Find out the subsequential limits of a sequence Let $$x_n=(-1)^n \left(2+\frac{3^n}{n!}+\frac{4}{n^2}\right)$$ and find the upper and lower limits of the sequence $\{x_n\}_{n=1}^\infty$.


We put $n=1, 2, 3 \dots $ then $x_1=-9, x_2 = \frac{15}{2}, \dots $ after some stage we see that limit superior is $x_2$ and inferior is $x_1$ is it right?
Please explain.
Notice 

$\underset{n \to \infty}{\lim \sup}\, x_n = \inf \{x_{2n} : n \in \mathbb{N}\} $,

and

$\underset{n \to \infty}{\lim \inf}\, x_n = \sup \{x_{2n-1} : n \in \mathbb{N}\} $.

Moreover, $2$ is a lower bound for the set $\left\{2+\frac{3^{2n}}{(2n)!}+\frac{1}{n^2} : n \in \mathbb{N}\right\}$, and $-2$ is an upper bound for the set $\left\{-2-\frac{3^{2n-1}}{(2n-1)!}-\frac{4}{(2n-1)^2}  : n \in \mathbb{N}\right\} $.

Also observe that we have $\frac{3^n}{n!} &lt; 2^{6-n}$ whenever $n&gt;6$.

Let $\varepsilon&gt;0$ be given. By the Archimedean Property we may find positive integers $N_1$ and $N_2$ such that
\begin{aligned}&amp;N_1  &gt; 7 - \frac{\log\varepsilon}{\log 2}, \text{ and} \\&amp;
N_2\varepsilon &gt; 8 .
\end{aligned}
Set $N=\max\{7, N_1, N_2\}$. So if $n \geq N$, then we have
\begin{equation}\frac{3^n}{n!}+\frac{4}{n^2}&lt;\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon .
\end{equation}
So there is $N \in \mathbb{N}$ such that 
\begin{aligned} &amp;x_{2N}&lt;2+\varepsilon, \text{ and} \\&amp; x_{2N-1}&gt;-2-\varepsilon.
\end{aligned}
Therefore, $\underset{n \to \infty}{\lim \sup}\, x_n =2$ and $\underset{n \to \infty}{\lim \inf}\, x_n =-2$.

Note: Since $N \geq 7$, we know that $2N-1&gt;N$ and $2N&gt;N$.
Here is a less rigorous but quicker way to look at the problem.

We have 
$$x_n=(-1)^n \left(2+\frac{3^n}{n!}+\frac{4}{n^2}\right)$$

Let's look at 
$$\lvert x_n\rvert=2+\frac{3^n}{n!}+\frac{4}{n^2}$$

We know that
$\dfrac4{n^2}$ is decreasing and $2$ is constant.

The first few terms of $\dfrac{3^n}{n!}$ are
$$\dfrac{3}{1},\,
\dfrac{3\cdot3}{2\cdot1},\,
\dfrac{3\cdot3\cdot3}{3\cdot2\cdot1},\,
\dfrac{3\cdot3\cdot3\cdot3}{4\cdot3\cdot2\cdot1},\,
\dfrac{3\cdot3\cdot3\cdot3\cdot3}{5\cdot4\cdot3\cdot2\cdot1},\cdots$$

which has 2nd and 3rd terms equal and 4th terms onwards decreasing.

All in all, $\lvert x_n\rvert$ will be decreasing from 3rd terms onwards. Hence, $x_1$ and $x_2$ are the lower and upper limits respectively.