Superior limit of a certain sequence Let $(x_n)$ be the sequence:
$$\{1, 2, 1 + \frac12, 2 + \frac12, 1 + \frac13, 2 + \frac13, \ldots \}$$
I (think I) understand why $\lim \inf x_n = 1$, but I'm not sure why $\lim \sup x_n =2$. Any clarification?
Thank you.
 A: More generally, if $\{a_n\}$ and $\{b_n\}$ are sequences with limit $a$ and $b$ respectively, then for $c_n=\{a_1,b_1,a_2,b_2,\dots\}$ we get $\limsup c_n=\max(a,b)$ and $\liminf c_n=\min(a,b)$.
A: Hint: If you have two sequences $\{x_n\}$ and $\{y_n\}$ such that $\{x_n\}$ is convergente then $\limsup(x_n+y_n)=\lim x_n+\limsup y_n$.
A: $$sup\left\{ { x }_{ n } \right\} =2+\frac { 1 }{ n } ,n=1,2,...\\ inf\left\{ { x }_{ n } \right\} =1+\frac { 1 }{ n } ,n=1,2,...\\ \lim _{ n\rightarrow \infty  }{ sup\left\{ { x }_{ n } \right\} = } \lim _{ n\rightarrow \infty  }{ \left( 2+\frac { 1 }{ n }  \right) = } 2\\ \lim _{ n\rightarrow \infty  }{ inf\left\{ { x }_{ n } \right\} = } \lim _{ n\rightarrow \infty  }{ \left( 1+\frac { 1 }{ n }  \right) = } 1$$
A: Go back to the definition of the $\limsup $:
If we set $a_{k}=\sup_{n\geq k}\left \{ x_{n} \right \}$, then 
$a_{1}=2$
$a_{2}=2$
$a_{3}=2+1/2$
$a_{4}=2+1/3$ and in general 
$a_{k}=2+1/(k-1)$ if $k>2$
Now by definition, 
$\limsup _{n\to \infty}\left \{ x_{n} \right \}=\lim_{k\to \infty}a_{k}=2$
