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This afternoon I tried to read and understand some sections of the paper Some Applications of Diophantine Approximation by R. Tijdeman procedding from Number Theory for the Millenium III, A K Peters (2002).

I wondered about the transcendence of infinite sums that involve the Möbius function, denoted in this post as $\mu(n)$ for integers $n\geq 1$ (see if you want the corresponding Wikipedia Möbius function or the MathWorld encyclopedia that also is very good). My background on linear forms in logarithms isn't good, and neither I have a good intuition about these theories. I know that certain series involving the Möbius function have a good and deep mathematical content.

Question. My belief is that for each polynomial $Q(x)$ with integer coefficients, that is $Q(x)\in\mathbb{Z}[X]$, such that for all $x>1$ is positive $Q(x)>0$ with degree $\deg(Q)\geq 2$ one has that the sum of the series $$\sum_{n=1}^{\infty}\frac{\mu(n)}{Q(n)}$$ is transcendent. Can you find a counterexample or provide some reasoning about the veracity of the statement? Many thanks.

If it is very difficult to find the counterexample, then please add what work can be done to know what about the veracity of my belief. I hope that my question is interesting, and that you can provide some answer.

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    $\begingroup$ Many thanks upvoter, sincerely was a bad day! $\endgroup$ – user142929 Aug 21 '19 at 20:03
  • $\begingroup$ Your question doesn't make sense because $\sum_n \frac{\mu(n)}{Q(n)}$ is understood only for $Q(n) =n^k$. Even if that $f(x) = \sum_n \mu(n) e^{-n x}$ is the inverse Mellin transform of $\Gamma(s)/\zeta(s)$ means there is some little hope to say something about $g(a)=\int_0^\infty f(x) e^{-ax} dx= \sum_n \frac{\mu(n)}{n+a}$ (which does converge by the PNT), if $Q$ is separable then $\sum_n \frac{\mu(n)}{Q(n)}=\sum_j \frac{g(a_j) }{Q'(a_j)}$ $\endgroup$ – reuns Aug 27 '19 at 19:14
  • $\begingroup$ Many thanks for your remarks @reuns , my hope is that some user can to answer my question. Notice that you think this problem cannot be solved does not mean that other mathematicians can't solve it: I am asking about find a counterexample or provide reasonings about the veracity of my statement. $\endgroup$ – user142929 Aug 28 '19 at 5:43
  • $\begingroup$ No this is not how mathematics work. For most sequences the mathematics don't care if $\lim c(k)$ is rational algebraic or transcendent. Thus to hope say something about it you need to find alternate (simpler) expressions. For $c(k)=\sum_{n \le k} \mu(n)/Q(n)$ it is obvious there are not (because if there was one it would be useful in the theory of $\zeta(s)$) $\endgroup$ – reuns Aug 28 '19 at 17:36

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