This question is related to the question Is every cauchy sequence bounded? The sequence $\{a_n\}$ used in that question $$a_n=\frac{1}{n-1}$$
has the first element $a_1\rightarrow\infty$. As shown in the answer of that question, $a_1=\infty$ is not well-defined, so we can not say it is an unbounded Cauchy sequence.
However, in an unbounded sequence $$a_n=n,$$ we also have $$\lim_{n\rightarrow\infty} a_n\rightarrow\infty.$$ Why $\infty$ works here?
$\lim_{n \to +\infty} a_n = +\infty$ means that for each $M \in \mathbb{R}$, there exists a $n_0 \in \mathbb{N}$ such that for all $n \geq n_0$ $$a_n > M.$$
This means that the terms of the sequence get arbitrarily large.
Note that in your first sequence you somehow claim that $\infty$ is an element of that sequence, but it is a concept to denote unboundedness, it can not be an element of a sequence.
edit: proof that $\lim_{n\to \infty}a_n = + \infty$ with $a_n = n \in \mathbb{N}$ :
Choose $M \in \mathbb{R}$ arbitrarily, then there exists a natural number $n_0$ with $n_0 > M$ (fundamental property of $\mathbb{R}$. Let $n \geq n_0$, then $$a_n = n \geq n_0 > M.$$ This proves that $\lim_{n\to \infty}a_n = + \infty$.
However, it can be 'the limit of a sequence', as the definition above states.