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(Sorry for my poor english..)

I have a question in number theory. (Just my curiosity)

Let $f(z)\in \mathbb{Z}[x]$ be an irreducible polynomial with degree is larger than or equal to 2. Then is the number \begin{equation} \sum_{n=1}^{\infty} \frac{1}{f(n)} \end{equation} irrational? Or there is an irreducible polynomial $f(x)\in \mathbb{Z}[x]$ such that this number is rational?

  1. More generally, if for all $n\in \mathbb{N}$, $f(x)$ and $f(x+n)$ are relativlely prime in $\mathbb{Z}[x]$, then is it true..?

I also know that proving the irrationality is very difficult, for example $\zeta(2k+1)$...

I could not find references about this problem.. Has this problem been proved? Or, is there a name for this problem?

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    $\begingroup$ To give an example showing that this is probably a very deep question in general: when $f(n)=n^2+1$, the series evaluates to $\big( \pi(e^{2\pi}+1)/(e^{2\pi}-1) - 1\big)/2$ (says Mathematica); and I don't believe that number is known to be irrational...! $\endgroup$
    – Greg Martin
    Commented Sep 26, 2018 at 17:53
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    $\begingroup$ @GregMartin You would think so, but you would actually be wrong! Algebraic relations between $\pi$ and $e^\pi$, or rather lack thereof, are much better understood that those between $\pi$ and $e$ - indeed, they are algebraically independent, rendering that number (provably!) transcendental. See Wikipedia $\endgroup$
    – Wojowu
    Commented Sep 26, 2018 at 18:01
  • $\begingroup$ Is there any expectation that every absolutely convergent series of this form is an (exponential) period? Or a trick to show this? $\endgroup$
    – Marty
    Commented Sep 26, 2018 at 21:07
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    $\begingroup$ @Marty: For $f$ even and free of multiple roots, I think all is explained by the partial fraction expansion of the cotangent function. On all of the complex plane $\mathbb{C} \ni z$, we have $\pi \cot{\pi z = \lim_{N \to \infty}} \sum_{n = -N}^N \frac{1}{n+z}$. $\endgroup$
    – Vesselin Dimitrov
    Commented Sep 26, 2018 at 21:51
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    $\begingroup$ So $\sum_{n=-\infty}^\infty \frac{1}{f(n)}$ instead of $\sum_{n=1}^\infty \frac{1}{f(n)}$ @VesselinDimitrov $\endgroup$
    – reuns
    Commented Sep 26, 2018 at 23:22

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