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Suppose a number $a=\sum_{r\in R,s\in S}r^{-s}$ where R is a subset of natural numbers with positive density and S is a subset of natural numbers of density 0. Is $a$ transcendental?

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I would guess that there is a missing $-$ sign, that $a = \sum r^{-s}$. – Douglas Zare Apr 11 '13 at 5:02
If so, this question asks for something stronger than the well-known open problem of whether $\zeta(2n+1)$ is transcendental. – Douglas Zare Apr 11 '13 at 5:05
Thanks. Yes, $a=\sum_{r\in R,s\in S}r^{-s}$. – Huichi Huang Apr 11 '13 at 12:31
up vote 7 down vote accepted

If you allow $S$ to be finite, then the answer is no: any real number $x\in(0,\frac{\pi^2}6-1)$ can be written as $\sum_{r\in R} r^{-2}$ for some set $R$ of positive integers with positive density.

To see this, first choose $m\ge2$ such that $m^2x > \frac{\pi^2}6$, and let $R_1 = m\mathbb N$, so that $\sum_{r\in R_1} r^{-2} = \frac{\pi^2}{6m^2} < x$. Then use the greedy algorithm: recursively define $n_k$ to be the least positive integer not in $R_1 \cup \lbrace n_1,\dots,n_{k-1}\rbrace$ such that $\sum_{j=1}^k n_j^{-2} < x - \frac{\pi^2}{6m^2}$. Setting $R_2 = \lbrace n_1,n_2,\dots \rbrace$ and $R=R_1\cup R_2$, we see that $\sum_{r\in R} r^{-2} = x$. One can show that $R_2$ has density 0: if $x_k = x - \frac{\pi^2}{6m^2} - \sum_{j=1}^k n_j^{-2}$, then the fact that $n_k^{-2} < x_{k-1} \le (n_k-1)^{-2}$ show that $x_k \le (n_k-1)^{-2} - n_k^{-2} < 2n_k^{-3} < 2x_{k-1}^{3/2}$, and so the $n_k$ form a very sparse set.

I believe a similar proof will show the following: given any subset $S$ of $\lbrace2,3,4,\dots\rbrace$, there exists $x_0(S)>0$ such that any real number $x\in(0,x_0(S))$ can be written as $\sum_{r\in R} \sum_{s\in S} r^{-s}$ for some set $R$ of positive density. (Basically, one replaces the function $t^{-2}$ in the previous proof with the function $\sum_{s\in S} t^{-s}$.)

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