MathDB

Original in Hungarian; translated with Google translate; polished by myself.
Prove that, the signs

\varepsilon_n = \pm 1

can be chosen such that the function

f(s) = \sum_{n = 1}^\infty\frac{\varepsilon_n}{n^s}\colon \{s\in\Bbb C:\operatorname{Re}s > 1\}\to \Bbb C

converges to every complex value at every point

\xi \in \{s\in\Bbb C:\operatorname{Re}s = 1\}

(i.e. for every

\xi \in \{s\in\Bbb C:\operatorname{Re}s = 1\}

and every

z \in \Bbb C

, there exists a sequence

s_n \to \xi

\operatorname{Re}s_n > 1

, for which

f(s_n) \to z

Problem Statement