Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|cite|improve this question
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

1 Answer 1

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 $\sum_{s\in S} t^{-s}$.)

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.