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Found problems: 1

2022 Miklós Schweitzer, 8
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 $f(s_n) \to z$).
dirichlet series convergence algebra sequences