# Limit Superior and Inferior of a Sequence

## Definition 2.5.1

Let $\{s_n\}$ be a sequence in $\mathbb{R}$. The limit superior of $\{s_n\}$, denoted as $\limsup_{n\to\infty}s_n$, is defined as

\begin{align*} \limsup_{n\to\infty}s_n:=\lim_{k\to\infty}b_k:=\inf_{k\in\mathbb{N}}\sup\{s_n: n\geq k \} \in \mathbb{R} \cup\{ \pm \infty\} \end{align*}

The limit inferior of $\{s_n\}$, denoted as $\liminf_{n\to\infty}s_n$, defined as

\begin{align*} \liminf_{n\to\infty}s_n:=\lim_{k\to\infty}a_k :=\sup_{k\in\mathbb{N}}\inf\{s_n: n\geq k\} \in \mathbb{R} \cup\{ \pm \infty\} \end{align*}

## Theorem 2.5.3

Let $\{s_n\}$ be a sequence in $\mathbb{R}$.

(a) $\beta=\limsup_{n\to\infty}s_n$ is a real number if and only if

\begin{align} \forall \epsilon >0, \exists n_0\in\mathbb{N} \text{ such that } n\geq n_0 \Rightarrow s_n < \beta +\epsilon \label{eq:1} \\ \forall \epsilon >0, \forall n\in\mathbb{N}, \exists k\in\mathbb{N} \text{ with } k\geq n \text{ such that } s_k > \beta-\epsilon \label{eq:2} \end{align}

(b) $\limsup_{n\to\infty}s_n=\infty$ if and only if

\begin{align} \forall M\in\mathbb{R}, \forall n\in\mathbb{R}, \exists k\in\mathbb{N} \text{ with } k\geq n \text { such that } s_k\geq M \end{align}

(c) $\limsup_{n\to\infty}s_n = -\infty$ if and only if $\lim_{n\to\infty}s_n =-\infty$

<Proof>

$\Rightarrow$ Let $\beta= \limsup_{n\to\infty}s_n \in \mathbb{R}$ and let $a_k = \sup_{n\geq k} s_n$. Since $\lim_{k\to\infty}a_k=\beta$, there exists a $n_0\in\mathbb{N}$ such that

\begin{align*} n\geq n_0 \Rightarrow a_n < \beta + \epsilon. \end{align*}

Since $a_{n_0}=\sup\{s_k: k\geq n_0 \}$,

\begin{align*} n\geq n_0 \Rightarrow s_n \leq a_{n_0}<\beta + \epsilon. \end{align*}

On the other hand, since the sequence $\{a_k \}$ is decreasing, $a_n \geq \beta$ for all $n\in\mathbb{N}$. That is

\begin{align*} \sup\{s_k: k\geq n\}=a_n\geq \beta > \beta-\epsilon. \end{align*}

Therefore for each $n\in\mathbb{N}$, there is $k\in\mathbb{N}$ with $k\geq n$ such that $s_k > \beta-\epsilon$.

$\Leftarrow$ Conversely, suppose that Equation $\ref{eq:1}$ and $\ref{eq:2}$ hold.

Let $\epsilon >0$ be given. By Equation $\ref{eq:1}$, there exists a $n_0\in\mathbb{N}$ such that $n\geq n_0 \Rightarrow s_n < \beta + \epsilon$. Since $\beta+\epsilon$ is an upper bound of $\{s_n: n\geq n_0 \}$,

\begin{align*} a_{n_0}=\sup\{s_n: n\geq n_0 \} \leq \beta+\epsilon. \end{align*}

Since the sequence $\{ a_n \}$ is decreasing,

\begin{align*} a_n \leq a_{n_0} \leq \beta +\epsilon \text{ if } n\geq n_0. \end{align*}

Thus,

\begin{align*} \limsup_{n\to\infty}s_n=\lim_{n\to\infty}a_n \leq \beta+\epsilon. \end{align*}

Let $\beta^\prime := \limsup_{n\to\infty}s_n$. Since the choice of $\epsilon$ is an arbitrary, $\beta^\prime \leq \beta$. Now we want to show that $\beta^\prime = \beta$.

Suppose that $\beta^\prime \lneq \beta$. Then we take $\epsilon_0 > 0$ such that

\begin{align*} \beta^\prime + \epsilon < \beta -\epsilon_0 \end{align*}

For an instance, $\epsilon_0 := (\beta-\beta^\prime) / 3$. Since $\limsup_{n\to\infty}s_n=\beta^\prime\in\mathbb{R}$, there exits $n_0\in\mathbb{N}$ such that

\begin{align*} n\geq n_0 \Rightarrow s_n<\beta^\prime +\epsilon < \beta-\epsilon_0 \end{align*}

by Equation $\ref{eq:1}$, which contradicts to Equation $\ref{eq:2}$.

$\therefore \beta=\beta^\prime$.

(b) $\Rightarrow$ Suppose that $\limsup_{n\to\infty}s_n = \infty$. Let $a_k = \sup_{n\geq k}s_n$. Since $\{a_n \}$ is decreasing sequence, $a_n=\infty$ for all $n\in\mathbb{N}$. Since $a_n = \sup \{s_k : k\geq n \}$, the set $\{s_k : k\geq n \}$ is not bounded above.

$\therefore$ For all $M\in\mathbb{R},$ there is $k\in\mathbb{N}$ with $k\geq n$ such that $s_k > M$.

$\Leftarrow$ Suppose that for all $M\in\mathbb{R}$ and for all $n\in\mathbb{N}$, there is a $k\in\mathbb{N}$ with $k\geq n$ such that $s_k \geq M$. Then the set $E_n :=\{s_k : k\geq n \}$ is not bounded above and thus $a_n=\sup E_n=\infty$ for all $n\in\mathbb{N}$.

$\therefore \limsup_{n\to\infty}s_n = \inf_{n\in\mathbb{N}}a_n=\infty$

(c) $\Rightarrow$ Suppose that $\limsup_{n\to\infty}s_n = -\infty$. Let $a_n=\sup\{ s_k: k\geq n\}$.

Let $M<0$ be given. Since $a_n\to -\infty$ as $n\to\infty$, there exists $n_0\in\mathbb{N}$ such that $n\geq n_0 \Rightarrow a_n \leq M.$ Since $s_n\leq a_n, s_n \leq a_n <M$ for all $n\geq n_0$.

$\therefore \limsup_{n\to\infty}s_n = -\infty$.

$\Leftarrow$ Suppose that $\lim_{n\to\infty}s_n = -\infty$. Let $M<0$ be given. Then there is $n_0\in\mathbb{N}$ such that $n\geq n_0\Rightarrow s_n \leq M$.

Since $a_{n_0}$ is the least upper bound of $\{ s_n: n\geq n_0\}, a_{n_0}\leq M$ and $s_n \leq M$ for all $n\geq n_0$,

\begin{align*} s_n \leq a_{n_0} \leq M \text{ for all } n\geq n_0. \end{align*}

Since $\{ a_n \}$ is monotone decreasing, $n\geq n_0 \Rightarrow a_n \leq a_{n_0}\leq M$.

$\therefore \limsup_{n\to\infty}s_n = \lim_{n\to\infty}a_n = -\infty$.

$\tag*{\square}$

## Theorem 2.5.4

Let $\{s_n\}$ be a sequence in $\mathbb{R}$.

(a) $\liminf_{n\to\infty}s_n=\alpha\in\mathbb{R}$ if and only if

\begin{align} \forall\epsilon >0, \exists n_0\in\mathbb{N} \text{ s.t. } n\geq n_0\Rightarrow s_n>\alpha-\epsilon \label{eq:4} \\ \forall\epsilon>0, \forall n\in\mathbb{N}, \exists k\in\mathbb{N} \text{ with } k\geq n \text{ s.t. } s_k < \alpha + \epsilon \label{eq:5} \end{align}

<Proof>

$\Rightarrow$ Let $\epsilon >0$ be given. Define $b_k :=\inf\{s_n: n\geq k \}$. Since $\liminf_{n\to\infty}=\lim_{k\to\infty}b_k=\alpha \in\mathbb{R}$, there exists $n_0\in\mathbb{N}$ such that

\begin{align*} k\geq n_0 \Rightarrow b_k > \alpha-\epsilon \end{align*}

Since $b_k$ is infimum of $\{ s_n: n\geq k\}$,

\begin{align*} s_n\geq b_k > \alpha-\epsilon \text{ for all } n \geq k \geq n_0. \end{align*}

$\therefore s_n > \alpha- \epsilon$ for $n\geq n_0$.

On the other hand, $\{b_k \}$ is increasing sequence,

\begin{align*} \inf\{s_k: k\geq n\}=b_n \leq \alpha < \alpha + \epsilon. \end{align*}

Since $\alpha$ is the greatest lower bound of $\{b_n: n=1,2,\ldots \}, \alpha+\epsilon$ is not a lower bound anymore.

$\therefore s_k < \alpha +\epsilon$ for some $k\geq n$.

$\Leftarrow$ Conversely, suppose that Equation $\ref{eq:4}$ and $\ref{eq:5}$ hold.

Let $\epsilon >0$ be given. By Equation $\ref{eq:4}$, there is a $n_0\in\mathbb{N}$ such that

\begin{align*} n\geq n_0 \Rightarrow s_n > \alpha-\epsilon. \end{align*}

In other words, $\alpha-\epsilon$ is a lower bound of $\{s_n: n\geq n_0\}$ and thus

\begin{align*} b_{n_0}=\inf\{s_n: n\geq n_0\}\geq \alpha-\epsilon \end{align*}

Since $\{b_k\}$ is monotone increasing, $b_n\geq b_{n_0}$ for all $n\geq n_0$, i.e., $b_n \geq \alpha -\epsilon$ for $n\geq n_0$.

Thus, $\liminf_{n\to\infty}s_n = \lim_{n\to\infty}b_n \geq \alpha-\epsilon$. Since the choice of $\epsilon >0$ is an arbitrary, $\alpha^\prime :=\liminf_{n\to\infty}s_n \geq \alpha$.

Now we want to show that $\alpha^\prime = \alpha$. Suppose that $\alpha^\prime \gneq\alpha$. Then we can take $\epsilon_0>0$ such that $\alpha^\prime -\epsilon_0 > \alpha +\epsilon_0$.

Since $\alpha^\prime = \liminf_{n\to\infty}s_n$, there exits a $n_0\in\mathbb{N}$ such that

\begin{align*} n\geq n_0 \Rightarrow s_n > \alpha^\prime-\epsilon > \alpha+\epsilon_0, \end{align*} which contradicts to the inequality $\ref{eq:5}$.

$\therefore \alpha=\liminf_{n\to\infty}s_n$.

$\tag*{\square}$

(b) $\liminf_{n\to\infty}s_n=-\infty$ if and only if

\begin{align} \forall M\in\mathbb{R}, \forall n\in\mathbb{N}, \exists k\in\mathbb{N} \text{ with } k\geq n \text{ such that } s_k \leq M \label{eq:6} \end{align}

<Proof>

$\Rightarrow$ Suppose that $\liminf_{n\to\infty}s_n=-\infty$. Let $b_k :=\inf \{s_n: n\geq k\}$. Since $\lim\inf_{n\to\infty}s_n = \sup_{k\in\mathbb{N}}b_k=-\infty$ $b_n=-\infty$ for all $n\in\mathbb{N}$.

Since $\inf\{s_k: k\geq n \}=b_n$, the set $\{s_k: k\geq n \}$ is not bounded below. Thus, for any $M\in\mathbb{R}$, there is $k\in\mathbb{N}$ with $k\geq n$ such that $s_k > M$.

$\therefore$ for all $M\in\mathbb{R}$ for all $n\in\mathbb{N}$, there is $k\in\mathbb{N}$ with $k\geq n$ such that $s_k > M$.

$\Leftarrow$ Let $b_n =\inf\{s_k: k\geq n \}$. For all $M\in\mathbb{R},$ and for all $n\in\mathbb{N}$, there is $k\in\mathbb{N}$ with $k\geq n$ such that $s_k \leq M$ by the assumption $\ref{eq:6}$. It implies that the set $E_n:=\{s_k: k\geq n \}$ is not bounded below for all $n\in\mathbb{N}$ and thus $b_n = -\infty$ for all $n\in\mathbb{N}$.

$\therefore \liminf_{n\to\infty}=\sup_{n\in\mathbb{N}}b_n=-\infty$

(c) $\liminf_{n\to\infty}s_n =\infty$ if and only if $s_n\to\infty$ as $n\to\infty$.

<Proof>

$\Rightarrow$ Suppose that $\liminf_{n\to\infty}s_n = \infty$. Let $b_n :=\inf \{s_k : k\geq n\}$. Then $s_n \geq b_n$.

Let $M>0$ be given. Since $\liminf_{n\to\infty}s_n=\lim_{n\to\infty}b_n=\infty$, there exits a $n_0\in\mathbb{N}$ such that $n\geq n_0 \Rightarrow b_n >M$.

Since $s_n \geq b_n, s_n > M$ for all $n\geq n_0$.

$\therefore \lim_{n\to\infty}s_n = \infty$.

$\Leftarrow$ Conversely, suppose that $\lim_{n\to\infty}s_n=\infty$. Let $M>0$ be given. Then there exists $n_0\in\mathbb{N}$ such that $n\geq n_0 \Rightarrow s_n > M$. Then $M$ is a lower bound of the set $\{s_n : n\geq n_0\}$. Then $b_{n_0} = \inf \{s_n: n\geq n_0\}\geq M$.

Since $\{b_n\}$ is monotone increasing, $b_n \geq M$ for all $n\geq n_0$.

$\therefore \lim\inf_{n\to\infty}=\lim_{n\to\infty}b_n = \infty$.

$\tag*{\square}$

## Theorem 2.5.6

Let $\{a_n\}, \{b_n\}$ be bounded sequences in $\mathbb{R}$. Then

\begin{align} \lim\inf_{n\to\infty}a_n + \liminf_{n\to\infty}b_n &\leq \lim\inf_{n\to\infty}(a_n+b_b) \\ &\leq\liminf_{n\to\infty}a_n + \limsup_{n\to\infty}b_n \\ &\leq \limsup_{n\to\infty}(a_n+b_n) \\ &\leq \limsup_{n\to\infty}a_n + \limsup_{n\to\infty}b_n \end{align}

## Theorem 2.5.7

Let $\{s_n \}$ be a sequence in $\mathbb{R}$ and let $E$ be the set of subsequential limits of $\{s_n\}$ in $\mathbb{R}$ with $\pm\infty$.

Then $\limsup_{n\to\infty}, \liminf_{n\to\infty}\in E$ and

\begin{align*} \limsup_{n\to\infty}s_n = \sup E, \quad \liminf_{n\to\infty}s_n = \inf E. \end{align*}

<Proof>

1) Suppose that $s=\limsup_{n\to\infty}\in\mathbb{R}$.

We want to show that there is a subsequence $\{s_{n_k}\}$ such that $\lim_{k\to\infty}s_{n_k}=s$.

Take $\epsilon =1$. Then there is $N_1\in\mathbb{N}$ such that $n\geq N_1 \Rightarrow s_n < s+1$ by $\ref{eq:1}$. Moreover there is $k\in\mathbb{N}$ with $k \geq N_1$ such that $s_{k} > s+1$ by $\ref{eq:2}$. Then take $n_1$ with the smallest integer such that $s-1<s_{n_1} < s+1$.

For the next step, take $\epsilon=1/2$. Then there is $M_2 \in\mathbb{N}$ such that $n\geq M_2 \Rightarrow s_n < s+1$ by $\ref{eq:1}$. Take $N_2:=\max\{n_1, M_2 \}+1$. Moreover there is $k\in\mathbb{N}$ with $k \geq N_2$ such that $s_{k} > s+1$ by $\ref{eq:2}$. Then take $n_2$ with the smallest integer such that $s-\frac{1}{2}<s_{n_2} < s+\frac{1}{2}$.

Repeat this process and we get strictly increasing sequence $\{n_k\}$ and $s-\frac{1}{k} < s_{n_k} < s+\frac{1}{k}$.

$\therefore s_{n_k}\to s$ as $k\to\infty$

$\therefore \limsup_{n\to\infty}=s\in E$.

2) Suppose that $\limsup_{n\to\infty}s_n=\infty$. By Theorem 2.5.3, for all $M\in\mathbb{R}$, for all $k\in\mathbb{N}$, there is a $n\in\mathbb{N}$ with $n\geq k$ such that $s_n \geq M$.

Take $M=1$ and $k=1$. Then there is $n_1 \geq 1$ such that $s_{n_1} \geq 1$.

For the next step, take $M:=s_{n_1}+1$ and $k=n_1 +1$. Then there is $n_2 \geq n_1 +1$ such that $s_{n_2} \geq s_{n_1}+1$.

Repeat this process. Then we get the subsequence $\{s_{n_k}\}$ which is monotone increasing and not bounded above, i.e., $\lim_{k\to\infty}s_{n_k}=\infty$.

$\therefore \limsup_{n\to\infty}=\infty \in E$.

Similarly we can prove that $\liminf_{n\to\infty}s_n \in E$.

Next, we want to show that $s=\sup E$.

Since $s\in E, s\leq \sup E =:\beta$. If $s=\infty, s= \sup E$. Otherwise, suppose that $s \lneq \beta$.

Since $\beta$ is the least upper bound, there is $\alpha \in E$ such that $s<\alpha \leq \beta$. Take $\epsilon>0$ such that $s+\epsilon < \alpha -\epsilon$. By $\ref{eq:1}$, there is $n_0\in\mathbb{N}$ such that

\begin{align*} n\geq n_0 \Rightarrow s_n < s_n+\epsilon < \alpha -\epsilon. \end{align*}

It means that there are finitely many $s_n$ such that $s_n > \alpha -\epsilon$. It implies that there is no subsequence $\{ s_{n_k}\}$ such that $s_{n_k}\to\alpha$ as $k\to\infty$. But $\alpha \in E$ is a subsequential limit of of $\{s_n\}$, which leads to a contradiction.

$\therefore s=\sup E$.

Finally, we want to show that $\liminf_{n\to\infty} s_n=\inf E.$

Let $s:=\liminf_{n\to\infty}s_n\in\mathbb{R}$ (If $s=-\infty$, it is trivial to show that $s=\inf E$.) and let $\beta=\inf E$. Then $s\geq \beta$.

Suppose that $s \gneq\beta$. We can take $\epsilon>0$ such that $\alpha+\epsilon < \beta -\epsilon$. By $\ref{eq:4}$, there is $n_0\in\mathbb{N}$ such that $n\geq n_0 \Rightarrow \alpha + \epsilon < \beta-\epsilon < s_n$. It means that there are only finitely many $s_n$ such that $s_n < \alpha +\epsilon$, which implies that there is no subsequence $\{s_{n_k}\}$ such that $s_{n_k}\to\alpha$ as $k\to\infty$. But it is a contradiction to the assumption that $\alpha\in E$ is a subsequential limit of $\{s_n\}$.

$\therefore \liminf_{n\to\infty}s_n =\inf E$.

$\tag*{\square}$

## Theorem

Let $X$ be a metric space and let $\{p_n\}$ be a sequence in $X$. Then the set of all the subsequential limits of $\{p_n\}$ is closed.

<Proof>

Let $E$ be the set of subsequential limits of $\{p_n\}$ and let $p\in E^\prime$. It suffices to show that $p\in E$.

Since $p$ is a limit point of $E$, for every $\varepsilon_n=1/2^{n+1}$, there is a subsequential limit $l_n\in N^\prime_{\varepsilon_n}(p)\cap E$. Since $l_n$ is a subsequential limit of $\{p_n\}$, there is $N\in\mathbb{N}$ such that $k\geq N \Rightarrow p_{n_k} \in (l_n-\varepsilon_n, l_n +\varepsilon_n)$. In other words there are infinitely many $p_{n}$ in $N_{\varepsilon_n}(l_n)$.

So, we can find a strictly increasing sequence $\{n_k\}$ such that $p_{n_k}\in N_{\varepsilon_k}(l_k)$ for all $k\in\mathbb{N}$. Then with the triangular inequality,

\begin{align*} d(p_{n_k}, p) &\leq d(p_{n_k}, l_k) + d(l_k, p) \\ &=2\varepsilon_k \\ &= \frac{1}{2^k} \end{align*}

$\therefore p_{n_k}\to p$ as $k\to\infty$.

$\tag*{\square}$

## Reference

• Manfred Stoll, Introduction to Real Analysis, Pearson

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