Topology on the set $\mathbb N$, $U$ is open iff either $1\not\in U$ or else $\sum_{n\not\in U}\frac1n\lt\infty$

Question

Define a topology on the set $\mathbb N$ of all natural numbers by calling a set $U$ open if either $1\not\in U$ or else $\sum_{n\not\in U}\frac1n\lt\infty$. Take $A=\mathbb N\setminus\{1\}$. Then, show that there is no sequence with values in $A$ converges to $1$.

---

I find this question in here. (Loot at 'answer' part)

[My attempt]

Since ,for all $N \in \mathbb N$, $\{1,N,N+1,N+2,N+3,...\}$ is open set containing $1$, if $\{x_n\}_{n=1}^{\infty}$ is convergent sequence to $1$, then the range of $\{x_n\}_{n=1}^{\infty}$ is unbounded. So, what is next step?

Answer 1

Since $x_n$ is unbounded we can find a subsequence $x_{n_k}$ such that $x_{n_k}>k^{2}$ for all $k$. Let $U=\{1\}\cup (\{x_{n_1},x_{n_2},...\})^{c}$. Then $U$ is an open set containing $1$. Since $x_i \to 1$ we must have $x_i \in U$ for all $i$ sufficiently large but this is clearly false.