Mathematics Program · Real Analysis II · Basic Analysis I & II · Jiří Lebl · sources and attribution

Section 2.3 Limit superior, limit inferior, and Bolzano–Weierstrass

Note: 1–2 lectures, alternative proof of BW optional
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In this section, we study bounded sequences and their subsequences. We define what we call limit superior and limit inferior of a bounded sequence, which are the largest and smallest possible limits of subsequences. Furthermore, we prove the Bolzano–Weierstrass theorem
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Named after the Czech mathematician Bernhard Placidus Johann Nepomuk Bolzano (1781–1848), and the German mathematician Karl Theodor Wilhelm Weierstrass (1815–1897).
, an indispensable tool in analysis, showing the existence of convergent subsequences.
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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.
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Subsection 2.3.1 Upper and lower limits

There are ways of creating monotone sequences out of any sequence, and in this fashion we get limit superior and limit inferior. These limits always exist for bounded sequences.
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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.
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Definition 2.3.1.

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,
\begin{equation} \limsup_{n \to \infty} x_n \coloneqq \lim_{n \to \infty} a_n , \qquad \liminf_{n \to \infty} x_n \coloneqq \lim_{n \to \infty} b_n . \end{equation}
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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.
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Proof.

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.
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The second item follows as the sequences \(\{ a_n \}_{n=1}^\infty\) and \(\{ b_n \}_{n=1}^\infty\) are monotone and bounded.
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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} \lim_{n\to \infty} b_n \leq \lim_{n\to \infty} a_n. \qedhere \end{equation}
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A graph of an example sequence x sub n in dots for 50 points, that is, n from 1 to 50. In the middle there are two horizontal dashed lines, the upper one is marked as lim sup as n goes to infinity of x sub n and the lower one is marked as lim inf as n goes to infinity of x sub n. The dots land both above and below the lines but they seem to congregate closer to the region between the two lines on the right hand side. Two extra sequences are marked, on the bottom with diamonds, is a sequence that is less than or equal to all the dots and constant, until it hits a dot. Then it goes up and is constant until it hits a dot, and this keeps repeating. So it looks like a series of horizontal rows ending where the diamond as at the same place as a dot, and all the other dots are above the diamonds. All the diamonds are below the lower dashed line. A similar pattern but reversed is with circles from above, that is a row of circles until it hits a dot, then it goes down and repeats, and all the circles are above or on the dots. All the circles are above the upper dashed line.
Figure 2.5. First 50 terms of an example sequence. Terms \(x_n\) of the sequence are marked with dots (\(\boldsymbol{\cdot}\)), \(a_n\) are marked with circles (\(\circ\)), and \(b_n\) are marked with diamonds (\(\diamond\)).
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Example 2.3.3.

Let \(\{ x_n \}_{n=1}^\infty\) be defined by
\begin{equation} x_n \coloneqq \begin{cases} \frac{n+1}{n} & \text{if } n \text{ is odd,} \\ 0 & \text{if } n \text{ is even.} \end{cases} \end{equation}
Let us compute the \(\liminf\) and \(\limsup\) of this sequence. See also Figure 2.6. First the limit inferior:
\begin{equation} \liminf_{n\to\infty} x_n = \lim_{n\to\infty} \bigl( \inf \{ x_k : k \geq n \} \bigr) = \lim_{n\to\infty} 0 = 0 . \end{equation}
For the limit superior, we write
\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,
\begin{equation} \limsup_{n\to\infty} x_n = 1 . \end{equation}
Do note that the sequence \(\{ x_n \}_{n=1}^\infty\) is not a convergent sequence.
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The same marking as in the previous figure but now for a very specific sequence. The lower dashed line (the lim inf) is the horizontal axis and the upper dashed line is roughly in the middle of the picture. The first dot is high up above the upper dashed line and a circle is at the same place, while the diamond is on the horizontal axis. In fact, all the diamonds are on the horizontal axis. The second dot is on the horizontal axis, but the circle is above the upper dashed line but lower than the first dot. The third dot is at the same height as the second circle and the third circle is at the same spot. The pattern now keeps repeating as the dots for odd n and the circles get closer and closer to the upper dashed line from above.
Figure 2.6. First 20 terms of the sequence in Example 2.3.3. The marking is as in Figure 2.5.
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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 need they consist of the same numbers. For example, if the sequence is \(\{ \nicefrac{1}{n} \}_{n=1}^\infty\text{,}\) then \(b_n = 0\) for all \(n \in \N\text{.}\)
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Proof.

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
\begin{equation} a_{(n_{k-1}+1)} - x_m < \frac{1}{k} . \end{equation}
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 \(x\text{.}\)
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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 \(k \geq 2\text{,}\)
\begin{equation} \begin{split} \sabs{a_{n_k} - x_{n_k}} & = a_{n_k} - x_{n_k} \\ & \leq a_{(n_{k-1}+1)} - x_{n_k} \\ & < \frac{1}{k} . \end{split} \end{equation}
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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
\begin{equation} \sabs{a_{n_k} - x} < \frac{\epsilon}{2} . \end{equation}
Find an \(M_2 \in \N\) such that
\begin{equation} \frac{1}{M_2} \leq \frac{\epsilon}{2}. \end{equation}
Take \(M \coloneqq \max \{M_1 , M_2 , 2 \}\text{.}\) For all \(k \geq M\text{,}\)
\begin{equation} \begin{split} \sabs{x- x_{n_k}} & = \sabs{a_{n_k} - x_{n_k} + x - a_{n_k}} \\ & \leq \sabs{a_{n_k} - x_{n_k}} + \sabs{x - a_{n_k}} \\ & < \frac{1}{k} + \frac{\epsilon}{2} \\ & \leq \frac{1}{M_2} + \frac{\epsilon}{2} \leq \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon . \end{split} \end{equation}
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We leave the statement for \(\liminf\) as an exercise.
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Subsection 2.3.2 Using 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.
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Proof.

Let \(a_n\) and \(b_n\) be as in Definition 2.3.1. In particular, for all \(n \in \N\text{,}\)
\begin{equation} b_n \leq x_n \leq a_n . \end{equation}
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
\begin{equation} \lim_{n\to \infty} b_n = \lim_{n\to \infty} x_n = \lim_{n\to \infty} a_n . \end{equation}
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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, \(\liminf_{n\to\infty} x_n = x\text{.}\)
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Limit superior and limit inferior behave nicely with subsequences.
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Proof.

The middle inequality has been proved already. We will prove the third inequality, and leave the first inequality as an exercise.
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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}
Lemma 2.2.3 gives
\begin{equation} \lim_{n\to\infty} c_n \leq \lim_{n\to\infty} a_n , \end{equation}
which is the desired conclusion.
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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,
\begin{equation} \liminf_{n\to\infty} x_n \leq \lim_{k\to\infty} x_{n_k} \leq \limsup_{n\to\infty} x_n . \end{equation}
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Similarly, we get the following useful test for convergence of a bounded sequence. We leave the proof as an exercise.
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Subsection 2.3.3 Bolzano–Weierstrass theorem

While it is not true that every 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.
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Proof.

Theorem 2.3.4 says that there exists a subsequence whose limit is \(\limsup_{n\to\infty} x_n\text{.}\)
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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.
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As the theorem is so important to analysis, we present an explicit proof. The idea of the following proof also generalizes to different contexts.
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Proof.

(Alternate proof of Bolzano–Weierstrass) As the sequence is bounded, 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 \(\{ x_{n_i} \}_{i=1}^\infty\) converges then follows by the squeeze lemma.
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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 \(x_{n_{k+1}} \in [y,b_k]\text{.}\)
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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,
\begin{equation} b_i - a_i = \frac{b_1-a_1}{2^{i-1}} . \end{equation}
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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 \(i\text{,}\) we have \(x \leq y\text{.}\) As the sequences are monotone, we have (why?) that for all \(i\text{,}\)
\begin{equation} y-x \leq b_i-a_i = \frac{b_1-a_1}{2^{i-1}} . \end{equation}
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.
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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.
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Subsection 2.3.4 Infinite 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.
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Definition 2.3.9.

We say \(\{ x_n \}_{n=1}^\infty\) diverges to infinity
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Sometimes it is said that \(\{ x_n \}_{n=1}^\infty\) converges to infinity.
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
\begin{equation} \lim_{n \to \infty} x_n \coloneqq \infty . \end{equation}
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
\begin{equation} \lim_{n \to \infty} x_n \coloneqq -\infty . \end{equation}
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With this definition and allowing \(\infty\) and \(-\infty\text{,}\) we can write \(\lim_{n\to\infty} x_n\) for any monotone sequence.
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Proof.

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, \(\lim_{n\to\infty} x_n = -\infty\text{.}\)
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Example 2.3.11.

\begin{equation} \lim_{n\to \infty} n = \infty, \qquad \lim_{n\to \infty} n^2 = \infty, \qquad \lim_{n\to \infty} -n = -\infty. \end{equation}
We leave verification to the reader.
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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.
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Definition 2.3.12.

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
\begin{equation} \limsup_{n \to \infty} x_n \coloneqq \inf \{ a_n : n \in \N \}, \qquad \text{and} \qquad \liminf_{n \to \infty} x_n \coloneqq \sup \{ b_n : n \in \N \}. \end{equation}
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This definition agrees with the definition for bounded sequences.
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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.
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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 \(\{ b_n \}_{n=1}^\infty\text{.}\)
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The definition behaves as expected with \(\limsup\) and \(\liminf\text{,}\) see exercises 2.3.13 and 2.3.14.
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Example 2.3.14.

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 \(\{ x_k : k \geq n \}\) consists of \(0\) and positive numbers. So,
\begin{equation} \lim_{n\to \infty} x_n \quad \text{does not exist}, \qquad \limsup_{n\to \infty} x_n = \infty , \qquad \liminf_{n\to \infty} x_n = 0. \end{equation}
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Exercises 2.3.5 Exercises

2.3.1.

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.
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2.3.2.

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.
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2.3.3.

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 \(\displaystyle \liminf_{n\to\infty} x_n \leq \liminf_{k\to\infty} x_{n_k}\text{.}\)
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2.3.5.

  1. Let \(x_n \coloneqq \dfrac{{(-1)}^n}{n}\text{.}\) Find \(\limsup\limits_{n\to\infty} x_n\) and \(\liminf\limits_{n\to\infty} x_n\text{.}\)
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  2. Let \(x_n \coloneqq \dfrac{(n-1){(-1)}^n}{n}\text{.}\) Find \(\limsup\limits_{n\to\infty} x_n\) and \(\liminf\limits_{n\to\infty} x_n\text{.}\)
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2.3.6.

Let \(\{ x_n \}_{n=1}^\infty\) and \(\{ y_n \}_{n=1}^\infty\) be bounded sequences such that \(x_n \leq y_n\) for all \(n\text{.}\) Show
\begin{equation} \limsup_{n\to\infty} x_n \leq \limsup_{n\to\infty} y_n \qquad \text{and} \qquad \liminf_{n\to\infty} x_n \leq \liminf_{n\to\infty} y_n . \end{equation}
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2.3.7.

Let \(\{ x_n \}_{n=1}^\infty\) and \(\{ y_n \}_{n=1}^\infty\) be bounded sequences.
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  1. Show that \(\{ x_n + y_n \}_{n=1}^\infty\) is bounded.
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  2. Show that
    \begin{equation} \Bigl(\liminf_{n\to \infty} x_n\Bigr) + \Bigl(\liminf_{n\to \infty} y_n\Bigr) \leq \liminf_{n\to \infty} \, (x_n+y_n) . \end{equation}
    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.
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  3. Find an explicit \(\{ x_n \}_{n=1}^\infty\) and \(\{ y_n \}_{n=1}^\infty\) such that
    \begin{equation} \Bigl(\liminf_{n\to \infty} x_n\Bigr) + \Bigl(\liminf_{n\to \infty} y_n\Bigr) < \liminf_{n\to \infty} \, (x_n+y_n) . \end{equation}
    Hint: Look for examples that do not have a limit.
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2.3.8.

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).
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  1. Show that
    \begin{equation} \Bigl(\limsup_{n\to \infty} x_n\Bigr) + \Bigl(\limsup_{n\to \infty} y_n\Bigr) \geq \limsup_{n\to \infty} \, (x_n+y_n) . \end{equation}
    Hint: See previous exercise.
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  2. Find an explicit \(\{ x_n \}_{n=1}^\infty\) and \(\{ y_n \}_{n=1}^\infty\) such that
    \begin{equation} \Bigl(\limsup_{n\to \infty} x_n\Bigr) + \Bigl(\limsup_{n\to \infty} y_n\Bigr) > \limsup_{n\to \infty} \, (x_n+y_n) . \end{equation}
    Hint: See previous exercise.
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2.3.9.

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 cluster point \(0\text{,}\) but \(0 \notin S\text{.}\) Prove the following version of the Bolzano–Weierstrass theorem:
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Theorem. Let \(S \subset \R\) be a bounded infinite set, then there exists at least one cluster point of \(S\).
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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 \(S\text{.}\)
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2.3.10.

(Challenging)  
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  1. Prove that every sequence contains a monotone subsequence. Hint: Call \(n \in \N\) a peak of the sequence \(\{ x_n \}_{n=1}^\infty\) if \(x_m \leq x_n\) 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.
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  2. Conclude the Bolzano–Weierstrass theorem.
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2.3.11.

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 \(x\text{.}\)
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  1. First show that \(\{ x_n \}_{n=1}^\infty\) is bounded.
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  2. Now show that \(\{ x_n \}_{n=1}^\infty\) converges to \(x\text{.}\)
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2.3.12.

Let \(\{x_n\}_{n=1}^\infty\) be a bounded sequence.
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  1. 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 \(x_n < r\text{.}\)
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  2. If \(s\) is a number as in a), then prove \(\limsup\limits_{n\to\infty} x_n \leq s\text{.}\)
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  3. Show that if \(S\) is the set of all \(s\) as in a), then \(\limsup\limits_{n\to\infty} x_n = \inf \, S\text{.}\)
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2.3.13.

(Easy)   Suppose \(\{ x_n \}_{n=1}^\infty\) is such that \(\liminf\limits_{n\to\infty} x_n = -\infty\text{,}\) \(\limsup\limits_{n\to\infty} x_n = \infty\text{.}\)
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  1. 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.
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  2. Find an example of such a sequence.
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2.3.14.

Let \(\{ x_n \}_{n=1}^\infty\) be a sequence.
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  1. Show that \(\lim\limits_{n\to\infty} x_n = \infty\) if and only if \(\liminf\limits_{n\to\infty} x_n = \infty\text{.}\)
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  2. Then show that \(\lim\limits_{n\to\infty} x_n = - \infty\) if and only if \(\limsup\limits_{n\to\infty} x_n = -\infty\text{.}\)
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  3. 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, \(\lim_{n\to\infty} x_n = \sup \{ x_n : n \in \N \}\text{.}\)
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2.3.15.

Prove the following stronger version of Lemma 2.2.12, the ratio test. Suppose \(\{ x_n \}_{n=1}^\infty\) is a sequence such that \(x_n \neq 0\) for all \(n\text{.}\)
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  1. Prove that if
    \begin{equation} \limsup_{n \to \infty} \frac{\sabs{x_{n+1}}}{\sabs{x_n}} < 1 , \end{equation}
    then \(\{ x_n \}_{n=1}^\infty\) converges to \(0\text{.}\)
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  2. Prove that if
    \begin{equation} \liminf_{n \to \infty} \frac{\sabs{x_{n+1}}}{\sabs{x_n}} > 1 , \end{equation}
    then \(\{ x_n \}_{n=1}^\infty\) is unbounded.
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2.3.16.

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 \(\limsup\limits_{n\to\infty} x_n = a_\ell\text{.}\)
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2.3.17.

Suppose \(\{ x_n \}_{n=1}^\infty\) is a sequence, and \(a_n \coloneqq \sup \{ x_k : k \geq n \}\) and \(b_n \coloneqq \inf \{ x_k : k \geq n \}\) as before.
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  1. Prove that if \(a_\ell = \infty\) for some \(\ell \in \N\text{,}\) then \(\limsup\limits_{n\to\infty} x_n = \infty\text{.}\)
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  2. Prove that if \(b_\ell = -\infty\) for some \(\ell \in \N\text{,}\) then \(\liminf\limits_{n\to\infty} x_n = -\infty\text{.}\)
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2.3.18.

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.
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2.3.19.

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 \(k \geq M\text{,}\)
\begin{equation} x_k - \Bigl(\limsup_{n\to\infty} x_n\Bigr) < \epsilon \qquad \text{and} \qquad \Bigl(\liminf_{n\to\infty} x_n\Bigr) - x_k < \epsilon . \end{equation}
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2.3.20.

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 \(\liminf\text{.}\)
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