Are the sums $\sum_{n=1}^{\infty} \frac{1}{(n!)^k}$ transcendental?

Question

This question is inspired by my answer to the question How to compute $\prod_{n=1}^\infty\left(1+\frac{1}{n!}\right)$? .

The sums $f(k) = \sum_{n=1}^{\infty} \frac{1}{(n!)^k}$ (for positive integer $k$) came up, and I noticed that $f(1) = e-1$ was transcendental and $f(2) = I_0(2)-1$ (modified Bessel function) was probably transcendental since $J_0(1)$ (Bessel function) is transcendental.

So, I made the conjecture that $all$ the $f(k)$ are transcendental, and I am here presenting it as a question.

The only progress I have made is to show that all the $f(k)$ are irrational.

This follows the standard proof that $e$ is irrational: if $f(k) = \frac{a}{b} =\sum_{n=1}^{\infty} \frac{1}{(n!)^k}$, multiplying by $(b!)^k$ gives $a (b!)^{k-1}(b-1)! =\sum_{n=1}^{b} \frac{(b!)^k}{(n!)^k} +\sum_{n=b+1}^{\infty} \frac{(b!)^k}{(n!)^k} $ and the left side is an integer and the right side is an integer plus a proper fraction (easily proved).

I have not been able to prove anything more, but it somehow seems to me that it should be possible to prove that $f(k)$ is not the root of a polynomial of degree $\le k$.

Answer 1

Write $$f(x) = \sum_{n=0}^{\infty}f_n \frac{x^n}{n!}$$ for the formal divided power series with coefficients in $\mathbb{Q}$ as well as the holomorphic function near $0$ it defines (if it defines one). We say that $f$ is

• holonomic if there exist nonzero polynomials $p_0(x),p_1(x),\ldots,p_M(x)$ with coefficients in $\mathbb{Q}$ such that $$\sum_{m=0}^Mp_m(x)f^{(m)}(x)=0\text{,}$$ and
• a Siegel $E$-function if $\lvert f_n \rvert$ and the lcm of the denominators of the first $n$ terms of $f_0,f_1,\ldots$ are $O(c^n)$ for some $c$.

With these definitions, consider in particular $$f(x) \stackrel{\triangle}{=}\sum_{n=0}^{\infty}\frac{(nk)!}{(n!)^{k}}\frac{x^{nk}}{(nk)!}\text{.}$$

• $f(x)$ is holonomic: making the substitution $x \to x^k$ in the generalized hypergeometric differential equation (DLMF 16.8.3) and expanding, one finds $$\sum_{j=1}^k\left\{\begin{array} & k\\ j\end{array}\right\}x^{j-1}f^{(j)}(x) = k^k x^{k-1}f$$
• $f(x)$ is an $E$-function; in fact it is a special case of the Siegel hypergeometric $E$-function $$S_{pq}(a_1,a_2,\ldots,a_p;b_1,b_2,\ldots,b_q;x) =\sum_{n=0}^{\infty}\frac{(a_1)_n(a_2)_n\cdots (a_p)_n ((q-p+1)n)!}{(b_1)_n(b_2)_n\cdots (b_q)_n}\frac{x^{(q-p+1)n}}{((q-p+1)n)!}\text{;}$$ cf. Siegel (1949, ch. II, sec. 9), Waldschmidt (2016).

Shidlovsky (1954; cited by Beukers and Wolfart (1988, p.69)) showed:

Suppose that $f(x)$ is a holonomic $E$-function that is not a polynomial. If $\xi p_M(\xi)\neq 0$ ($\xi \in \mathbb{C}$), then $\xi$ and $f(\xi)$ are not both algebraic.

Again, $p_M$ is the polynomial that is the coefficient of the leading term $f^{(M)}$ in the differential equation that $f$ satisfies (for our particular $f$, it's $x^{k-1}$). $\xi p_M(\xi)\neq 0$ means that for our particular $f$, we must only consider $x\neq 0$. And since $x$ is algebraic if and only if $x^{1/k}$ is, we conclude

If $x\in \bar{\mathbb{Q}}^*$ then $\sum_{n=0}^{\infty}\frac{x^{n}}{(n!)^k}$ is not algebraic.

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Beukers, F., and J. Wolfart. 1988. “Algebraic Values of Hypergeometric Functions.” In New Advances in Transcendence Theory, edited by Alan Baker, 1st ed., 68–81. Cambridge University Press.

Siegel, Carl Ludwig. 1949. Transcendental Numbers. Annals of Mathematics Studies 16. Princeton University Press.

Waldschmidt, Michel. 2016. “Irrationality and Transcendence of Values of Special Functions.” In Proceedings of the XVth Annual Conference of Society for Special Functions and Their Applications.

Answer 2

Suppose we have an irreducible polynomial $p(X)=a_dX^d+\ldots+a_1X+a_0\in\Bbb Z[X]$ with $p(\alpha)=0$ and $a_d\ne0$.

For each $N$, we can combine the first $N$ summands and find an integer $A_N$ such that $$\left|\alpha-\frac{A_N}{(N!)^k}\right| =\sum_{n>N}\frac 1{(n!)^k}<\frac 2{((N+1))!^k}.$$ For $x$ close enough to $\alpha$, we have $|p(x)|\le 2|a_d|\cdot|x-\alpha|^d$, hence for $N\gg 0$ and by the MWT, $$\left|p\left(\frac {A_n}{(N!)^k}\right)\right|\le\frac{2^{d+1}|a_d|}{((N+1)!)^{kd}}.$$ On the other hand, $$ (N!)^{kd}p\left(\frac {A_n}{(N!)^k}\right)$$ is a non-zero integer, hence $\ge1$ or $\le -1$. We conclude $$ \frac1{(N!)^{kd}}\le \frac{2^{d+1}|a_d|}{((N+1)!)^{kd}},$$ or $$ (N+1)^{k}\le2\sqrt[d]{|2a_d|}.$$ As this inequality cannot hold for infinitely many $N$, we arrive at a contradiction. We conclude that $p$ as above does not exist and so $\alpha$ is transcendental.