Unlock the Secrets of Math: How Sequences and Series Hold the Key to Analysis

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### 1.4 Sequences and Series

Most concepts in analysis can be reduced to statements about the behavior of sequences and series. Moreover, understanding infinite series depends on understanding of sequences. A _sequence_ of real numbers can be defined as a mapping from \(\mathbb{N}\) into \(\mathbb{R}\), a real-valued function defined on the positive integers. Although there are different notations for describing sequences, they are usually written as

\[\{x_{n}\}=\{x_{1},x_{2},\dots\},\]

where \(x_{n}\in\mathbb{R}\) for each \(n\in\mathbb{N}\).

**Example 20**.: Each of the following is ways to describe a sequence.

1. \(\left\{1,\ \frac{1}{2},\ \frac{1}{3},\ \frac{1}{4},\ \cdots\right\}.\)
2. \(x_{n}=(-1)^{n+1}\) or \(\{x_{n}\}=\{1,\,-1,\,1,\,-1,\ \dots\}.\)
3. \(\left\{\frac{1+n}{n}\right\}_{1}^{\infty}=\frac{2}{1},\ \frac{3}{2},\ \frac{4}{3},\ \dots.\)
4. \(\{x_{n}\}\), where \(x_{n}=\sin n\) for each \(n\in\mathbb{N}\).
5. \(x_{1}=1;\ \ x_{2}=1;\ \ x_{n}=x_{n-2}+x_{n-1}\) if \(n\geq 3\) (_Fibonacci sequence_, defined recursively).

**Definition 10** (Convergence of a Sequence).: The sequence \(\{x_{n}\}\) is said to _converge to a limit \(L\in\mathbb{R}\)_ if for each \(\epsilon>0\) there corresponds a number \(N=N(\epsilon)\) such that

\[|x_{n}-L|<\epsilon\quad\text{for all}\quad n\geq N.\]

In this case \(L\) is the _limit_, and we write \(\lim_{n\to\infty}x_{n}=L\) or just \(x_{n}\to L\). If no such \(L\) exists, we say \(\{x_{n}\}\)_diverges_. The whole definition can also be written in a compact way as follows (Figure 1.15):

\[\forall\epsilon>0,\quad\exists N\in\mathbb{N}\quad\text{such that}\quad n\geq N \Longrightarrow|x_{n}-L|<\epsilon.\]

The above definition is subtle and requires an explanation. The Greek letter \(\epsilon\) measures the “nearness” of the \(x_{n}\)’s to \(L\), \(N\) is the “stopping” place, and \(|x_{n}-L|\) measures the distance between \(x_{n}\) and \(L\). Thus the convergence of \(x_{n}\) to \(L\) means that for each prescribed discrepancy \(\epsilon\), the numbers \(x_{n}\) are within distance \(\epsilon\) of \(L\) for all \(n\geq N\). Thus \(N\) is where you must stop for this desirable outcome to emerge. Notice that \(|x_{n}-L|<\epsilon\) for all but a _finite_ number of indices \(n\). There is another issue in this definition that we must be aware of it. From some point on _every_ element of the sequence approximates the limit \(L\) to any desired accuracy. Therefore we could consider only values of \(\epsilon\) of the form \(\frac{1}{2}10^{-k}\). Then the statement \(|x_{n}-L|<\frac{1}{2}10^{-k}\) means that \(x_{n}\) and \(L\) agree to \(k\) decimal places. In other words, a sequence \(\{x_{n}\}\) converges to \(L\) precisely when eventually all the terms of the sequence agree with \(L\) to \(k\) decimal places.

**Example 21**.: Let \(x_{n}=\frac{2n+3}{n+5}\). Show that \(\{x_{n}\}\) converges.

First we need a candidate for \(L\). By plugging in numbers for \(n\), it is not hard to see that when \(n\to\infty\), we have \(L=2\), thus \(\lim_{n\to\infty}x_{n}=2\). Before we try to prove this by using the \(\epsilon-N\) “game,” we must explore the relationship between them. We manipulate the desired inequality in search of a suitable \(N\). Now,

\[|x_{n}-L|=\left|\frac{2n+3}{n+5}-2\right|=\left|\frac{2n+3-2(n+5)}{n+5}\right| =\left|\frac{-7}{n+5}\right|=\frac{7}{n+5}\]

Figure 1.15: Convergence to a limit \(L\)and \(\dfrac{7}{n+5}<\epsilon\) holds true if and only if \(\dfrac{7}{\epsilon}

Proof

: Let \(\epsilon>0\) be given. Setting \(N=\dfrac{7}{\epsilon}-5\) and \(n\in\mathbb{N}\) with \(n\geq N\), we obtain

\[|x_{n}-L|=\left|\dfrac{2n+3}{n+5}-2\right|=\dfrac{7}{n+5}<\dfrac{7}{N+5}= \dfrac{7}{\dfrac{7}{\epsilon}-5+5}=\epsilon.\qed\]

Note that if \(\epsilon=\frac{1}{10}\), then the stopping place is \(N=65\). Thus for this particular \(\epsilon\), the terms of the sequence \(x_{n}\) are within \(\frac{1}{10}\) if \(n\geq 65\). It should be apparent that _the value \(N\) depends on the choice of \(\epsilon\)_.

**Definition 11**.: Given a real number \(L\in\mathbb{R}\) and a positive number \(\epsilon>0\), the set

\[V_{\epsilon}(L)=\{x\in\mathbb{R}:\ |x-L|<\epsilon\}\]

is called the \(\epsilon\)-_neighborhood_ of \(L\).

Notice that \(V_{\epsilon}(L)\) consists of all those points whose distance from \(L\) is less than \(\epsilon\). Equivalently \(V_{\epsilon}(L)=(L-\epsilon,L+\epsilon)\) is an interval since

\[|x-L|<\epsilon\Leftrightarrow L-\epsilon

Using the above terminology we can make a topological version of the convergence of sequences. Namely, we say a sequence \(\{x_{n}\}\) converges to \(L\) if, given every \(\epsilon\)-neighborhood \(V_{\epsilon}(L)\) of \(L\), there exists a point in the sequence after which all the terms are in \(V_{\epsilon}(L)\). In other words \(V_{\epsilon}(L)\) contains all but a finite number of the terms of \(\{x_{n}\}\). The natural number \(N\) in the above definition is the last stopping place, where the sequence enters \(V_{\epsilon}(L)\) never to leave (Figure 16).

**Example 22**.: Show that the sequence \(\{x_{n}\}=(-1)^{n+1}=\{1,\,-1,\,1,\,-1,\,\cdots\}\) diverges.

Proof

: Clearly this sequence oscillates between \(1\) and \(-1\). As in Example 21 we invoke the definition for \(\epsilon\)-\(N\) convergence. We prove that the limit does not exist by contradiction. Suppose the sequence \(\{x_{n}\}\) converges to \(L\). From the

Figure 16: Convergence of a sequence

definition for every \(\epsilon>0\), we must find \(N\). Thus let \(\epsilon=\dfrac{1}{100}>0\), and choose \(N\) as in the definition. Then for any even integer \(n_{1}\) with \(n_{1}\geq N\), we have

\[|x_{n_{1}}-L|=|-1-L|<0.01,\]

which means \(L\in(-1.01,-0.99)\). Similarly if \(n_{2}\) is any odd integer with \(n_{2}\geq N\), then we have

\[|x_{n_{2}}-L|=|1-L|<0.01\]

implying \(L\in(0.99,1.01)\). Since \(L\) cannot lie in two disjoint intervals, this is a contradiction to our assumption.

A sequence \(\{x_{n}\}\) is _bounded above_ if there exists \(M\) such that \(x_{n}\leq M\) for all \(n\in\mathbb{N}\); it is _bounded below_ if there exists \(m\) such that \(x_{n}\geq m\) for all \(n\). The _range_ of a sequence is its set of values \(\{x_{n}:\ n\in\mathbb{N}\}\). We say a sequence \(\{x_{n}\}\) is **bounded** if its range is a bounded set. In other words it is both bounded above and below. We now give two important properties of convergent sequences.

**Theorem 10**.: _Let \(\{x_{n}\}\) be a sequence of real numbers._

* _If_ \(\{x_{n}\}\) _converges to_ \(L\) _and_ \(L^{*}\)_, then_ \(L=L^{*}\) _(limit is unique)._
* _If_ \(\{x_{n}\}\) _is convergent, then_ \(\{x_{n}\}\) _is bounded._

Proof

:
* Let \(\epsilon>0\) be given. There exist integers \(N\) and \(N^{\prime}\) such that \[n\geq N\quad\text{implies}\quad|x_{n}-L|<\dfrac{\epsilon}{2}\] \[n\geq N^{\prime}\quad\text{implies}\quad|x_{n}-L^{*}|<\dfrac{\epsilon}{2}.\] Hence if \(n\geq\max\{N,N^{\prime}\}\), we have \[|L-L^{*}|\leq|L-x_{n}|+|x_{n}-L^{*}|<\epsilon.\] Since \(\epsilon\) was arbitrary, we have \(|L-L^{*}|=0\) and therefore \(L=L^{*}\). * Suppose \(\{x_{n}\}\) converges to \(L\). For \(\epsilon=1\), there is an integer \(N\) such that \(n\geq N\) implies that \(|x_{n}-L|<1\). Setting \[K=\max\{1,|x_{1}-L|,|x_{2}-L|,\dots,|x_{N}-L|\}\] will give \(|x_{n}-L|\leq K\) for \(n=1,2,3,\dots\).

There are ways to create new sequences from “known” sequences as described in the following theorem.

**Theorem 11** (Algebraic Limit Theorem).: _Suppose \(\{x_{n}\}\) and \(\{y_{n}\}\) are sequences such that \(x_{n}\to L\) and \(y_{n}\to M\), then_

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**Remark 9**.: Part d) of the above theorem implies that if \(\{x_{n}\}\) and \(\{y_{n}\}\) are sequences such that \(x_{n}\to L\) and \(y_{n}\to M\), then

\[\frac{x_{n}}{y_{n}}\to\frac{L}{M}\]

provided \(y_{n}\neq 0\) for all \(n\) and \(M\neq 0\). This is the case because the result for product of sequences implies

\[x_{n}\frac{1}{y_{n}}\to L\frac{1}{M}.\]

**Theorem 12** (The Squeeze Principle).: _Let \(\{x_{n}\}\), \(\{y_{n}\}\), and \(\{z_{n}\}\) be sequences such that_

\[x_{n}\leq y_{n}\leq z_{n}\qquad\text{for all }\,n.\]

_If \(x_{n}\to L\) and \(z_{n}\to L\), then \(y_{n}\to L\) too._

Proof

: Since \(x_{n}\to L\) and \(z_{n}\to L\), for a given \(\epsilon>0\), we can choose \(N_{1}\) such that \(n\geq N_{1}\) implies that \(|x_{n}-L|<\epsilon\); similarly, there is \(N_{2}\) such that \(n\geq N_{2}\), which implies that \(|z_{n}-L|<\epsilon\). Now let \(N=\max\{N_{1},N_{2}\}\). Then for the sequence \(y_{n}\), since \(n\geq N\), we have that

\[L-\epsilon

which implies for a given \(\epsilon>0\) there is an \(N\) such that \(n\geq N\) implies

\[|y_{n}-L|<\epsilon.\qed\]