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.
 A: 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 $\liminf$.
A: Limsup and liminf are the smallest and the bigest (maximum and minimum) values of limit points of the sequence. The given sequences has two limit points, than the biger one will be limsup and the smaller one liminf. So $limsupx_n=1$ and $liminfx_n=0$.
