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Section 2.1 Sequences

Subsection 2.1.1 Definitions

An infinite sequence is a function whose domain is the ordered set \((\N, \le)\text{.}\) In this chapter we only consider sequences of real numbers, so we simply call them sequences. If \(\vx\) is a sequence, we write \(x_n\) for its value at \(n\) and call \(x_n\) the \(n\)-th term of \(\vx\text{.}\) We also denote the sequence \(\vx\) by \((x_n)\text{,}\) and we identify a real number \(r\) with the constant sequence \(n \mapsto r\text{.}\)

Example 2.1.1.

Since sequences are in particular real-valued functions, adjectives for real-valued functions apply to them as well:
  • \((1/n)\) is decreasing and bounded.
  • \((n)\) is increasing and bounded below, but not bounded above.
  • \(((-1)^n)\) is bounded, but neither increasing nor decreasing.
A sequence \(\vx\) is null if for every \(\varepsilon \gt 0\) there exists \(N \in \N\) such that
\begin{equation*} |x_n| \lt \varepsilon \quad\text{for all } n \ge N. \end{equation*}
In words, a sequence \(\vx = (x_n)\) is null if the absolute value of its terms \(|x_n|\) can be arbitrarily close to zero provided that the index \(n\) of the terms is sufficiently large.
It often makes an argument more intuitive by replacing "there exists \(N \in \N\) such that for all \(n \ge N\)" with either of the following phrases:
  1. for all sufficiently large \(n\text{,}\)
  2. for all but finitely many \(n\text{.}\)
We may abbreviate "\(n\) is sufficiently large" by \(n \gg 0\) or \(n \approx \infty\text{.}\) For example, the statement \(\vx \approx 0\) means that for every \(\varepsilon \gt 0\text{,}\) we have \(|x_n| \lt \varepsilon\) for all \(n \gg 0\text{.}\)
We say that a sequence \(\vx\)converges to \(r \in\R\) or \(r\) is a limit of \(\vx\text{,}\) if the sequence \(\vx-r\) is null. We denote this by
\begin{equation*} \lim x_n = r, \qquad x_n \to r, \qquad \text{or} \qquad \vx \approx r. \end{equation*}
The last notation captures the idea that a convergent sequence can be viewed as an approximation to its limit.
A sequence is convergent if it has a limit, otherwise it is divergent. We also distinguish two common types of divergent behavior. We write \(x_n \to +\infty\) if for every \(M \in \R\) we have \(x_n \gt M\) for all sufficiently large \(n\text{.}\) Similarly, we write \(x_n \to -\infty\) if for every \(M \in \R\) we have \(x_n \lt M\) for all sufficiently large \(n\text{.}\)

Subsection 2.1.2 Fundamental Results

Proof.
For (1), let \(\varepsilon \gt 0\text{.}\) Choose \(N_1,N_2\) so that \(|x_n| \lt \varepsilon/2\) for all \(n \ge N_1\) and \(|y_n| \lt \varepsilon/2\) for all \(n \ge N_2\text{.}\) Then for \(n \ge \max\{N_1,N_2\}\text{,}\) the triangle inequality gives \(|x_n+y_n| \le |x_n|+|y_n| \lt \varepsilon\text{.}\)
For (2), by Proposition 1.1.2 there exists \(M \gt 0\) such that \(|c_n| \lt M\) for every \(n\text{.}\) Let \(\varepsilon \gt 0\text{,}\) then for all \(n \gg 0\text{,}\) \(|x_n| \lt \varepsilon\) and hence \(|c_n x_n| \lt M\varepsilon\text{.}\) This shows that \((c_nx_n)\) is a null sequence because \(M\varepsilon\) is an arbitrary positive number as \(M \ge 0\) and \(\varepsilon \ge 0\) is arbitrary.
The proof of (1) illustrates an important general point: if a statement \(P\) holds for all sufficiently large \(n\) and a statement \(Q\) also holds for all sufficiently large \(n\text{,}\) then the conjunction "\(P\) and \(Q\)" holds for all sufficiently large \(n\text{.}\)

Proof.
Suppose \(x_n \to r\) and \(x_n \to r'\text{.}\) Then both \((r-x_n)\) and \((x_n-r')\) are null sequences. By Proposition 2.1.2, their sum \((r-r')\) is also null. A constant sequence is null only if its constant value is zero, so \(r-r'=0\text{.}\) Therefore \(r=r'\text{.}\)

Proof.
Suppose \(x_n \to r\text{.}\) Then there exists \(N\) such that \(|x_n-r| \lt 1\) for all \(n \ge N\text{.}\) Hence \(|x_n| \le |x_n-r|+|r| \lt 1+|r|\) for all \(n \ge N\text{.}\) It follows that
\begin{equation*} |x_n| \le \max\{|x_1|,\ldots,|x_N|,1+|r|\} \end{equation*}
for every \(n\text{.}\) Thus \((x_n)\) is bounded.
The proof above shows that "eventually bounded" and "bounded" are the same for sequences.

Proof.
First suppose \((x_n)\) is increasing and bounded. Then its set of terms \(S:=\{x_n \colon n \in \N\}\) is nonempty and bounded above, so \(s:=\sup S\) exists.
Let \(\varepsilon \gt 0\text{.}\) Since \(s-\varepsilon\) is not an upper bound of \(S\text{,}\) there exists \(N\) such that \(s-\varepsilon \lt x_N \le s\text{.}\) As \((x_n)\) is increasing, for every \(n \ge N\) we have
\begin{equation*} s-\varepsilon \lt x_N \le x_n \le s \lt s+\varepsilon. \end{equation*}
Hence \(|x_n-s| \lt \varepsilon\) for all \(n \ge N\text{,}\) so \(x_n \to s\text{.}\)
If \((x_n)\) is decreasing and bounded, then \((-x_n)\) is increasing and bounded. By the increasing case, \((-x_n)\) converges, and then Proposition 2.1.6 implies that \((x_n)\) converges as well.
The next proposition shows that the usual algebraic operations on convergent sequences again produce convergent sequences, and that taking limits commutes with these operations.

Proof.
Both \((x_n-x)\) and \((y_n-y)\) are null, so Proposition 2.1.2 shows that their sum \((x_n-x)+(y_n-y)=(x_n+y_n)-(x+y)\) is null. Therefore \(x_n+y_n \to x+y\text{.}\)
Also \(x_ny_n-xy = x_n(y_n-y) + (x_n-x)y\text{.}\) By Proposition 2.1.4, the sequence \((x_n)\) is bounded. Hence Proposition 2.1.2 implies that \((x_n(y_n-y))\) is null. Since the constant sequence \(y\) is bounded, the same proposition shows that \(((x_n-x)y)\) is null. Therefore \((x_ny_n-xy)\) is null, and so \(x_ny_n \to xy\text{.}\)
For the reciprocal, assume \(y \neq 0\text{.}\) Since \(y_n \to y\text{,}\) there exists \(N\) such that \(|y_n-y| \lt |y|/2\) for all \(n \ge N\text{.}\) For such \(n\text{,}\) \(|y_n| \ge |y|-|y_n-y| \gt |y|/2\text{,}\) so in particular \(y_n \neq 0\text{.}\) Moreover, for all \(n \ge N\text{,}\)
\begin{equation*} \left| \frac{1}{y_n} - \frac{1}{y}\right| = \left|\frac{y-y_n}{y y_n}\right| \le \frac{2|y_n-y|}{|y|^2}. \end{equation*}
Since \((y_n-y)\) is null and \(2/|y|^2\) is a fixed number, Proposition 2.1.2 implies that \((1/y_n - 1/y)\) is null. Thus \(1/y_n \to 1/y\text{.}\)

Proof.
Let \(\varepsilon \gt 0\text{.}\) Since \(x_n \to r\) and \(y_n \to r\text{,}\) we have \(|x_n-r| \lt \varepsilon\) and \(|y_n-r| \lt \varepsilon\) for all sufficiently large \(n\text{.}\) For such \(n\text{,}\)
\begin{equation*} -\varepsilon \lt x_n-r \le z_n-r \le y_n-r \lt \varepsilon. \end{equation*}
Hence \(|z_n-r| \lt \varepsilon\) for all sufficiently large \(n\text{,}\) so \(z_n \to r\text{.}\)

Proof.
Suppose instead that \(r \lt 0\text{.}\) Then for all \(n \gg 0\text{,}\) \(x_n-r \ge -r = |r|\text{,}\) so \(|x_n-r|\ge |r| \gt 0\text{.}\) This contradicts \(x_n \to r\text{.}\) Therefore \(r \ge 0\text{.}\)

Proof.
A subsequence of a sequence \((x_n)\) is a sequence of the form \((x_{n_k})\text{,}\) where \((n_k)\) is a strictly increasing sequence of natural numbers. In other words, a subsequence of \((x_n)\) is a sequence of the form \((x_{\sigma(n)})\text{,}\) where \(\sigma\) is a strictly increasing function from \(\N\) to itself.

Proof.
Suppose \(x_n \to L\) and \((x_{n_k})\) is a subsequence. Since \((n_k)\) is strictly increasing, we have \(n_k \ge k\) for every \(k\text{.}\)
Let \(\varepsilon \gt 0\text{.}\) Choose \(N\) such that \(|x_n-L| \lt \varepsilon\) for all \(n \ge N\text{.}\) If \(k \ge N\text{,}\) then \(n_k \ge k \ge N\text{,}\) so \(|x_{n_k}-L| \lt \varepsilon\text{.}\) Hence \(x_{n_k} \to L\text{.}\)
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