Compute the limsup and the liminf of the following sequence

Question

Define the following sequence $a:\mathbb{N}\to \mathbb{R}$ by, $$ a_n = \frac{(-1)^nn + 1}{\sin\left( \frac{\pi n}{3} \right) n + 1} $$

I feel like the answer is $\infty$ but I am not sure.

Answer 1

If $n = 3k$ is even for $k \in \Bbb Z$, then

$$a_n = \frac{(-1)^nn+1}{\sin(k\pi)n+1} = \frac{1\cdot n+1}{0\cdot n+1}= n+1.$$

If $n = 3k$ is odd, then

$$a_n = \frac{(-1)^nn+1}{\sin(k\pi)n+1} = \frac{-1\cdot n+1}{0\cdot n+1}= -n+1.$$

Thus the $\limsup$ goes to $\infty$ and the $\liminf$ goes to $-\infty$ regardless of what the other terms do.

Answer 2

If you rewrite it as $\frac{(-1)^n + \frac{1}{n}}{\sin \frac{\pi n}{3} + \frac{1}{n}}$ you'll see that the limit takes very large values for $n = 3k$ for some $k$ and something else for other values. Hence limit does not exist.