# Are the sums ∑∞n=11(n!)k\sum_{n=1}^{\infty} \frac{1}{(n!)^k} transcendental?

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.

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 $\le k$.

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

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,

On the other hand,

is a non-zero integer, hence $\ge1$ or $\le -1$. We conclude

or

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.