Section2.3Limit superior, limit inferior, and Bolzano–Weierstrass

Note: 1–2 lectures, alternative proof of BW optional

In this section we study bounded sequences and their subsequences. In particular, we define the so-called limit superior and limit inferior of a bounded sequence and talk about limits of subsequences. Furthermore, we prove the Bolzano–Weierstrass theorem^{ 1 }, an indispensable tool in analysis, showing the existence of convergent subsequences.

We proved that every convergent sequence is bounded; nevertheless, there exist many bounded divergent sequences. For instance, the sequence \(\bigl\{ {(-1)}^n \bigr\}_{n=1}^\infty\) is bounded, but divergent. All is not lost, however, and we can still compute certain limits with a bounded divergent sequence.

Subsection2.3.1Upper and lower limits

There are ways of creating monotone sequences out of any sequence, and in this fashion we get the so-called limit superior and limit inferior. These limits always exist for bounded sequences.

If a sequence \(\{ x_n \}_{n=1}^\infty\) is bounded, then the set \(\{ x_k : k \in \N \}\) is bounded. For every \(n\text{,}\) the set \(\{ x_k : k \geq n \}\) is also bounded (as it is a subset), so we take its supremum and infimum.

Definition2.3.1.

Let \(\{ x_n \}_{n=1}^\infty\) be a bounded sequence. Define the sequences \(\{ a_n \}_{n=1}^\infty\) and \(\{ b_n \}_{n=1}^\infty\) by \(a_n \coloneqq \sup \{ x_k : k \geq n \}\) and \(b_n \coloneqq \inf \{ x_k : k \geq n \}\text{.}\) Define, if the limits exist,

For a bounded sequence, liminf and limsup always exist (see below). It is possible to define liminf and limsup for unbounded sequences if we allow \(\infty\) and \(-\infty\text{,}\) and we do so later in this section. It is not hard to generalize the following results to include unbounded sequences; however, we first restrict our attention to bounded ones.

Proposition2.3.2.

Let \(\{ x_n \}_{n=1}^\infty\) be a bounded sequence. Let \(a_n\) and \(b_n\) be as in the definition above.

The sequence \(\{ a_n \}_{n=1}^\infty\) is bounded monotone decreasing and \(\{ b_n \}_{n=1}^\infty\) is bounded monotone increasing. In particular, \(\liminf\limits_{n\to\infty} x_n\) and \(\limsup\limits_{n\to\infty} x_n\) exist.

\(\displaystyle \limsup_{n \to \infty} x_n = \inf \{ a_n : n \in \N \}\) and \(\displaystyle \liminf_{n \to \infty} x_n = \sup \{ b_n : n \in \N \}\text{.}\)

Let us see why \(\{ a_n \}_{n=1}^\infty\) is a decreasing sequence. As \(a_n\) is the least upper bound for \(\{ x_k : k \geq n \}\text{,}\) it is also an upper bound for the subset \(\{ x_k : k \geq n+1 \}\text{.}\) Therefore \(a_{n+1}\text{,}\) the least upper bound for \(\{ x_k : k \geq n+1 \}\text{,}\) has to be less than or equal to \(a_n\text{,}\) the least upper bound for \(\{ x_k : k \geq n) \}\text{.}\) That is, \(a_n \geq a_{n+1}\) for all \(n\text{.}\) Similarly (an exercise), \(\{ b_n \}_{n=1}^\infty\) is an increasing sequence. It is left as an exercise to show that if \(\{ x_n \}_{n=1}^\infty\) is bounded, then \(\{ a_n \}_{n=1}^\infty\) and \(\{ b_n \}_{n=1}^\infty\) must be bounded.

The second item follows as the sequences \(\{ a_n \}_{n=1}^\infty\) and \(\{ b_n \}_{n=1}^\infty\) are monotone and bounded.

For the third item, note that \(b_n \leq a_n\text{,}\) as the \(\inf\) of a nonempty set is less than or equal to its \(\sup\text{.}\) The sequences \(\{ a_n \}_{n=1}^\infty\) and \(\{ b_n \}_{n=1}^\infty\) converge to the limsup and the liminf respectively. Apply Lemma 2.2.3 to obtain

\begin{equation*}
\limsup_{n\to\infty} x_n =
\lim_{n\to\infty}
\bigl(
\sup \{ x_k : k \geq n \}
\bigr) .
\end{equation*}

It is not hard to see that

\begin{equation*}
\sup \{ x_k : k \geq n \} =
\begin{cases}
\frac{n+1}{n} & \text{if } n \text{ is odd,} \\
\frac{n+2}{n+1} & \text{if } n \text{ is even.}
\end{cases}
\end{equation*}

We leave it to the reader to show that the limit is 1. That is,

Do note that the sequence \(\{ x_n \}_{n=1}^\infty\) is not a convergent sequence.

We associate certain subsequences with \(\limsup\) and \(\liminf\text{.}\) It is important to notice that \(\{ a_n \}_{n=1}^\infty\) and \(\{ b_n \}_{n=1}^\infty\) are not subsequences of \(\{ x_n \}_{n=1}^\infty\text{,}\) nor do they have to even consist of the same numbers. For example, for the sequence \(\{ \nicefrac{1}{n} \}_{n=1}^\infty\text{,}\)\(b_n = 0\) for all \(n \in \N\text{.}\)

Theorem2.3.4.

If \(\{ x_n \}_{n=1}^\infty\) is a bounded sequence, then there exists a subsequence \(\{ x_{n_k} \}_{k=1}^\infty\) such that

Define \(a_n \coloneqq \sup \{ x_k : k \geq n \}\text{.}\) Write \(x \coloneqq \limsup_{n\to\infty} x_n = \lim_{n\to\infty} a_n\text{.}\) We define the subsequence inductively. Let \(n_1 \coloneqq 1\text{,}\) and suppose \(n_1,n_2,\ldots,n_{k-1}\) are already defined for some \(k \geq 2\text{.}\) Pick an \(m \geq n_{k-1} + 1\) such that

Such an \(m\) exists as \(a_{(n_{k-1}+1)}\) is a supremum of the set \(\{ x_\ell : \ell \geq n_{k-1} + 1 \}\) and hence there are elements of the sequence arbitrarily close (or even possibly equal) to the supremum. Set \(n_{k} \coloneqq m\text{.}\) The subsequence \(\{ x_{n_k} \}_{k=1}^\infty\) is defined. Next, we must prove that it converges to \(x\text{.}\)

For all \(k \geq 2\text{,}\) we have \(a_{(n_{k-1}+1)} \geq a_{n_k}\) (why?) and \(a_{n_{k}} \geq x_{n_k}\text{.}\) Therefore, for every \(k \geq 2\text{,}\)

Let us show that \(\{ x_{n_k} \}_{k=1}^\infty\) converges to \(x\text{.}\) Note that the subsequence need not be monotone. Let \(\epsilon > 0\) be given. As \(\{ a_n \}_{n=1}^\infty\) converges to \(x\text{,}\) the subsequence \(\{ a_{n_k} \}_{k=1}^\infty\) converges to \(x\text{.}\) Thus there exists an \(M_1 \in \N\) such that for all \(k \geq M_1\text{,}\) we have

We leave the statement for \(\liminf\) as an exercise.

Subsection2.3.2Using limit inferior and limit superior

The advantage of \(\liminf\) and \(\limsup\) is that we can always write them down for any (bounded) sequence. If we could somehow compute them, we could also compute the limit of the sequence if it exists, or show that the sequence diverges. Working with \(\liminf\) and \(\limsup\) is a little bit like working with limits, although there are subtle differences.

Proposition2.3.5.

Let \(\{ x_n \}_{n=1}^\infty\) be a bounded sequence. Then \(\{ x_n \}_{n=1}^\infty\) converges if and only if

First suppose \(\liminf_{n\to\infty} x_n = \limsup_{n\to\infty} x_n\text{.}\) Then \(\{ a_n \}_{n=1}^\infty\) and \(\{ b_n \}_{n=1}^\infty\) both converge to the same limit. By the squeeze lemma (Lemma 2.2.1), \(\{ x_n \}_{n=1}^\infty\) converges and

Now suppose \(\{ x_n \}_{n=1}^\infty\) converges to \(x\text{.}\) By Theorem 2.3.4, there exists a subsequence \(\{ x_{n_k} \}_{k=1}^\infty\) converging to \(\limsup_{n\to\infty} x_n\text{.}\) As \(\{ x_n \}_{n=1}^\infty\) converges to \(x\text{,}\) every subsequence converges to \(x\) and so \(\limsup_{n\to\infty} x_n = \lim_{k\to\infty} x_{n_k} = x\text{.}\) Similarly, \(\liminf_{n\to\infty} x_n = x\text{.}\)

Limit superior and limit inferior behave nicely with subsequences.

Proposition2.3.6.

Suppose \(\{ x_n \}_{n=1}^\infty\) is a bounded sequence and \(\{ x_{n_k} \}_{k=1}^\infty\) is a subsequence. Then

The middle inequality has been proved already. We will prove the third inequality, and leave the first inequality as an exercise.

We want to prove that \(\limsup_{k\to\infty} x_{n_k} \leq \limsup_{n\to\infty} x_n\text{.}\) Define \(a_n \coloneqq \sup \{ x_k : k \geq n \}\) as usual. Also define \(c_n \coloneqq \sup \{ x_{n_k} : k \geq n \}\text{.}\) It is not true that \(\{ c_n \}_{n=1}^\infty\) is necessarily a subsequence of \(\{ a_n \}_{n=1}^\infty\text{.}\) However, as \(n_k \geq k\) for all \(k\text{,}\) we have \(\{ x_{n_k} : k \geq n \} \subset \{ x_k : k \geq n \}\text{.}\) A supremum of a subset is less than or equal to the supremum of the set, and therefore

\begin{equation*}
c_n \leq a_n \qquad \text{for all $n$}.
\end{equation*}

Limit superior and limit inferior are the largest and smallest subsequential limits. If the subsequence \(\{ x_{n_k} \}_{k=1}^\infty\) in the previous proposition is convergent, then \(\liminf_{k\to\infty} x_{n_k} = \lim_{k\to\infty} x_{n_k} =
\limsup_{k\to\infty} x_{n_k}\text{.}\) Therefore,

Similarly, we get the following useful test for convergence of a bounded sequence. We leave the proof as an exercise.

Proposition2.3.7.

A bounded sequence \(\{ x_n \}_{n=1}^\infty\) is convergent and converges to \(x\) if and only if every convergent subsequence \(\{ x_{n_k} \}_{k=1}^\infty\) converges to \(x\text{.}\)

Subsection2.3.3Bolzano–Weierstrass theorem

While it is not true that a bounded sequence is convergent, the Bolzano–Weierstrass theorem tells us that we can at least find a convergent subsequence. The version of Bolzano–Weierstrass we present in this section is the Bolzano–Weierstrass for sequences of real numbers.

Theorem2.3.8.Bolzano–Weierstrass.

Suppose a sequence \(\{ x_n \}_{n=1}^\infty\) of real numbers is bounded. Then there exists a convergent subsequence \(\{ x_{n_i} \}_{i=1}^\infty\text{.}\)

Proof.

Theorem 2.3.4 says that there exists a subsequence whose limit is \(\limsup_{n\to\infty} x_n\text{.}\)

The reader might complain right now that Theorem 2.3.4 is strictly stronger than the Bolzano–Weierstrass theorem as presented above. That is true. However, Theorem 2.3.4 only applies to the real line, but Bolzano–Weierstrass applies in more general contexts (that is, in \(\R^n\)) with pretty much the exact same statement.

As the theorem is so important to analysis, we present an explicit proof. The idea of the following proof also generalizes to different contexts.

Proof.

(Alternate proof of Bolzano–Weierstrass) As the sequence is bounded, then there exist two numbers \(a_1 < b_1\) such that \(a_1 \leq x_n \leq b_1\) for all \(n \in \N\text{.}\) We will define a subsequence \(\{ x_{n_i} \}_{i=1}^\infty\) and two sequences \(\{ a_i \}_{i=1}^\infty\) and \(\{ b_i \}_{i=1}^\infty\text{,}\) such that \(\{ a_i \}_{i=1}^\infty\) is monotone increasing, \(\{ b_i \}_{i=1}^\infty\) is monotone decreasing, \(a_i \leq x_{n_i} \leq b_i\) and such that \(\lim_{i\to\infty} a_i =
\lim_{i\to\infty} b_i\text{.}\) That \(x_{n_i}\) converges then follows by the squeeze lemma.

We define the sequences inductively. We will define the sequences so that for all \(i\text{,}\) we have \(a_i < b_i\text{,}\) and that \(x_n \in [a_i,b_i]\) for infinitely many \(n \in \N\text{.}\) We have already defined \(a_1\) and \(b_1\text{.}\) We take \(n_1 \coloneqq 1\text{,}\) that is \(x_{n_1} = x_1\text{.}\) Suppose that up to some \(k \in \N\text{,}\) we have defined the subsequence \(x_{n_1}, x_{n_2}, \ldots,
x_{n_k}\text{,}\) and the sequences \(a_1,a_2,\ldots,a_k\) and \(b_1,b_2,\ldots,b_k\text{.}\) Let \(y \coloneqq \frac{a_k+b_k}{2}\text{.}\) Clearly \(a_k < y < b_k\text{.}\) If there exist infinitely many \(j \in \N\) such that \(x_j \in [a_k,y]\text{,}\) then set \(a_{k+1} \coloneqq a_k\text{,}\)\(b_{k+1}
\coloneqq y\text{,}\) and pick \(n_{k+1} > n_{k}\) such that \(x_{n_{k+1}} \in [a_k,y]\text{.}\) If there are not infinitely many \(j\) such that \(x_j \in [a_k,y]\text{,}\) then it must be true that there are infinitely many \(j \in
\N\) such that \(x_j \in [y,b_k]\text{.}\) In this case pick \(a_{k+1} \coloneqq y\text{,}\)\(b_{k+1}
\coloneqq b_k\text{,}\) and pick \(n_{k+1} > n_{k}\) such that \(x_{n_{k+1}} \in [y,b_k]\text{.}\)

We now have the sequences defined. What is left to prove is that \(\lim_{i\to\infty} a_i = \lim_{i\to\infty} b_i\text{.}\) The limits exist as the sequences are monotone. In the construction, \(b_i - a_i\) is cut in half in each step. Therefore, \(b_{i+1} - a_{i+1} = \frac{b_i-a_i}{2}\text{.}\) By induction,

Let \(x \coloneqq \lim_{i\to\infty} a_i\text{.}\) As \(\{ a_i \}_{i=1}^\infty\) is monotone,

\begin{equation*}
x = \sup \{ a_i : i \in \N \} .
\end{equation*}

Let \(y \coloneqq \lim_{i\to\infty} b_i = \inf \{ b_i : i \in \N \}\text{.}\) Since \(a_i < b_i\) for all \(i\text{,}\) then \(x \leq y\text{.}\) As the sequences are monotone, then for all \(i\text{,}\) we have (why?)

Because \(\frac{b_1-a_1}{2^{i-1}}\) is arbitrarily small and \(y-x \geq 0\text{,}\) we have \(y-x = 0\text{.}\) Finish by the squeeze lemma.

Yet another proof of the Bolzano–Weierstrass theorem is to show the following claim, which is left as a challenging exercise. Claim: Every sequence has a monotone subsequence.

Subsection2.3.4Infinite limits

Just as for infima and suprema, it is possible to allow certain limits to be infinite. That is, we write \(\lim_{n\to\infty} x_n = \infty\) or \(\lim_{n\to\infty} x_n = -\infty\) for certain divergent sequences.

Definition2.3.9.

We say \(\{ x_n \}_{n=1}^\infty\)diverges to infinity^{ 2 } if for every \(K \in
\R\text{,}\) there exists an \(M \in \N\) such that for all \(n \geq M\text{,}\) we have \(x_n >
K\text{.}\) In this case we write

Similarly, if for every \(K \in \R\) there exists an \(M \in \N\) such that for all \(n \geq M\text{,}\) we have \(x_n < K\text{,}\) we say \(\{ x_n \}_{n=1}^\infty\)diverges to minus infinity and we write

The case of monotone increasing follows from Exercise 2.3.14 part c) below. Suppose \(\{x_n\}_{n=1}^\infty\) is decreasing and unbounded. That the sequence is unbounded means that for every \(K \in \R\text{,}\) there is an \(M \in \N\) such that \(x_M < K\text{.}\) By monotonicity, \(x_n \leq x_M < K\) for all \(n \geq M\text{.}\) Therefore, \(\lim_{n\to\infty} x_n = -\infty\text{.}\)

We may also allow \(\liminf\) and \(\limsup\) to take on the values \(\infty\) and \(-\infty\text{,}\) so that we can apply \(\liminf\) and \(\limsup\) to absolutely any sequence, not just a bounded one. Unfortunately, the sequences \(\{ a_n \}_{n=1}^\infty\) and \(\{ b_n \}_{n=1}^\infty\) are not sequences of real numbers but of extended real numbers. In particular, \(a_n\) can equal \(\infty\) for some \(n\text{,}\) and \(b_n\) can equal \(-\infty\text{.}\) So we have no definition for the limits. But since the extended real numbers are still an ordered set, we can take suprema and infima.

Definition2.3.12.

Let \(\{ x_n \}_{n=1}^\infty\) be an unbounded sequence of real numbers. Define sequences of extended real numbers by \(a_n \coloneqq \sup \{ x_k : k \geq n \}\) and \(b_n \coloneqq \inf \{ x_k : k \geq n \}\text{.}\) Define

This definition agrees with the definition for bounded sequences.

Proposition2.3.13.

Let \(\{ x_n \}_{n=1}^\infty\) be an unbounded sequence. Define \(\{ a_n \}_{n=1}^\infty\) and \(\{ b_n \}_{n=1}^\infty\) as above. Then \(\{ a_n \}_{n=1}^\infty\) is decreasing, and \(\{ b_n \}_{n=1}^\infty\) is increasing. If \(a_n\) is a real number for every \(n\text{,}\) then \(\limsup_{n\to\infty} x_n = \lim_{n\to\infty} a_n\text{.}\) If \(b_n\) is a real number for every \(n\text{,}\) then \(\liminf_{n\to\infty} x_n = \lim_{n\to\infty} b_n\text{.}\)

Proof.

As before, \(a_n = \sup \{ x_k : k \geq n \} \geq \sup \{ x_k : k \geq n+1 \} =
a_{n+1}\text{.}\) So \(\{ a_n \}_{n=1}^\infty\) is decreasing. Similarly, \(\{ b_n \}_{n=1}^\infty\) is increasing.

If the sequence \(\{ a_n \}_{n=1}^\infty\) is a sequence of real numbers, then \(\lim_{n\to\infty} a_n = \inf \{ a_n : n \in \N \}\text{.}\) This follows from Theorem 2.1.10 if \(\{ a_n \}_{n=1}^\infty\) is bounded and Proposition 2.3.10 if \(\{a_n \}_{n=1}^\infty\) is unbounded. We proceed similarly with \(\{ b_n \}_{n=1}^\infty\text{.}\)

The definition behaves as expected with \(\limsup\) and \(\liminf\text{,}\) see exercises 2.3.13 and 2.3.14.

Example2.3.14.

Suppose \(x_n \coloneqq 0\) for odd \(n\) and \(x_n \coloneqq n\) for even \(n\text{.}\) Then \(a_n = \infty\) for all \(n\text{,}\) since for every \(M\text{,}\) there exists an even \(k\) such that \(x_k = k \geq M\text{.}\) On the other hand, \(b_n = 0\) for all \(n\text{,}\) as for every \(n\text{,}\) the set \(\{ b_k : k \geq n \}\) consists of \(0\) and positive numbers. So,

Suppose \(\{ x_n \}_{n=1}^\infty\) is a bounded sequence. Define \(a_n\) and \(b_n\) as in Definition 2.3.1. Show that \(\{ a_n \}_{n=1}^\infty\) and \(\{ b_n \}_{n=1}^\infty\) are bounded.

Exercise2.3.2.

Suppose \(\{ x_n \}_{n=1}^\infty\) is a bounded sequence. Define \(b_n\) as in Definition 2.3.1. Show that \(\{ b_n \}_{n=1}^\infty\) is an increasing sequence.

Exercise2.3.3.

Finish the proof of Proposition 2.3.6. That is, suppose \(\{ x_n \}_{n=1}^\infty\) is a bounded sequence and \(\{ x_{n_k} \}_{k=1}^\infty\) is a subsequence. Prove \(\displaystyle \liminf_{n\to\infty} x_n \leq
\liminf_{k\to\infty} x_{n_k}\text{.}\)

Hint: One proof is to find a subsequence \(\{ x_{n_m}+y_{n_m} \}_{m=1}^\infty\) of \(\{ x_n + y_n \}_{n=1}^\infty\) that converges. Then find a subsequence \(\{ x_{n_{m_i}} \}_{i=1}^\infty\) of \(\{ x_{n_m} \}_{m=1}^\infty\) that converges.

Find an explicit \(\{ x_n \}_{n=1}^\infty\) and \(\{ y_n \}_{n=1}^\infty\) such that

Let \(\{ x_n \}_{n=1}^\infty\) and \(\{ y_n \}_{n=1}^\infty\) be bounded sequences (by the previous exercise, \(\{ x_n + y_n \}_{n=1}^\infty\) is bounded).

If \(S \subset \R\) is a set, then \(x \in \R\) is a cluster point if for every \(\epsilon > 0\text{,}\) the set \((x-\epsilon,x+\epsilon) \cap S
\setminus \{ x \}\) is not empty. That is, if there are points of \(S\) arbitrarily close to \(x\text{.}\) For example, \(S \coloneqq \{ \nicefrac{1}{n} : n \in \N \}\) has a unique (only one) cluster point \(0\text{,}\) but \(0 \notin S\text{.}\) Prove the following version of the Bolzano–Weierstrass theorem:

Theorem. Let \(S \subset \R\) be a bounded infinite set, then there exists at least one cluster point of \(S\).

Hint: If \(S\) is infinite, then \(S\) contains a countably infinite subset. That is, there is a sequence \(\{ x_n \}_{n=1}^\infty\) of distinct numbers in \(S\text{.}\)

Exercise2.3.10.

(Challenging)

Prove that every sequence contains a monotone subsequence. Hint: Call \(n \in \N\) a peak if \(a_m \leq a_n\) for all \(m \geq n\text{.}\) There are two possibilities: Either the sequence has at most finitely many peaks, or it has infinitely many peaks.

Conclude the Bolzano–Weierstrass theorem.

Exercise2.3.11.

Prove a stronger version of Proposition 2.3.7. Suppose \(\{ x_n \}_{n=1}^\infty\) is a sequence such that every subsequence \(\{ x_{n_m} \}_{m=1}^\infty\) has a subsequence \(\{ x_{n_{m_i}} \}_{i=1}^\infty\) that converges to \(x\text{.}\)

First show that \(\{ x_n \}_{n=1}^\infty\) is bounded.

Now show that \(\{ x_n \}_{n=1}^\infty\) converges to \(x\text{.}\)

Exercise2.3.12.

Let \(\{x_n\}_{n=1}^\infty\) be a bounded sequence.

Prove that there exists an \(s\) such that for every \(r > s\text{,}\) there exists an \(M \in \N\) such that for all \(n \geq M\text{,}\) we have \(x_n < r\text{.}\)

If \(s\) is a number as in a), then prove \(\limsup\limits_{n\to\infty} x_n \leq s\text{.}\)

Show that if \(S\) is the set of all \(s\) as in a), then \(\limsup\limits_{n\to\infty} x_n = \inf \, S\text{.}\)

Exercise2.3.13.

(Easy) Suppose \(\{ x_n \}_{n=1}^\infty\) is such that \(\liminf\limits_{n\to\infty} x_n = -\infty\text{,}\)\(\limsup\limits_{n\to\infty} x_n = \infty\text{.}\)

Show that \(\{ x_n \}_{n=1}^\infty\) is not convergent, and also that neither \(\lim\limits_{n\to\infty} x_n = \infty\) nor \(\lim\limits_{n\to\infty} x_n = -\infty\) is true.

Find an example of such a sequence.

Exercise2.3.14.

Let \(\{ x_n \}_{n=1}^\infty\) be a sequence.

Show that \(\lim\limits_{n\to\infty} x_n = \infty\) if and only if \(\liminf\limits_{n\to\infty} x_n = \infty\text{.}\)

Then show that \(\lim\limits_{n\to\infty} x_n = - \infty\) if and only if \(\limsup\limits_{n\to\infty} x_n = -\infty\text{.}\)

If \(\{ x_n \}_{n=1}^\infty\) is monotone increasing, show that either \(\lim_{n\to\infty} x_n\) exists and is finite or \(\lim_{n\to\infty} x_n = \infty\text{.}\) In either case, \(\lim_{n\to\infty} x_n = \sup \{ x_n : n \in \N \}\text{.}\)

Exercise2.3.15.

Prove the following stronger version of Lemma 2.2.12, the ratio test. Suppose \(\{ x_n \}_{n=1}^\infty\) is a sequence such that \(x_n \not= 0\) for all \(n\text{.}\)

Suppose \(\{ x_n \}_{n=1}^\infty\) is a bounded sequence, \(a_n \coloneqq \sup \{ x_k : k \geq n \}\) as before. Suppose that for some \(\ell \in \N\text{,}\)\(a_\ell \notin \{ x_k : k \geq \ell \}\text{.}\) Then show that \(a_j = a_\ell\) for all \(j \geq \ell\text{,}\) and hence \(\limsup\limits_{n\to\infty} x_n = a_\ell\text{.}\)

Exercise2.3.17.

Suppose \(\{ x_n \}_{n=1}^\infty\) is a sequence, and \(a_n \coloneqq \sup \{ x_k : k \geq n \}\) and \(b_n \coloneqq \sup \{ x_k : k \geq n \}\) as before.

Prove that if \(a_\ell = \infty\) for some \(\ell \in \N\text{,}\) then \(\limsup\limits_{n\to\infty} x_n = \infty\text{.}\)

Prove that if \(b_\ell = -\infty\) for some \(\ell \in \N\text{,}\) then \(\liminf\limits_{n\to\infty} x_n = -\infty\text{.}\)

Exercise2.3.18.

Suppose \(\{ x_n \}_{n=1}^\infty\) is a sequence such that both \(\liminf_{n\to\infty} x_n\) and \(\limsup_{n\to\infty} x_n\) are finite. Prove that \(\{ x_n \}_{n=1}^\infty\) is bounded.

Exercise2.3.19.

Suppose \(\{ x_n \}_{n=1}^\infty\) is a bounded sequence, and \(\epsilon > 0\) is given. Prove that there exists an \(M\) such that for all \(k \geq M\text{,}\)

Extend Theorem 2.3.4 to unbounded sequences: Suppose that \(\{ x_n \}_{n=1}^\infty\) is a sequence. If \(\limsup_{n\to\infty} x_n = \infty\text{,}\) then prove that there exists a subsequence \(\{ x_{n_i} \}_{i=1}^\infty\) converging to \(\infty\text{.}\) Then prove the same result for \(-\infty\text{,}\) and then prove both statements for \(\liminf\text{.}\)