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)=e−1 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!)k−1(b−1)!=∑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+a0∈Z[X] with p(α)=0 and ad≠0.

For each N, we can combine the first N summands and find an integer AN such that

|α−AN(N!)k|=∑n>N1(n!)k<2((N+1))!k.

For x close enough to α, we have |p(x)|≤2|ad|⋅|x−α|d, hence for N≫0 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!)kd≤2d+1|ad|((N+1)!)kd,

or

(N+1)k≤2d√|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*