# I cant understand this explanation of limsup

1. Jan 8, 2009

### transgalactic

regarding this defintion

http://img515.imageshack.us/img515/5666/47016823jz1.gif

i was told that
Remember that if $x_n$ is bounded then $\limsup x_n = \lim \left( \sup \{ x_k | k\geq n\} \right)$.
The sequence, $\sup \{ x_k | k\geq n\}$ is non-increasing, therefore its limits is its infimum.
Thus, $\limsup x_n = \inf \{ \sup\{ x_k | k\geq n\} | n\geq 0 \}$[/quote]

i cant understand the first part

why he is saying that
$\sup \{ x_k | k\geq n\}$
is not increasing.
you are taking a bounded sequence and you get one number
which is SUP (its least upper bound)
thats it.
no more members

??

2. Jan 8, 2009

### HallsofIvy

Staff Emeritus
But $\left{x_k|k\ge n}$ is not a single sequence- it is a different sequence for every different n.

For example, if ${x_n= (-1)^n/n}= {-1, 1/2, -1/3, 1/4, -1/5, ...} then [itex]sup{x_k|k\ge 1}$ is the largest of {-1, 1/2, -1/3, 1/4, -1/5, ...} which is 1/2. $sup{x_k|k\ge 2}$ is the largest of {1/2, -1/3, 1/4, -1/5, ...}, again 1/2. $sup{x_k|k\ge 3}$ is the largest of {-1/3, 1/4, -1/5, ...}, which is 1/4. Similarly, $sup{x_k|k\ge 4}$ is also 1/4 but $sup{x_k|k\ge 5}$ is 1/6, etc.