I want to compute liminf and limsup of $\left( (-1)^{n^3} \left( 1+\frac{1}{n}\right)^n\right)$.
I have thought the following so far:
From definition we have that $\lim \inf x_n=\lim_{n \to \infty} \left( inf_{k \geq n} x_k \right)$ and $\lim \sup x_n=\lim_{n \to \infty} \left( \sup_{k \geq n} x_k \right)$.
If $k$ is odd, then $(-1)^{k^3} \left( 1+\frac{1}{k}\right)^k=-\left( 1+\frac{1}{k}\right)^k$.
If $k$ is even , then $(-1)^{k^3} \left( 1+\frac{1}{k}\right)^k=\left( 1+\frac{1}{k}\right)^k$.
It holds that $-\left( 1+\frac{1}{k}\right)^k \geq - \left( 1+\frac{1}{n}\right)^k$ and $\left( 1+\frac{1}{k}\right)^k \leq \left( 1+\frac{1}{n}\right)^k$.
But we cannot bound $\left( 1+\frac{1}{n}\right)^k$ and $- \left( 1+\frac{1}{n}\right)^k$ by an expression of $n$, can we? (Thinking)
If not, how can we compute liminf and limsup?
We have that
$$\left| (-1)^{n^3} \left( 1+\frac{1}{n}\right)^n\right|\le e$$
and
$n^3$ even $\implies (-1)^{n^3}\left( 1+\frac{1}{n}\right)^n \to e$
$n^3$ odd $\implies (-1)^{n^3}\left( 1+\frac{1}{n}\right)^n \to -e$
then we can conclude that $\limsup=e$ and $\liminf=-e$.