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?



Transcendency of  numbers of a special form. 
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 $\sum\_{s\in S} t^{-s}$.)


