6.1). Then, the singular point of this equation is $x=0$. For more rational values of confluent hypergeometric functions, see [19] (Proposition 4.1).

In general, the differential system for $f\left(x\right)$ can be obtained as follows. We use notation $f(b,s,t;x)$ for the dependence of f on parameters $b,s,t$, and $f(b,s,s-1;x)=e^x$ by convention. Then, we have

$$(t+bx{\displaystyle \frac{d}{dx}})f(b,s,t;x)=f(b,s,t-1;x)$$

By induction, we find that $f(b,s,t;x)$ satisfies a linear ordinary differential equation of order $t-s+2$. The only possible singular point is $x=0$.

9. Related Results

In conclusion, we prove the irrationality of values for the special kind of generalized hypergeometric functions (see Theorem 1):

$$f\left(x\right)=\sum _{k=0}^{\infty }{\displaystyle \frac{x^k}{k!(bk+s)(bk+s+1)\cdots (bk+t)}},$$

if $x\in \mathbb{Q}-\left\{0\right\}$ is not a root of a rational polynomial depending on $b,s,t$; estimates of its irrationality measure can be found as follows:

$$2\le \mu \left(f\left(x_0\right)\right)\le b+2.$$

The mod-p method could be interesting to generalize to a p-adic version and obtain more information about rational approximations of $f\left(x\right)$.

After the paper was finished, we learned of a recent preprint by Fischler–Rivoal [20]. They claimed they had solved a long-standing problem in the theory of Type-E functions, which states that if a Type-E function has an irrational value at a rational point, then the irrationality measure of the value is 2. So their result combined with the irrationality part of our main theorem actually implies that the irrationality measure of those values of $f\left(x\right)$ is exactly 2, which greatly improves the irrationality measure estimate part of our main theorem.