lobht98

Answered

2022-06-28

I am trying to find a conjecture apparently made by Erdős and Straus. I say apparently because I have had so much trouble finding anything information about it that I'm beginning to doubt its existence. Here it is:

Let $\varphi (X)$ be a rational function over $\mathbb{Q}$ that is defined at every positive integer. If the sum $\sum _{n=1}^{\mathrm{\infty}}\varphi (n)$ converges, it is either rational or transcendental, i.e., it is never an irrational algebraic number.

Has anyone heard of this conjecture? I was told about it by my supervisor, but he doesn't remember where he heard about it.

Let $\varphi (X)$ be a rational function over $\mathbb{Q}$ that is defined at every positive integer. If the sum $\sum _{n=1}^{\mathrm{\infty}}\varphi (n)$ converges, it is either rational or transcendental, i.e., it is never an irrational algebraic number.

Has anyone heard of this conjecture? I was told about it by my supervisor, but he doesn't remember where he heard about it.

Answer & Explanation

massetereqe

Expert

2022-06-29Added 21 answers

The introduction discusses variations on the following conjecture of Erdős, which apparently goes back to 1949 but did not appear in print in Erdős's intended form until a 1965 paper of Livingston.

Conjecture (Erdős). Let f be a number-theoretic function with period $q>0$ such that $|f(n)|=1$ for $1\le n<q$ and $f(q)=0$. Then

$\sum _{n=1}^{\mathrm{\infty}}\frac{f(n)}{n}\ne 0$

whenever the series converges.

In 1973, Baker, Birch, and Wirsing attribute the following problem to Chowla:

Problem. Does there exist a rational-valued function $f(n)$, periodic with prime period $p$, such that

$\sum _{n=1}^{\mathrm{\infty}}\frac{f(n)}{n}=0?$

Baker, Birch, and Wirsing disprove the claim by proving the following theorem.

Theorem: Suppose $f:\mathbb{Z}\to \overline{\mathbb{Q}}$ is a nonvanishing function with period $q$. If (i) $f(n)=0$ whenever $1<gcd(r,q)<q$, and (ii) the qth cyclotomic polynomial is irreducible over $\mathbb{Q}(f(1),\dots ,f(q))$, then

$\sum _{n=1}^{\mathrm{\infty}}\frac{f(n)}{n}\ne 0.$

This result and a related result of Okada are used as a basis for Adhikari, Saradha, Shorey, and Tijdeman to prove the following theorem.

Theorem. Suppose $f:\mathbb{Z}\to \overline{\mathbb{Q}}$ is periodic with period $q$. If the series

$\sum _{n=1}^{\mathrm{\infty}}\frac{f(n)}{n}$

converges to some number $S$, then either $S=0$ or $S$ is transcendental.

Applying Baker's theorem tends to lead to dichotomies of the form "S is either rational or transcendental"; beyond that I don't find direct motivation for the claim or prior conjectures of the form you indicate. (But I'm no number theorist.)

Conjecture (Erdős). Let f be a number-theoretic function with period $q>0$ such that $|f(n)|=1$ for $1\le n<q$ and $f(q)=0$. Then

$\sum _{n=1}^{\mathrm{\infty}}\frac{f(n)}{n}\ne 0$

whenever the series converges.

In 1973, Baker, Birch, and Wirsing attribute the following problem to Chowla:

Problem. Does there exist a rational-valued function $f(n)$, periodic with prime period $p$, such that

$\sum _{n=1}^{\mathrm{\infty}}\frac{f(n)}{n}=0?$

Baker, Birch, and Wirsing disprove the claim by proving the following theorem.

Theorem: Suppose $f:\mathbb{Z}\to \overline{\mathbb{Q}}$ is a nonvanishing function with period $q$. If (i) $f(n)=0$ whenever $1<gcd(r,q)<q$, and (ii) the qth cyclotomic polynomial is irreducible over $\mathbb{Q}(f(1),\dots ,f(q))$, then

$\sum _{n=1}^{\mathrm{\infty}}\frac{f(n)}{n}\ne 0.$

This result and a related result of Okada are used as a basis for Adhikari, Saradha, Shorey, and Tijdeman to prove the following theorem.

Theorem. Suppose $f:\mathbb{Z}\to \overline{\mathbb{Q}}$ is periodic with period $q$. If the series

$\sum _{n=1}^{\mathrm{\infty}}\frac{f(n)}{n}$

converges to some number $S$, then either $S=0$ or $S$ is transcendental.

Applying Baker's theorem tends to lead to dichotomies of the form "S is either rational or transcendental"; beyond that I don't find direct motivation for the claim or prior conjectures of the form you indicate. (But I'm no number theorist.)

Yahir Tucker

Expert

2022-06-30Added 8 answers

Assume that Erdos theorem is true. Then ${f}_{s}(n)=\frac{1}{{n}^{2s+1}}$, $s=1,2,\dots $ are clearly a rational functions and

$\sum _{n=1}^{\mathrm{\infty}}{f}_{s}(n)=\sum _{n=1}^{\mathrm{\infty}}\frac{1}{{n}^{2s+1}}=\zeta (2s+1),$

will be rationals or transcendentals, which I think is highly considerable statement. Hence I expect the proof to your answer must be quite hard.

Note also that rational function is $\frac{{n}^{2}-2}{{n}^{5}+1}$ but ${F}_{n}$ is not, where ${F}_{n}$ is the $n-$th Fibonacci number.

$\sum _{n=1}^{\mathrm{\infty}}{f}_{s}(n)=\sum _{n=1}^{\mathrm{\infty}}\frac{1}{{n}^{2s+1}}=\zeta (2s+1),$

will be rationals or transcendentals, which I think is highly considerable statement. Hence I expect the proof to your answer must be quite hard.

Note also that rational function is $\frac{{n}^{2}-2}{{n}^{5}+1}$ but ${F}_{n}$ is not, where ${F}_{n}$ is the $n-$th Fibonacci number.

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