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 ϕ ( X ) be a rational function over Q that is defined at every positive integer. If the sum ∑ n = 1 ∞ ϕ ( 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.

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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&#039;m beginning to doubt its existence. Here it is:Let   ϕ  (  X  ) be a rational function over       Q   that is defined at every positive integer. If the sum       ∑          n      =      1        ∞    ϕ  (  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&#039;t remember where he heard about it.

massetereqe · Accepted Answer

The introduction discusses variations on the following conjecture of Erdős, which apparently goes back to 1949 but did not appear in print in Erdős&#039;s intended form until a 1965 paper of Livingston.Conjecture (Erdős). Let f be a number-theoretic function with period   q  &amp;gt;  0 such that       |    f  (  n  )      |    =  1 for   1  ≤  n  &amp;lt;  q and   f  (  q  )  =  0. Then      ∑          n      =      1              ∞                  f      (      n      )        n    ≠  0whenever 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      ∑          n      =      1              ∞                  f      (      n      )        n    =  0  ?Baker, Birch, and Wirsing disprove the claim by proving the following theorem.Theorem: Suppose   f  :      Z    →            Q        ¯   is a nonvanishing function with period   q. If (i)   f  (  n  )  =  0 whenever   1  &amp;lt;  gcd  (  r  ,  q  )  &amp;lt;  q, and (ii) the qth cyclotomic polynomial is irreducible over       Q    (  f  (  1  )  ,  …  ,  f  (  q  )  ), then      ∑          n      =      1              ∞                  f      (      n      )        n    ≠  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  :      Z    →            Q        ¯   is periodic with period   q. If the series      ∑          n      =      1              ∞                  f      (      n      )        n  converges to some number   S, then either   S  =  0 or   S is transcendental.Applying Baker&#039;s theorem tends to lead to dichotomies of the form &quot;S is either rational or transcendental&quot;; beyond that I don&#039;t find direct motivation for the claim or prior conjectures of the form you indicate. (But I&#039;m no number theorist.)

Yahir Tucker · Accepted Answer

Assume that Erdos theorem is true. Then       f    s    (  n  )  =      1          n              2        s        +        1            ,   s  =  1  ,  2  ,  … are clearly a rational functions and      ∑          n      =      1              ∞            f    s    (  n  )  =      ∑          n      =      1              ∞            1          n              2        s        +        1              =  ζ  (  2  s  +  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                     n        2            −      2                      n        5            +      1       but       F    n   is not, where       F    n   is the   n  −th Fibonacci number.

I am trying to find a conjecture apparently made by Erdős and Straus. I say apparently because I hav

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