Limit Supremum and Limit Infimum of an alternating sequence Alternate the terms of the sequence $ (1 + \frac {1}{n})$ and $(\frac {-1}{n})$ to obtain the sequence $(x_n)$ given by $(2,-1,3/2,-1/2, 4/3, -1/3, 5/4, -1/4, \ldots)$.
Determine the values of $\limsup(x_n)$, $\liminf(x_n)$. Also find $\sup\{x_n\}$ and $\inf\{x_n\}$.
I think that just from inspection if I were to break this into two subsequences namely $ (1 + \frac {1}{n})$ and $(\frac {-1}{n})$ I would see $(2,3/2,4/3,5/4\ldots)$ and $(-1,-1/2,-1/3, -1/4,\ldots)$, then I could see that the first is increasing and the second is decreasing.
$ (1 + \frac {1}{n}) \to 1$ as $n \to  \infty$ while  $(\frac {-1}{n}) \to 0$ as $n \to  \infty$. I suppose this would imply that $2 =\sup\{x_n\}$ and $ -1 = \inf\{x_n\}$, although I am probably wrong in that sense. So would it then follow that the limits are $\limsup(x_n) =2$ and $\liminf(x_n) = -1$? I'm very lost with the limit superior and limit inferior concept. Any help would be appreciated.

Notice that the positive terms are decreasing and the negative terms are increasing.
Hence $\sup_{k \ge n} x_k$ will be given by the first positive term in $x_n,x_{n+1},...$, and similarly,
$\inf_{k \ge n} x_k$ will be given by the first negative term in $x_n,x_{n+1},...$.
Since the first positive term of the whole sequence is $2$, we have $\sup_n x_n = 2$, and the $\inf$ is computed similarly.
We have $\lim_{n \to \infty} \sup_{k \ge n} x_k = \lim_{n \to \infty} \sup_{k \ge n} x_{2k-1} =  \lim_{n \to \infty} \sup_{k \ge n} (1+{1 \over k}) =
\lim_{n \to \infty} (1+{1 \over n}) = 1$.
The $\liminf$ is computed in a similar fashion.
Note: In general, the sequence $n \mapsto \sup_{k \ge n} x_k$ is non-increasing, the $\limsup$ is the limit of this non-increasing sequence. Similarly
for $\liminf$.
