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3.2 The Limsup and the Liminf

We have already seen that if $\{x_{n}\}^{\infty}_{n=1}$ is a bounded sequence then the limitpoints of the sequence are in between the supremum and the infinum of the sequence. We actually have a much more precise picture.

Definition 3.2.1

Let $\overline{x}=\{x_{n}\}^{\infty}_{n=1}$ be a bounded sequence of real numbers.
Then $\limsup_{n\to\infty}x_{n}$ (the Limsup or the Limit Superior of $\overline{x}$ ) is defined as the Least Upper Bound of the limitpoint set $\mbox{LIM}(\overline{x})$ . Similarly, $\liminf_{n\to\infty}x_{n}$ (the Liminf or the Limit Inferior of $\overline{x}$ ) is defined as the Largest Lower Bound of the limitpoint set $\mbox{LIM}(\overline{x})$ .

Question 3.2.1

How much is $\limsup_{n\to\infty}\frac{1}{n}?$

Solution: The supremum of the sequence is of course $1$ , but somewhat surprisingly, $\limsup_{n\to\infty}\frac{1}{n}=0$ . Indeed, the set of limitpoints of the sequence $\{\frac{1}{n}\}^{\infty}_{n=1}$ consists of one single element the zero. So the $0$ is both the limsup and the liminf of this sequence. In general, if a sequence convergent then its limsup and liminf is nothing but the limit of the sequence.

Proposition 3.2.1 (Limsups are limitpoints)

Let $\overline{x}=\{x_{n}\}^{\infty}_{n=1}$ be a bounded sequence of real numbers. Then $\limsup_{n\to\infty}x_{n}\in\mbox{LIM}(\overline{x})$ and $\liminf_{n\to\infty}x_{n}\in\mbox{LIM}(\overline{x})$ .

Remember, the supremum of an infinite set $S$ is not necessarily an element of $S$ . So, the statement of the proposition is not at all trivial. Since $\limsup_{n\to\infty}x_{n}\in\mbox{LIM}(\overline{x})$ is the Least Upper Bound of the set of limitpoints, there exists a sequence of limitpoints $\{l_{k}\}^{\infty}_{k=1}\subset\mbox{LIM}(\overline{x})$ such that $l_{k}\to\limsup_{n\to\infty}x_{n}$ . Now we will do the same trick as in the solution of Question 3.1.4. Although the numbers $l_{m}$ are not elements of our sequence but they are limitpoints of our sequence so we can get as close to them as we want. That is, there exist numbers $n_{1}<n_{2}<n_{3}\dots$ such that $|x_{n_{k}}-l_{k}|\leq\frac{1}{k}$ . Then by the Sum Rule, $\lim_{k\to\infty}x_{n_{k}}=\limsup_{n\to\infty}x_{n}$ . The same goes for the liminf as well. $\square$

Proposition 3.2.2 (How to imagine the Limsup)

Let $\overline{x}=\{x_{n}\}^{\infty}_{n=1}$ be a bounded sequence. Let $L=\limsup_{n\to\infty}x_{n}$ . Then:

1.

There exists a subsequence $\{x_{n_{k}}\}^{\infty}_{k=1}$ tending to $L$ .
2.

For any $\varepsilon>0$ there exists an $N>0$ such that if $n\geq N$ , then $x_{n}\leq L+\varepsilon\,.$

On the other hand, if a certain real number satisfies the two conditions above, then it equals to the Limsup of the sequence.

Proof: First note that the first condition means that $L$ is a limitpoint and we just proved that the Limsup is a limitpoint. Now, if the second condition is not satisfied, then there exists a subsequence $\{x_{n_{k}}\}^{\infty}_{k=1}$ such that for any $k\geq 1$ $x_{n_{k}}>\varepsilon+L$ . By the Bolzano-Weierstrass Theorem this subsequence contains a convergent subsequence, and its limit would be strictly larger than $L$ , in contradiction with the fact that $L$ is the largest limitpoint. Finally, suppose that a certain number $T$ satisfies the two conditions. Then, $T$ is a limit point of $\overline{x}$ . On the other hand, by the second condition any limitpoint of $\overline{x}$ must be less or equal than $T$ . Hence $T$ is the Limsup of the sequence. $\square$

Remark 3.2.1 (Fun fact)

When we prove the Bolzano-Weierstrass Theorem from the Least Upper Bound Principle and the Largest Lower Bound Principle, we actually found the limsup and the liminf as limit points.

Observation 3.2.1

Let $\overline{x}=\{x_{n}\}^{\infty}_{n=1}$ be a bounded sequence of real numbers. Then for any limitpoint $l$ of $\overline{x}$ :

\inf(\overline{x})\leq\liminf(\overline{x})\leq l\leq\limsup(\overline{x})\leq% \sup{\overline{x}}\,.

Example 3.2.1

Let $x_{n}=(-1)^{n}+\frac{1}{n}$ . Then $\limsup x_{n}=1$ , $\liminf x_{n}=-1$ .

When we added two convergent sequences then their limits were added as well. The Limsup is a bit trickier.

Proposition 3.2.3

Let $\overline{x}=\{x_{n}\}^{\infty}_{n=1}$ and $\overline{y}=\{y_{n}\}^{\infty}_{n=1}$ be bounded sequences of real numbers. Then $\limsup_{n\to\infty}(x_{n}+y_{n})\leq\limsup_{n\to\infty}x_{n}+\limsup_{n\to% \infty}y_{n}$ .
Similarly, $\liminf_{n\to\infty}x_{n}+\liminf_{n\to\infty}y_{n}\leq\liminf_{n\to\infty}(x_% {n}+y_{n})\,.$

Proof: We prove only the first inequality. Let $L$ be a limitpoint of $\{x_{n}+y_{n}\}^{\infty}_{n=1}$ . It is enough to prove that

L\leq\limsup_{n\to\infty}x_{n}+\limsup_{n\to\infty}y_{n}\,.

(3.1)

Indeed, $(\ref{oct11a})$ means that $\limsup_{n\to\infty}x_{n}+\limsup_{n\to\infty}y_{n}$ is an upper bound for the set of limitpoints of $\{x_{n}+y_{n}\}^{\infty}_{n=1}$ . Hence it is greater or equal than the least upper bound. The least upper bound is exactly $\limsup_{n\to\infty}(x_{n}+y_{n})$ .

$L$ is a limitpoint, so $L=\lim_{k\to\infty}(x_{n_{k}}+y_{n_{k}})$ for a certain subsequence. The problem is that even if $\{x_{n_{k}}+y_{n_{k}}\}^{\infty}$ is convergent it is possible that neither $\{x_{n_{k}}\}^{\infty}$ nor $\{y_{n_{k}}\}^{\infty}$ are convergent. Yes indeed, if $\{x_{n_{k}}\}^{\infty}$ is any non-convergent sequence and $y_{n_{k}}=1-x_{n_{k}}$ then $\{x_{n_{k}}+y_{n_{k}}\}^{\infty}$ is the most obviously convergent (constant) sequence of the world. The simple looking Proposition 3.2.2 comes to the rescue. This proposition explains what limsup is. By this proposition, the limsup EVENTUALLY looks more and more like a supremum even it is actually an infimum. For any $\varepsilon>0$ there exists $N>0$ such that if $k\geq N$ , then $x_{n_{k}}\leq\limsup_{n\to\infty}x_{n_{k}}+\varepsilon$ and $y_{n_{k}}\leq\limsup_{n\to\infty}y_{n_{k}}+\varepsilon$ . Hence, if $k\geq N$ then $x_{n_{k}}+y_{n_{k}}\leq\limsup_{n\to\infty}x_{n_{k}}+\limsup_{n\to\infty}y_{n_% {k}}+2\varepsilon$ . Therefore

L\leq\limsup_{n\to\infty}x_{n_{k}}+\limsup_{n\to\infty}y_{n_{k}}+2\varepsilon\,.

(3.2)

OK, (3.2) is not as good as the inequality (3.1) because of the $\varepsilon$ in the formula. However, (3.2) holds for ANY $\varepsilon>0$ so it must hold for the zero as well, that is (3.1) holds. $\square$

Example 3.2.2

Let $x_{n}=12+\frac{1}{n}$ if $n$ is in the form of $3k$ . Let $x_{n}=5-\frac{1}{n}$ if $n$ is in the form of $3k+1$ . Let $x_{n}=3+2\frac{1}{n^{2}}$ if $n$ is in the form of $3k+2$ . Then eventually $x_{n}$ is very close to $12$ or $5$ or $3$ . Hence, the limitpoints of $\{x_{n}\}^{\infty}_{n=1}$ are $12,5,3$ . That is, $\limsup_{n\to\infty}x_{n}=12$ , $\liminf_{n\to\infty}x_{n}=3\,.$