2.5: Limit Superior and Limit Inferior
- Page ID
- 49103
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)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.
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. \(\square\)
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).
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. \(\square\)
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
\[\liminf _{n \rightarrow \infty} a_{n}=-\infty.\]
Let \(\left\{a_{n}\right\}\) be a sequence and \(\ell \in \mathbb{R}\). The following are equivalent:
- \(\limsup _{n \rightarrow \infty} a_{n}=\ell\).
- 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.
Let \(\left\{a_{n}\right\}\) be a sequence and \(\ell \in \mathbb{R}\). The following are equivalent:
- \(\liminf _{n \rightarrow \infty} a_{n}=\ell\).
- 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
-
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The following corollary follows directly from Theorems 2.5.4 and 2.5.5.
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
Let \(\left\{a_{n}\right\}\) be a sequence.
- 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 \(\ell^{\prime} \leq \ell\).
- 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. \(\square\)
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
\[\limsup _{n \rightarrow \infty} a_{n}=\max A \text { and } \liminf _{n \rightarrow \infty} a_{n}=\min A.\]
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.
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
-
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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.
- \(a_{n}=(-1)^{n}\).
- \(a_{n}=\sin \left(\frac{n \pi}{2}\right)\).
- \(a_{n}=\frac{1+(-1)^{n}}{n}\).
- \(a_{n}=n \sin \left(\frac{n \pi}{2}\right)\).
Exercise \(\PageIndex{2}\)
For a sequence \(\left\{a_{n}\right\}\), prove that:
- \(\liminf _{n \rightarrow \infty} a_{n}=\infty\) if and only if \(\lim _{n \rightarrow \infty} a_{n}=\infty\).
- \(\limsup _{n \rightarrow \infty} a_{n}=-\infty\) 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:
- \(\sup _{k \geq n}\left(a_{n}+b_{n}\right) \leq \sup _{k \geq n} a_{k}+\sup _{k \geq n} b_{k}\).
- \(\inf _{k \geq n}\left(a_{n}+b_{n}\right) \geq \inf _{k \geq n} a_{k}+\inf _{k \geq n} b_{k}\).
Exercise \(\PageIndex{4}\)
Let \(\left\{a_{n}\right\}\) and \(\left\{b_{n}\right\}\) be bounded sequences.
- Prove that \(\limsup _{n \rightarrow \infty}\left(a_{n}+b_{n}\right) \leq \limsup _{n \rightarrow \infty} a_{n}+\limsup _{n \rightarrow \infty} b_{n}\).
- Prove that \(\liminf _{n \rightarrow \infty}\left(a_{n}+b_{n}\right) \geq \liminf _{n \rightarrow \infty} a_{n}+\liminf _{n \rightarrow \infty} b_{n}\).
- 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
- \(\limsup _{n \rightarrow \infty}\left(a_{n}+b_{n}\right)=\limsup _{n \rightarrow \infty} a_{n}+\limsup _{n \rightarrow \infty} b_{n}=\lim _{n \rightarrow \infty} a_{n}+\limsup _{n \rightarrow \infty} b_{n}\).
- \(\liminf _{n \rightarrow \infty}\left(a_{n}+b_{n}\right)=\liminf _{n \rightarrow \infty} a_{n}+\liminf _{n \rightarrow \infty} b_{n}=\lim _{n \rightarrow \infty} a_{n}+\liminf _{n \rightarrow \infty} b_{n}\).