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.

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 $liminfx_n=0$.
