@Jakobian: Thanks! Another question: if $a_n>b_n$ for each $n \in \mathbb{N}$, can we say that $\sup_{n\in\mathbb{N}}a_n > \sup_{n\in\mathbb{N}} b_n$? I think it is true because from $\sup_{n\in\mathbb{N}} a_n \ge a_n>b_n$ we deduce that $\sup_{n\in\mathbb{N}}a_n$ is an upper bound for $b_n$, hence $\sup_{n\in\mathbb{N}}a_n \ge \sup_{n\in\mathbb{N}} b_n$ because by definition the supremum is the least upper bound.
If by the sake of contradiction we assume $\sup_{n\in\mathbb{N}}a_n = \sup_{n\in\mathbb{N}}b_n$, we have $\sup_{n\in\mathbb{N}}b_n=\sup_{n\in\mathbb{N}}a_n \ge a_n>b_n$, so $\sup…
If by the sake of contradiction we assume $\sup_{n\in\mathbb{N}}a_n = \sup_{n\in\mathbb{N}}b_n$, we have $\sup_{n\in\mathbb{N}}b_n=\sup_{n\in\mathbb{N}}a_n \ge a_n>b_n$, so $\sup…