2.5: Limit Superior and Limit Inferior

We begin this section with a proposition which follows from Theorem 2.3.1. All sequences in this section are assumed to be of real numbers.

Proposition $\PageIndex{1}$

Let $\left\{a_{n}\right\}$ be a bounded sequence. Define

$$s_{n}=\sup \left\{a_{k}: k \geq n\right\}$$

and

$$t_{n}=\inf \left\{a_{k}: k \geq n\right\}.$$

Then $\left\{s_{n}\right\}$ and $\left\{t_{n}\right\}$ are convergent.

Proof

If $n \leq m$, then $\left\{a_{k}: k \geq m\right\} \subset\left\{a_{k}: k \geq n\right\}$. Therefore, it follows from Theorem 2.5.3 that $s_{n} \geq s_{m}$ and, so, the sequence $\left\{s_{n}\right\}$ is decreasing. Since $\left\{a_{n}\right\}$ is bounded, then so is $\left\{s_{n}\right\}$. In particular, $\left\{s_{n}\right\}$ is bounded below. Similarly, $\left\{t_{n}\right\}$ is increasing and bounded above. Therefore, both sequences are convergent by Theorem 2.3.1. $\square$

Definition $\PageIndex{1}$: Limit Superior

Let $\left\{a_{n}\right\}$ be a sequence. Then the limit superior of $\left\{a_{n}\right\}$\), denoted by $\limsup _{n \rightarrow \infty} a_{n}$, is defined by

$$\limsup _{n \rightarrow \infty} a_{n}=\lim _{n \rightarrow \infty} \sup \left\{a_{k}: k \geq n\right\}.$$

Note that $\limsup _{n \rightarrow \infty} a_{n}=\lim _{n \rightarrow \infty} s_{n}$, where $s_{n}$ is defined in (2.8).

Similarly, the limit inferior of $\left\{a_{n}\right\}$, denoted by $\liminf _{n \rightarrow \infty} a_{n}$, is defined by

$$\liminf _{n \rightarrow \infty} a_{n}=\liminf _{n \rightarrow \infty}\left\{a_{k}: k \geq n\right\}.$$

Note that $\liminf _{n \rightarrow \infty} a_{n}=\lim _{n \rightarrow \infty} t_{n}$, where $t_{n}$ is defined in (2.9).

Theorem $\PageIndex{2}$

If $\left\{a_{n}\right\}$ is not bounded above, then

$$\lim _{n \rightarrow \infty} s_{n}=\infty,$$

where $\left\{s_{n}\right\}$ is defined in (2.8).

Similarly, if $\left\{a_{n}\right\}$ is not bounded below, then

$$\lim _{n \rightarrow \infty} t_{n}=-\infty,$$

where $\left\{t_{n}\right\}$ is defined in (2.9).

Proof

Suppose $\left\{a_{n}\right\}$ is not bounded above. Then for any $k \in \mathbb{N}$, the set $\left\{a_{i}: i \geq k\right\}$ is also not bounded above. Thus, $s_{k}=\sup \left\{a_{i}: i \geq k\right\}=\infty$ for all $k$. Therefore, $\lim _{k \rightarrow \infty} s_{k}=\infty$. The proof for the second case is similar. $\square$

Remark $\PageIndex{3}$

By Theorem 2.5.2, we see that if $\left\{a_{n}\right\}$ is not bounded above, then

$$\limsup _{n \rightarrow \infty} a_{n}=\infty.$$

Similarly, if $\left\{a_{n}\right\}$ is not bounded below, then

$$\liminf _{n \rightarrow \infty} a_{n}=-\infty.$$

Theorem $\PageIndex{4}$

Let $\left\{a_{n}\right\}$ be a sequence and $\ell \in \mathbb{R}$. The following are equivalent:

1. $\limsup _{n \rightarrow \infty} a_{n}=\ell$.
2. For any $\varepsilon>0$, there exists $N \in \mathbb{R}$ such that

$$a_{n}<\ell+\varepsilon \text { for all } n \geq N,$$

and there exists a subsequence of $\left\{a_{n_{k}}\right\}$ of $\left\{a_{n}\right\}$ such that

$$\lim _{k \rightarrow \infty} a_{n_{k}}=\ell.$$

Proof

Suppose $\limsup _{n \rightarrow \infty} a_{n}=\ell$. Then $\lim _{n \rightarrow \infty} s_{n}=\ell$, where $S_{n}$ is defined as in (2.8). For any $\varepsilon>0$, there exists $N \in \mathbb{N}$ such that

$$\ell-\varepsilon<s_{n}<\ell+\varepsilon \text { for all } n \geq N.$$

This implies $s_{N}=\sup \left\{a_{n}: n \geq N\right\}<\ell+\varepsilon$. Thus,

$$a_{n}<\ell+\varepsilon \text { for all } n \geq N$$

Moreover, for $\varepsilon=1$, there exists $N_{1} \in \mathbb{N}$ such that

$$\ell-1<s_{N_{1}}=\sup \left\{a_{n}: n \geq N_{1}\right\}<\ell+1.$$

Thus, there exists $n_{1} \in \mathbb{N}$ such that

$$\ell-1<a_{n_{1}}<\ell+1.$$

For $\varepsilon=\frac{1}{2}$, there exists $N_{2} \in \mathbb{N}$ and $N_{2}>n_{1}$ such that

$$\ell-\frac{1}{2}<s_{N_{2}}=\sup \left\{a_{n}: n \geq N_{2}\right\}<\ell+\frac{1}{2}.$$

Thus, there exists $n_{2}>n_{1}$ such that

$$\ell-\frac{1}{2}<a_{n_{2}}<\ell+\frac{1}{2}.$$

In this way, we can construct a strictly increasing sequence $\left\{n_{k}\right\}$ of positive integers such that

$$\ell-\frac{1}{k}<a_{n_{k}}<\ell+\frac{1}{k}.$$

Therefore, $\lim _{k \rightarrow \infty} a_{n_{k}}=\ell$.

We now prove the converse. Given any $\varepsilon>0$, there exists $N \in \mathbb{N}$ such that

$$a_{n}<\ell+\varepsilon \text { and } \ell-\varepsilon<a_{n_{k}}<\ell+\varepsilon$$

for all $n \geq M$ and $k \geq N$. Let any $m \geq N$, we have

$$s_{m}=\sup \left\{a_{k}: k \geq m\right\} \leq \ell+\varepsilon.$$

By Lemma 2.1.8, $n_{m} \geq m$, so we also have

$$s_{m}=\sup \left\{a_{k}: k \geq m\right\} \geq a_{n_{m}}>\ell-\varepsilon.$$

Therefore, $\lim _{m \rightarrow \infty} s_{m}=\limsup _{m \rightarrow \infty} a_{n}=\ell$. $\square$

The following result is proved in a similar way.

Theorem $\PageIndex{5}$

Let $\left\{a_{n}\right\}$ be a sequence and $\ell \in \mathbb{R}$. The following are equivalent:

1. $\liminf _{n \rightarrow \infty} a_{n}=\ell$.
2. For any $\varepsilon>0$, there exists $N \in \mathbb{N}$ such that

$$a_{n}>\ell-\varepsilon \text { for all } n \geq N,$$

and there exists a subsequence of $\left\{a_{n_{k}}\right\}$ of $\left\{a_{n}\right\}$ such that

$$\lim _{k \rightarrow \infty} a_{n_{k}}=\ell.$$

Proof

Add proof here and it will automatically be hidden

The following corollary follows directly from Theorems 2.5.4 and 2.5.5.

Corollary $\PageIndex{6}$

Let $\left\{a_{n}\right\}$ be a sequence. Then

$$\lim _{n \rightarrow \infty} a_{n}=\ell \text { if and only if } \limsup _{n \rightarrow \infty} a_{n}=\liminf _{n \rightarrow \infty} a_{n}=\ell .$$

Proof

Add proof here and it will automatically be hidden

Corollary $\PageIndex{7}$

Let $\left\{a_{n}\right\}$ be a sequence.

1. Suppose $\limsup _{n \rightarrow \infty} a_{n}=\ell$ and $\left\{a_{n_{k}}\right\}$ is a subsequence of $\left\{a_{n}\right\}$ with

$$\lim _{k \rightarrow \infty} a_{n_{k}}=\ell^{\prime}.$$

Then $\ell^{\prime} \leq \ell$.

2. Suppose $\liminf _{n \rightarrow \infty} a_{n}=\ell$ and $\left\{a_{n_{k}}\right\}$ is a subsequence of $\left\{a_{n}\right\}$ with

$$\lim _{k \rightarrow \infty} a_{n_{k}}=\ell^{\prime}.$$

Then $\ell^{\prime} \geq \ell$.

Proof

We prove only (a) because the proof of (b) is similar. By Theorem 2.5.4 and the definition of limits, for any $\varepsilon>0$, there exists $N \in \mathbb{N}$ such that

$$a_{n}<\ell+\varepsilon \text { and } \ell^{\prime}-\varepsilon<a_{n_{k}}<\ell^{\prime}+\varepsilon$$

for all $n \geq N$ and $k \geq N$. Since $n_{N} \geq N$, this implies

$$\ell^{\prime}-\varepsilon<a_{n_{N}}<\ell+\varepsilon.$$

Thus, $\ell^{\prime}<\ell+2 \varepsilon$ and, hence, $\ell^{\prime} \leq \ell$ because $\mathcal{E}$ is arbitrary. $\square$

Remark $\PageIndex{8}$: Subsequential Limit

Let $\left\{a_{n}\right\}$ be a bounded sequence. Define

$$A=\left\{x \in \mathbb{R}: \text { there exists a subsequence }\left\{a_{n_{k}}\right\} \text { with } \lim a_{n_{k}}=x\right\}.$$

Each element of the set $A$ called a subsequential limit of the sequence $\left\{a_{n}\right\}$. It follows from Theorem 2.5.4, Theorem 2.5.5, and Corollary 2.5.7 that $A \neq \emptyset$ and

$$\limsup _{n \rightarrow \infty} a_{n}=\max A \text { and } \liminf _{n \rightarrow \infty} a_{n}=\min A.$$

Theorem $\PageIndex{9}$

Suppose $\left\{a_{n}\right\}$ is a sequence such that $a_{n}>0$ for every $n \in \mathbb{N}$ and

$$\limsup _{n \rightarrow \infty} \frac{a_{n+1}}{a_{n}}=\ell<1.$$

Then $\lim _{n \rightarrow \infty} a_{n}=0$.

Proof

Choose $\varepsilon>0$ such that $\ell+\varepsilon<1$. Then there exists $N \in \mathbb{N}$ such that

$$\frac{a_{n+1}}{a_{n}}<\ell+\varepsilon \text { for all } n \geq N.$$

Let $q=\ell+\varepsilon$. Then $0<q<1$. By induction,

$$0<a_{n} \leq q^{n-N} a_{N} \text { for all } n \geq N.$$

Since $\lim _{n \rightarrow \infty} q^{n-N} a_{N}=0$, one has $\lim _{n \rightarrow \infty} a_{n}=0$. $\square$

By a similar method, we obtain the theorem below.

Theorem $\PageIndex{10}$

Suppose $\left\{a_{n}\right\}$ is a sequence such that $a_{n}>0$ for every $n \in \mathbb{N}$ and

$$\liminf _{n \rightarrow \infty} \frac{a_{n+1}}{a_{n}}=\ell>1.$$

Then $\lim _{n \rightarrow \infty} a_{n}=\infty$.

Proof

Add proof here and it will automatically be hidden

Example $\PageIndex{1}$

Given a real number $\alpha$, define

$$a_{n}=\frac{\alpha^{n}}{n !}, n \in \mathbb{N}. \nonumber$$

Solution

When $\alpha =0$, it is obvious that $\lim _{n \rightarrow \infty} a_{n}=0$. Suppose $\alpha>0$. Then

$$\limsup _{n \rightarrow \infty} \frac{a_{n+1}}{a_{n}}=\lim _{n \rightarrow \infty} \frac{\alpha}{n+1}=0<1. \nonumber$$

Thus, $\lim _{n \rightarrow \infty} a_{n}=0$. In the general case, we can also show that $\lim _{n \rightarrow \infty} a_{n}=0$ by considering $\lim _{n \rightarrow \infty}\left|a_{n}\right|$ and using Exercise 2.1.3.

All sequences in this set of exercises are assumed to be in $\mathbb{R}$.

Exercise $\PageIndex{1}$

Find $\limsup _{n \rightarrow \infty} a_{n}$ and $\liminf _{n \rightarrow \infty} a_{n}$ for each sequence.

1. $a_{n}=(-1)^{n}$.
2. $a_{n}=\sin \left(\frac{n \pi}{2}\right)$.
3. $a_{n}=\frac{1+(-1)^{n}}{n}$.
4. $a_{n}=n \sin \left(\frac{n \pi}{2}\right)$.

Exercise $\PageIndex{2}$

For a sequence $\left\{a_{n}\right\}$, prove that:

1. $\liminf _{n \rightarrow \infty} a_{n}=\infty$ if and only if $\lim _{n \rightarrow \infty} a_{n}=\infty$.
2. $\limsup _{n \rightarrow \infty} a_{n}=-\infty$ if and only if $\lim _{n \rightarrow \infty} a_{n}=-\infty$.

Exercise $\PageIndex{3}$

Let $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ be bounded sequences. Prove that:

1. $\sup _{k \geq n}\left(a_{n}+b_{n}\right) \leq \sup _{k \geq n} a_{k}+\sup _{k \geq n} b_{k}$.
2. $\inf _{k \geq n}\left(a_{n}+b_{n}\right) \geq \inf _{k \geq n} a_{k}+\inf _{k \geq n} b_{k}$.

Exercise $\PageIndex{4}$

Let $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ be bounded sequences.

1. Prove that $\limsup _{n \rightarrow \infty}\left(a_{n}+b_{n}\right) \leq \limsup _{n \rightarrow \infty} a_{n}+\limsup _{n \rightarrow \infty} b_{n}$.
2. Prove that $\liminf _{n \rightarrow \infty}\left(a_{n}+b_{n}\right) \geq \liminf _{n \rightarrow \infty} a_{n}+\liminf _{n \rightarrow \infty} b_{n}$.
3. Find the two counter examples to show that the equalities may not hold in part (a) and part (b).

Exercise $\PageIndex{5}$

Let $\left\{a_{n}\right\}$ be a convergent sequence and let $\left\{b_{n}\right\}$ be an arbitrary sequence. Prove that

1. $\limsup _{n \rightarrow \infty}\left(a_{n}+b_{n}\right)=\limsup _{n \rightarrow \infty} a_{n}+\limsup _{n \rightarrow \infty} b_{n}=\lim _{n \rightarrow \infty} a_{n}+\limsup _{n \rightarrow \infty} b_{n}$.
2. $\liminf _{n \rightarrow \infty}\left(a_{n}+b_{n}\right)=\liminf _{n \rightarrow \infty} a_{n}+\liminf _{n \rightarrow \infty} b_{n}=\lim _{n \rightarrow \infty} a_{n}+\liminf _{n \rightarrow \infty} b_{n}$.