Confusion about definition of limsup and liminf

Question

The book that I am following gives the following definition for limsup:

A real number $\bar{a}$ is said to be the limit superior of a bounded sequence {$a_n$} iff for each $\epsilon > 0$, the following results hold:

1. $a_n$ > $\bar{a} - \epsilon$, for infinitely many values of n

2. $\exists$ a positive integer m such that

$a_n$ < $\bar{a} + \epsilon $ $\forall n \ge m$

I don't understand how 1 and 2 could both be true at the same time. Can someone explain this to me? Also why does the definition mention bounded sequences?

Answer 1

You’ve misstated (2): it should say that there is a positive integer $m$ such that $a_n<\bar a+\epsilon$ for all $n\ge m$. To see that (1) and (2) are not contradictory, consider any constant sequence: if $a_n=c$ for each $n\in\Bbb Z^+$, and $\epsilon>0$, then $a_n>c-\epsilon$ for every $n\ge 1$, and $a_n<c+\epsilon$ for every $n\ge 1$. In this case $\limsup_na_n=c$.

More generally, suppose that $\langle a_n:n\ge 1\rangle$ converges to some limit $c$, and let $\epsilon>0$. Then there is an $m\in\Bbb Z^+$ such that $|a_n-c|<\epsilon$ for all $n\ge m$, i.e., such that

$$c-\epsilon<a_n<c+\epsilon$$

for all $n\ge m$. Clearly, then, $a_n>c-\epsilon$ for infinitely many $n$, and $a_n<c+\epsilon$ for all $n\ge m$, so $\limsup_na_n=c$.

But a sequence doesn’t have to converge in order for the limit superior to make sense. Consider the sequence $\langle (-1)^n:n\ge 1\rangle$. If $\epsilon>0$, then $(-1)^n<1+\epsilon$ for all $n\ge 1$, and $(-1)^n>1-\epsilon$ for all even $n\ge 1$, so $\limsup_n(-1)^n=1$ even though the sequence does not converge. (In fact it turns out that $\limsup_na_n$ is the largest cluster point of the sequence.)

The requirment that the sequence to be bounded is to ensure that condition (2) can be satisfied. Specifically, we need the sequence to be bounded above, because if it isn’t, there is no real number $\bar a$ such that $a_n<\bar a+1$ for all sufficiently large $n$.

Answer 2

Last question answered first: If the sequence is unbounded such a real number cannot exist. So the $\limsup$ in that case is infinite.

The $\limsup$ can be seen as the largest accumulation point and property $1.$ ensures that it is indeed such an accumulation point from "below". So that in an area around $\overline{a}$ lay infinite many elements of the sequence.

The second property (which actually is missing a $\forall n\ge m$) ensures it is really the "largest" one. So if we add just a small number to $\overline{a}$ we find an element of the sequence s.t. all following ones are smaller then $\overline{a}$ plus the small number.

Answer 3

Why can't both be true? The two conditions are independent of each other. The second one says that there is some $m$ such that for all $n\geq m$ we have $a_n<\bar{a}+\epsilon$. So it deals only with an upper bound of the sequence and doesn't say anything about what happens below $a_n$. On the other hand the first condition deals with lower bounds: it says that we have $a_n>\bar{a}-\epsilon$ for infinitely many integers $n$.

For example, take $a_n=(-1)^n$. I claim the limsup is equal to $1$. Let's check the two conditions. Let $\epsilon>0$. First of all, for all $n\geq 1$ we have $(-1)^n<1+\epsilon$, so the second condition definitely holds. Also, we have $(-1)^n>1-\epsilon$ for all even integers, so in particular this is true for infinitely many integers. Thus the first condition holds as well.

The boundedness of the sequence is needed to assure that such a real number $\bar{a}$ exists. Note that if the sequence is not bounded from above then the second condition can't hold. And if it is not bounded from below then we might have problems with the first condition. The definition can be generalized though, limsup and liminf can be also infinite.

Answer 4

Do you understand the difference between "infinitely many" and "all"?

The first statement might be true for "all even numbers", so for "infinitely many numbers" while the second statement might be true for some odd number.