Let $f$ be a continuous complex-valued function on $\mathbb{N} \cup\{ \infty\}$, one-point compactification of $\mathbb{N}$.
Let $\{a_n \}$ be a sequence defined by $a_n =f(n)$ for each $n \in \mathbb{N}$. Then what can we say about the sequence $\{a_n \}$?
First, I read from a different post that $\mathbb{N} \cup\{ \infty\}$ is homeomorphic to the compact and bounded set $\{ \frac{1}{n}: n \in \mathbb{N} \} \cup \{0\}$, then can we say that the sequence must be bounded (because it is the image of compact set under a continuous function) and if so, can we say that the sequence is convergent to $0$?
Continuous functions send limits to limits, and since in $\mathbb{N} \cup \{\infty\}$, $\lim_{n \to \infty} n = \infty$, it is the case that $ \lim_{n \to \infty} a_n = \lim_{n \to \infty} f(n) = f(\lim_{n \to \infty} n) = f(\infty). $ Also as you note, $\mathbb{N} \cup \{\infty\}$ is compact, and the image of compact sets under continuous maps are compact (and thus bounded in $\mathbb{C}$), so the sequence is bounded. Convergent sequences are also bounded, so the above explanation is a bit overkill.