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