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 n=1(1+1n!)?“.

The sums
f(k)=n=11(n!)k
(for positive integer k)
came up,
and I noticed that
f(1)=e1 was transcendental
and f(2)=I0(2)1
(modified Bessel function)
was probably transcendental
since J0(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)=ab=n=11(n!)k,
multiplying by
(b!)k
gives
a(b!)k1(b1)!=bn=1(b!)k(n!)k+n=b+1(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 k.

Answer

Suppose we have an irreducible polynomial p(X)=adXd++a1X+a0Z[X] with p(α)=0 and ad0.

For each N, we can combine the first N summands and find an integer AN such that
|αAN(N!)k|=n>N1(n!)k<2((N+1))!k.
For x close enough to α, we have |p(x)|2|ad||xα|d, hence for N0 and by the MWT,
|p(An(N!)k)|2d+1|ad|((N+1)!)kd.
On the other hand,
(N!)kdp(An(N!)k)
is a non-zero integer, hence 1 or 1. We conclude
1(N!)kd2d+1|ad|((N+1)!)kd,
or
(N+1)k2d|2ad|.
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 α is transcendental.

Attribution
Source : Link , Question Author : marty cohen , Answer Author : Hagen von Eitzen

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