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Proof
: Suppose \(\{x_{n}\}\) is a Cauchy sequence. Lemma 4 guarantees that \(\{x_{n}\}\) is bounded. Thus we may use the Bolzano-Weierstrass Theorem to produce a convergent subsequence \(\{x_{n_{k}}\}\). Let \(x_{n_{k}}\to L\). The final step is to show that if a Cauchy sequence has a convergent subsequence, then in fact the full sequence \(\{x_{n}\}\) is convergent. Indeed, if \(x_{n_{k}}\to L\) as \(k\to\infty\), then for each \(\epsilon>0\) we have
\[|x_{n}-L|\leq|x_{n}-x_{n_{k}}|+|x_{n_{k}}-L|<|x_{n}-x_{n_{k}}|+\frac{\epsilon} {2}\]
for all sufficiently large indices \(n_{k}\). But because \(\{x_{n}\}\) is a Cauchy sequence,
\[|x_{n}-L|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon\]
for all sufficiently large \(n\). This shows that \(x_{n}\to L\) as \(n\to\infty\).
**Remark 12** (Completeness of \(\mathbb{R}\) Again).: In the previous section we discussed the Axioms of Completeness, which simply asserts that “every nonempty set bounded above has a supremum.” We then used this axiom to prove the Nested Intervals property and the Monotone Convergence Theorem. In the proof of Bolzano-Weierstrass, we utilized both the Monotone Convergence and Nested Intervals Theorems. Finally, the Bolzano-Weierstrass Theorem implies the Cauchy Criterion for convergent sequences. All of these fundamental results come from the same family, each describing the completeness of \(\mathbb{R}\) in a particular way.
### Lim Sup and Lim Inf
Note that many things we want to know about functions \(f:\mathbb{R}\to\mathbb{R}\) or the real line \(\mathbb{R}\) itself can be described in terms of sequences. For example, we will see that a continuous function \(f:\mathbb{R}\to\mathbb{R}\) can be defined as a function that “preserves” convergent sequences: \(f(\lim_{n\to\infty}x_{n})=\lim_{n\to\infty}f(x_{n})\). So far we have seen that monotone and bounded sequences converge, and a convergent sequence is necessarily bounded. Of course not every bounded sequence converges, but we can wonder how “far” is a bounded sequence from being convergent. Moreover, a sequence may or may not have limit, or it may have several converging subsequences. The following notions of limit superior and limit inferior will help to clarify these issues. We start with a convention for unbounded sequences; if \(\{x_{n}\}\) is not bounded above, we write
\[\limsup_{n\to\infty}x_{n}=+\infty.\]
Similarly, if \(\{x_{n}\}\) is not bounded below, we write
\[\liminf_{n\to\infty}x_{n}=-\infty.\]
If \(x_{n}\to+\infty\), we say \(\liminf_{n\to\infty}x_{n}=+\infty\), and if \(x_{n}\to-\infty\), we write \(\limsup_{n\to\infty}x_{n}=-\infty\).
**Definition 15**.: Let \(\{x_{n}\}\) be a sequence of real numbers. We define its limit superior by
\[\limsup_{n\to\infty}x_{n}:=\lim_{n\to\infty}(\sup\{x_{k}:\ k\geq n\})\]
and its limit inferior by
\[\liminf_{n\to\infty}x_{n}:=\lim_{n\to\infty}(\inf\{x_{k}:\ k\geq n\}).\]
Note that for a bounded sequence \(\{x_{n}\}\), we can define a new sequence \(\{a_{n}\}\) by
\[a_{1} =\sup\{x_{1},x_{2},x_{3},\dots\},\ a_{2}{=}\sup\{x_{2},x_{3},x_{4 },\dots\},\ \dots,\] \[a_{n} =\sup\{x_{n},x_{n+1},x_{n+2},\dots\}\]and call this sequence the _sup-sequence_ associated with \(\{x_{n}\}\). Then the above definition simply says \(\limsup\limits_{n\to\infty}x_{n}=\lim\limits_{n\to\infty}a_{n}\). Moreover, if \(\{x_{n}\}\) is bounded above and does not tend to \(-\infty\) and if we let \(A\) denote the set of subsequential limits of \(\{x_{n}\}\), the Bolzano-Weierstrass Theorem implies that \(A\neq\emptyset\). We then define \(\limsup\limits_{n\to\infty}x_{n}=\sup A\). Similarly, if \(\{x_{n}\}\) is bounded below and does not tend to \(+\infty\), one can set \(\liminf\limits_{n\to\infty}=\inf A\). It is worth mentioning that the number \(\sup A\) actually belongs to \(A\); thus \(\limsup\limits_{n\to\infty}x_{n}\) is actually the largest subsequential limit.
Similarly, \(\liminf\limits_{n\to\infty}x_{n}\) is the smallest subsequential limit if \(\{x_{n}\}\) is bounded below and does not tend to \(+\infty\). There are two simple consequences of the above definition. The first one is that
\[\limsup\limits_{n\to\infty}(-x_{n})=-\liminf\limits_{n\to\infty}x_{n},\]
i.e., upper and lower limits are kind of “reflections” of each other; thus properties of one will transfer to the other. The second consequence of the definition is the simple fact that
\[\liminf\limits_{n\to\infty}x_{n}\leq\limsup\limits_{n\to\infty}x_{n}.\]
**Remark 13**.: An alternate notation is \(\limsup\limits_{n\to\infty}=\overline{\lim}\) and \(\liminf\limits_{n\to\infty}=\underline{\lim}\).
**Theorem 17**.: _A sequence of real numbers \(\{x_{n}\}\) converges to a limit \(x\in\mathbb{R}\) if and only if_
\[\liminf\limits_{n\to\infty}x_{n}=\limsup\limits_{n\to\infty}x_{n}=x.\]
Proof
: In case that \(\{x_{n}\}\) converges to \(x\), we know that the sequence \(\{x_{n}\}\) is bounded and \(A=\{x\}\). From the explanation above,
\[\liminf\limits_{n\to\infty}x_{n}=\inf(A)=x=\sup(A)=\limsup\limits_{n\to\infty }x_{n}.\]
Now suppose
\[\liminf\limits_{n\to\infty}x_{n}=\limsup\limits_{n\to\infty}x_{n}=x.\]
Let \(A_{n}:=\sup\{x_{k}:\ k\geq n\}\) and \(B_{n}:=\inf\{x_{k}:\ k\geq n\}\). Note that our assumption is \(\lim\limits_{n\to\infty}A_{n}=\lim\limits_{n\to\infty}B_{n}\). Thus given \(\epsilon>0\), there is \(N\in\mathbb{N}\) such that \(|x-A_{N}|<\epsilon\) and \(|x-B_{N}|<\epsilon\). Since \(B_{N}\leq x_{n}\leq A_{N}\) for all \(n\geq N\), we also have \(-(x-B_{N})\leq x_{n}-x\leq A_{N}-x\) or \(|x_{n}-x|<\epsilon\) for all \(n\geq N\).
**Example 25**.: Let \(\{x_{n}\}\) be a sequence defined as \(\{x_{n}\}=(-1)^{n}(1+\frac{1}{n^{2}})\). To find its \(\liminf\limits_{n\to\infty}x_{n}\) and \(\limsup\limits_{n\to\infty}x_{n}\), notice that
\[\inf\{x_{k}:k\geq n\}=\left\{\begin{array}{ll}-\left(1+\frac{1}{n^{2}} \right)&\mbox{if \ \ n\ is odd}\\ -\left(1+\frac{1}{(n+1)^{2}}\right)&\mbox{if \ \ n\ is even,}\end{array}\right.\]hence \(\liminf_{n\to\infty}x_{n}=-1\). Similarly,
\[\sup\{x_{k}:k\geq n\}=\left\{\begin{array}{ll}1+\frac{1}{n^{2}}&\mbox{if \ \ n \ is even}\\ 1+\frac{1}{(n+1)^{2}}&\mbox{if \ \ n\ \ is odd,}\end{array}\right.\]
and hence \(\limsup_{n\to\infty}x_{n}=1\).
### Some Standard Sequences
The following are some special sequences whose limits can be found with the help of the binomial theorem and the “squeeze principle.” (See [42], p. 57.)
**Example 26**.:
* For each real number \(x\in\mathbb{R}\) with \(|x|<1\), \(\lim_{n\to\infty}x^{n}=0\).
* If \(a>0\), then \(\lim_{n\to\infty}\sqrt[n]{a}=1\).
* \(\lim_{n\to\infty}\sqrt[n]{n}=1\).
* For each pair of numbers \(a\) and \(x\) with \(|x|<1\), \(\lim_{n\to\infty}n^{a}x^{n}=0\).
* For each number \(a>0\), \(\lim_{n\to\infty}\frac{\log n}{n^{a}}=0\).
### Convergence in \(\mathbb{C}\)
A _complex sequence_\(\{z_{n}=x_{n}+iy_{n}\}\) is an assignment of a complex number \(z_{n}\) to each natural number \(n=1,2,\dots\), where \(\mbox{Re }z_{n}=x_{n}\) and \(\mbox{Im }z_{n}=y_{n}\), (for a detailed study of a complex number, we refer the reader to Chapter 3). Let us examine an example.
**Example 27**.: \(\{i^{n}\}\)_, \(\left\{\left(\frac{1-i}{1+i}\right)^{n}\right\}\), and \(\{(1+i)^{n}+(1-i)^{n}\}\) are all examples of complex sequences._
Note that for any given complex sequence \(\{z_{n}\}\), we can form two real sequences, namely \(\{\mbox{Re }z_{n}\}\) and \(\{\mbox{Im }z_{n}\}\), by considering just the real and imaginary parts of the sequence \(\{z_{n}\}\). For example, if \(\{z_{n}\}=\left\{\frac{1}{n}+i\left(\frac{n}{n+2}\right)\right\}\), then \(\{\mbox{Re }z_{n}\}=\left\{\frac{1}{n}\right\}\) and \(\{\mbox{Im }z_{n}\}=\left\{\frac{n}{n+2}\right\}\) are both real sequences.
**Definition 16**.: The sequence \(\{z_{n}\}\)_converges_ to limit \(z\) (in symbols: \(z_{n}\to z\)) as \(n\to\infty\) if, given \(\epsilon>0\), there is a natural number \(N=N(\epsilon)\) such that
\[|z_{n}-z|<\epsilon\quad\mbox{whenever}\quad n\geq N.\]
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converges. Now as a subsequence of \(\{\operatorname{Re}z_{n_{k}}\}\), the sequence \(\{\operatorname{Re}z_{m_{j}}\}\) must converge too. Note that we are taking a subsequence of a subsequence. Now apply Proposition 3 to combine real and imaginary parts of this sequence to conclude that \(\{z_{m_{j}}\}\) provides a convergent subsequence of \(\{z_{n}\}\).
**Corollary 3**.: _An infinite closed and bounded subset \(K\) of \(\mathbb{C}\) has a limit point in \(K\)._
Proof
: Select a sequence of distinct points \(z_{n}\) in \(K\). Since \(K\) is bounded, the above theorem asserts that \(\{z_{n}\}\) has a subsequence which converges. If \(z_{n}\to z\), then \(z\) is a limit point of \(K\); because \(K\) is closed it contains \(z\). Note that we will see later what we mean by “closed” sets in detail. Closed and bounded sets have a special name in \(\mathbb{C}\): compact sets.
### Infinite Series
An infinite series is a sum with infinitely many terms:
\[\sum_{k=1}^{\infty}a_{k}=a_{1}+a_{2}+a_{3}+\cdots,\]
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