If there's only one limit point, what's $\limsup_{n\to \infty} a_n$ and what's $\liminf_{n\to \infty} a_n$?

Question

$a_n:=(-1)^n\cdot 1/n-2n^2(1+(-1)^n)+n/(n+1)$ with $n\in \mathbb{N}$.

I came to the conclusion that there's just a single limit point. If $H$ is the set of Limit points, it's $H=\{1\}$. How do I determine lim inf and lim sup if there's just a single element in the set?

Answer 1

$\lim \inf a_n$ is the smallest limit point of the sequence and $\lim \sup a_n$ is the largest limit point. (In fact some books use this as the definition of $\lim \inf$ and $\lim \sup$). Hence $\lim \sup a_n =\lim \inf a_n=1$ which implies that $a_n \to 1$.

Answer 2

Since $(-1)^n \frac{1}{n} \rightarrow 0$ and $\frac{n}{n+1} \rightarrow 1$, you can focus on the sequence $b_n = -2n^2(1 + (-1)^n)$. Its only subsequential limits are $\lim b_{2n + 1}=0$ and $\lim b_{2n}=-\infty$. In fact, if $b_{n_k}$ converges, it must have only odd(or even) indexes $n_k$ for large $k$, and so it converges to $0$ or $-\infty$. So $1$ and $-\infty$ are the only subsequential limits of $a_n$.