whats the difference between below two limit notations
1) \[\lim \limits_{n\to \infty} (a_n)^{1/n}\] 2) \[\lim ~\sup~ (a_n)^{1/n}\]
2) always exists for a sequence bounded above, but 1) may not.
thanks, my textbook says something similar but i really dont get the difference yet
Just try a few examples and see what you get.
i think supremum is the limit of a sequence so they both give same numbers right ? \[\lim \limits_{n\to \infty} \frac{1}{n} = \lim \sup \frac{1}{n} = 0\]
another example \[\lim \limits_{n\to \infty}1 = \lim \sup 1= 1\]
take an oscillating sequence
another example \[\lim \limits_{n\to \infty} (-1)^n = DNE \\~\\\lim \sup (-1)^n= ?\] does this exist ?
yes
+1 ?
and inf = -1 ? Oh ooh
in fact, limsup is the largest limit point of a (bounded) sequence
i suppose below should not exist \[\lim\limits_{n\to \infty} \sup (-1)^n= ?\] does this exist ?
limsup exists for bounded sequences
limit may not
real quick question \(\lim\sup \frac{1}{n}\) equals 1 or 0 ?
if i consider the sequence : {1, 1/2, 1/3, ... } inf = 0 sup = 1
there is no difference in general , it only depend on the example or reason u have , the thing is sometimes when a sequence has a sup then its bounded above .. that make limit exist all the time
1 :)
whats diffference between sup and limsup ?
\[\limsup_{n\to\infty}x_n := \lim_{n\to\infty}\Big(\sup_{m\geq n}x_m\Big)\] \[\limsup_{n\to\infty}\frac{1}{n} = \lim_{n\to\infty}\Big(\sup_{m\geq n}\frac{1}{m}\Big)\] \[=\lim_{n\to\infty}\frac{1}{n}=0\]
\(\lim\limits_{n\to \infty} \sup~x_n\) considers only the tail of sequence @Zarkon ?
\[\limsup_{n\to\infty}(-1)^n= \lim_{n\to\infty}\Big(\sup_{m\geq n}(-1)^m\Big)\] \[=\lim_{n\to\infty}1=1\]
sup means the least greatest value of the sequence inf means the greatest smallest value , consider this open interval:- |dw:1415719920292:dw|
sup is the least upper bound. inf is the greatest lower bound.
Ohk, so the sequence need not converge to sup or inf right ?
it could converge to some number in between
|dw:1415720093084:dw|
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