lim sup iff conditions - please help explain

Question

Deciphering the definition of the upper limit, we see that $\limsup x_n=L$ is and only if the following two conditions are fulfilled:

• (a) $\forall\epsilon>0\;\;\exists N\in\mathbb{N}$ such that $\forall n\geq N\ x_n<L+\epsilon$;
• (b) $\forall\epsilon>0\;\;\forall N\in\mathbb{N} \exists n > N$ such that $x_n>L-\epsilon$.

Hi all, could you please explain this for me? What does condition (a) mean, and what does condition (b( mean?

thanks!

Answer 1

Very briefly condition (a) says that the limsup is not larger than $L$ and condition (b) says it's not smaller than $L$.

In a bit more detail, (a) says that if you take a number $M$ larger than $L$, then after maybe a finite number of exceptions the sequence is always less than $M$.

(b) says that if you take a number $M$ less than $L$, there are infinitely many members of the sequence that are greater than $M$.

Again, informally, I like to think of them as (a) kind of pushes the sequence down to $L$ as time goes on, but condition (b) always draws the sequence back up towards $L$, no matter how often they stray away. Note that (b) is kind of weaker. It lets the sequence go far below $L$ infinitely often, as long as it always comes back, whereas (a) says the sequence must stay not too high above $L$ eventually.

Answer 2

First fix an $\varepsilon>0$.

(a) means $L+\varepsilon$ is an upper bound for all $x$ of sufficiently high order.

(b) means $L-\varepsilon$ is not an upper bound for all such $x$.

It indeed implies $L=\limsup x_n$ since

• $\sup\limits_{n\ge N}x_n\le L+\frac1N\quad $ by (a), hence, passing to the limit, $\,\,\limsup x_n \le L$.
• $\sup\limits_{n\ge N}x_n>L-\frac1N\quad $ by (b), hence, passing to the limit, $\,\,\limsup x_n \ge L$.

Answer 3

Here's another way to write these conditions that may help:

(a) means that for every $\epsilon > 0$, every $x_n$ except for at most finitely many satisfy $x_n < L + \epsilon$. That is, there are no (infinite) subsequences that stay above $L + \epsilon$

(b) means that for every $\epsilon > 0$, there are infinitely many $x_n$ that satisfy $x_n > L - \epsilon$. That is, there exists a subsequence that stays above $L - \epsilon$.