I cant understand this explanation of limsup

[transgalactic]

regarding this defintion

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

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


i can't understand the first part

why he is saying that
[itex]\sup \{ x_k | k\geq n\}[/itex]
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

??

 

[HallsofIvy]

But [itex]\left{x_k|k\ge n}[/itex] is not a single sequence- it is a different sequence for every different n.

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