Upper and Lower Limits $a_n = (-1)^{n+1} \left(1+\frac{1}{n}\right) $

Question

Given that $a_n = (-1)^{n+1} \left(1+\frac{1}{n}\right)$ . Find the upper and lower limit of $a_n $

Using that $\\ K = \{x \in R | x\text{ is a limit point of }a_n \} $

sorry if my terminology is not on point ( english is not native language )

Answer 1

For the upper limit, note that $a_n \le|a_n|$ and $\lim |a_n|$ is easy to find.

For the lower limit, $a_n \ge -|a_n|$.

Note that both bounds are sharp for a half of the $a_n$'s...

Answer 2

We have

$$\lim_\infty a_{2n}=-1$$ $$\lim_\infty a_{2n+1}=1$$

thus $$\limsup a_n=1$$ and $$\liminf a_n=-1$$

Answer 3

First, draw a picture, your sequence is oscillating between a sequence getting closer to $1$ from above and a sequence getting closer to $-1$ from below, corresponding to odd and even $n$ respectively.

With this intuition, we may compute as follows $$ \lim_{k\to \infty}\sup_{n\geq k}a_k=\lim_{k\to \infty}(1+\frac{1}{2k+1})=1 $$ where the first inequality takes from noting that the the sequence above $1$ is decreasing to $1$. Similarly for the $\lim\inf$.