Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Possible Duplicate:
Limit Supremum and Infimum. Struggling the concept

Hey I'm trying to figure out what $\limsup{S_{n}}$ is compared to $\lim{S_{n}}$ as well as the difference of $\lim{S_{n}}$ and $\liminf{S_{n}}$

So for example (this is my current thinking process) if I have a monotone non increasing sequence $S_{n}:=1/n$ (where $n=1$ and goes to infinity). The $\limsup{S_{n}}$ is 1, and $\liminf{S_{n}}$ is 0. But we know the $\lim{S_{n}}$ is 0.

How does $\lim{S_{n}}=\liminf{S_{n}}=\limsup{S_{n}}?$

share|cite|improve this question

marked as duplicate by Ross Millikan, Brian M. Scott, Norbert, Noah Snyder, Marvis Oct 17 '12 at 18:06

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

See my answer from a previous question:… – Christopher A. Wong Oct 16 '12 at 23:40
up vote 2 down vote accepted

One definition of $\limsup s_n$ is $$\limsup s_n = \lim_{n \to \infty} \sup_{k \geq n} s_k$$ The corresponding definition of $\liminf s_n$ is $$\liminf s_n = \lim_{n \to \infty} \inf_{k \geq n} s_k$$ In your case, where $s_n = \dfrac1n$, we have $$\sup_{k \geq n} s_k = \sup_{k \geq n} \dfrac1k = \dfrac1n$$ Similarly, for $\liminf$. Hence, $$\limsup s_n = \lim_{n \to \infty} \sup_{k \geq n} s_k = \lim_{n \to \infty} \dfrac1n = 0$$

In general, if $\displaystyle \lim_{n \to \infty} s_n$ exists, then $$\limsup s_n = \lim s_n = \liminf s_n$$

Another way to define $\limsup$ and $\liminf$ is to look at the limit points of the sequence $s_n$ i.e. if $$S = \{\text{Limit points of the sequence }s_n\}$$ then $$\limsup s_n = \displaystyle \sup_{s \in S} S$$ and $$\liminf s_n = \displaystyle \inf_{s \in S} S$$ If $s_n = \dfrac1n$, then $S = \{0 \}$. Hence, $$\limsup s_n = 0 = \liminf s_n$$

share|cite|improve this answer

$limsupS_n$ of $x_n$ is the largest cluster point of $x_n$ if sequence is bounded above. $liminfS_n$ is the smallest cluster point if it is a bounded below sequence.

If a sequence converges to some $x$ its every subsequence converges to $x$. This is (most simpy) how lim$S_n$ = limsup$S_n$. The sequence of 1/n converges to 0, its only cluster point is 0. Thus lim$S_n$ = liminf$S_n$ = limsup$S_n$ = 0.

share|cite|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.