Finding the lim inf and lim sup of a sequence

Question

I'm working on the following problem:

Find $\lim \sup_n a_n$ and $\lim\inf_n a_n$ of the sequence given by $a_n = 1 + (-1)^n\frac{2n+3}n.$

I've done the following work so far:

Let $\{i_n\}_n$ be the sequence given by $i_n = \inf\{a_k:k\ge n\}.$ Then, $i_n = \{-4, -2, \frac {-8}5, \frac{-10}7,\cdots\}.$

I know that the $\lim \inf_na_n = \lim_n i_n$, but I can't figure out how to take the limit of $i_n$.

Answer 1

Let's first notice that $\frac{2n+3}{n}$ converges to $2$. Therefore, the sequence $$(-1)^n\frac{2n+3}{n}$$

has only two limit points, which are $2$ and $-2$. This implies that $(a_n)$ has only two limit points, which are $3$ and $-1$. Therefore, you get $$\liminf a_n = -1 \quad \text{and} \quad \limsup a_n = 3$$

Answer 2

Let's compute the even and odd terms. So, as $n\rightarrow \infty $, we have \begin{eqnarray*} a_{2n} &=&1+\frac{4n+3}{2n}=\frac{3n+3/2}{n}\to3, \\ a_{2n-1} &=&1-\frac{2\left( 2n-1\right) +3}{2n-1}=-\frac{n+1}{n-1/2}\to-1. \end{eqnarray*}