Find $\mathop{\limsup}\limits_{n\to\infty} a_n$ and $\mathop{\liminf}\limits_{n\to\infty} a_n$ if $a_n = n(2+(-1)^n)$

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Given the sequence $a_n = n(2+(-1)^n)$.

Find $\mathop{\overline{\lim}}\limits_{n\to\infty} a_n$ and $\mathop{\underline{\lim}}\limits_{n\to\infty} a_n$.

The following is how I approach this problem.

$\mathop{\lim}\limits_{n\to\infty} a_{2n} = \mathop{\lim}\limits_{n\to\infty} n(2+1) = \infty$

and

$\mathop{\lim}\limits_{n\to\infty} a_{2n+1} = \mathop{\lim}\limits_{n\to\infty} n(2-1) = \infty$

Could I conclude $\mathop{\overline{\lim}}\limits_{n\to\infty} a_n = \mathop{\underline{\lim}}\limits_{n\to\infty} a_n = \infty$?

I'm confused because $\infty$ is not a constant.

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What's the problem with the fact that $\infty$ is not a constant? For each $n\in\Bbb N$, $a_n\geqslant n$, and therefore $\lim_{n\to\infty}a_n=\infty$. So, $\limsup_{n\to\infty}a_n=\liminf_{n\to\infty}a_n=\infty$.

In general, if $l\in\Bbb R\cup\{\pm\infty\}$, then$$\lim_{n\to\infty}a_n=l\iff\limsup_{n\to\infty}a_n=\liminf_{n\to\infty}a_n=l.$$