Superior limit of a certain sequence

Question

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.

Answer 1

Hint: If you have two sequences $\{x_n\}$ and $\{y_n\}$ such that $\{x_n\}$ is convergente then $\limsup(x_n+y_n)=\lim x_n+\limsup y_n$.

Answer 2

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 $\liminf c_n=\min(a,b)$.

Answer 3

$$sup\left\{ { x }_{ n } \right\} =2+\frac { 1 }{ n } ,n=1,2,...\\ inf\left\{ { x }_{ n } \right\} =1+\frac { 1 }{ n } ,n=1,2,...\\ \lim _{ n\rightarrow \infty }{ sup\left\{ { x }_{ n } \right\} = } \lim _{ n\rightarrow \infty }{ \left( 2+\frac { 1 }{ n } \right) = } 2\\ \lim _{ n\rightarrow \infty }{ inf\left\{ { x }_{ n } \right\} = } \lim _{ n\rightarrow \infty }{ \left( 1+\frac { 1 }{ n } \right) = } 1$$

Answer 4

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,

$\limsup _{n\to \infty}\left \{ x_{n} \right \}=\lim_{k\to \infty}a_{k}=2$