Sequences

Section 5.1 Sequences

Subsection 5.1.1 The Basics

A sequence, for us, is a function with domain an ordered set \(I\) that has the same order type as a subset of the natural numbers. Often, \(I\) is a subset of \(\mathbb{Z}\) that has a minimum. We call the domain of a sequence its index set set and its range the set of terms. We write \(a_n\) instead of \(a(n)\) for the value of the sequence (whose name is \(a\)) at \(n\) and use \((a_n)_{n \in I}\text{,}\) or simply \((a_n)\) if its index set \(I\) is understood, to denote the sequence itself. A sequence is infinite (resp., finite) if its index set is infinite (resp., finite). We mainly focus on infinite sequences of real numbers in this book.

We think of a sequence \((a_n)\) as a list

\begin{equation*} a_1, a_2, a_3, \ldots \end{equation*}

So, \(a_1\) is the first term of the list, \(a_2\) is the second term and so on.

Example 5.1.

The simplest sequences are the constant sequences. For instance, the sequence with all terms \(0\text{:}\)

\begin{equation*} 0,0,0, \ldots \end{equation*}

A sequence of real numbers \((a_n)_{n \in I}\) is bounded above if there is some real number \(M\) such that \(a_n \le M\) for any \(n \in I\text{.}\) Likewise, \((a_n)_{n \in I}\) is bounded below if there is some real number \(m\) such that \(a_n \ge m\) for any \(n \in I\text{.}\) A sequence is bounded if it is both bounded above and below.

A sequence of real numbers is increasing (or decreasing). A sequence is monotone if it is either increasing or decreasing.

Example 5.2.

The sequence \(1,-1,1,\ldots \) has its \(n\)-th term equals \((-1)^{n+1}\text{.}\) It is neither increasing nor decreasing. Its set of terms is \(\{-1,1\}\) hence the sequence is bounded.

Example 5.3.

The sequence \((1/n)\) is decreasing and the sequence \((1-1/n)\) is increasing. So both of them as monotonic.

Given a sequence \((a_n)\text{.}\) It is often useful to consider the function \(a(x)\) where \(a(n)=a_n\) for all \(n\text{.}\) As we can study the behavior of \(a(x)\text{,}\) and deduce properties of \((a_n)\) using Calculus. Here is an example.

Example 5.4.

Let us determine whether the sequence \((a_n)\) where \(a_n = \dfrac{\sqrt{n+4}}{9n+4}\) is monotonic. The graph of the function \(a(x) = \dfrac{\sqrt{x+4}}{9x+4}\) suggests that \(a(n)=a_n\) indeed form a decreasing sequence. One can check that by examine the derivative of \(a(x)\text{.}\) Let \(f(x) = \sqrt{x+4}\) and \(g(x)=9x+4\text{.}\) Then \(a(x) =f(x)/g(x)\text{.}\) It follows from the quotient rule that the sign of \(a'(x)\) is the same as

\begin{equation*} f'(x)g(x)-f(x)g'(x) = -\dfrac{9x+68}{2\sqrt{x+4}} \lt 0. \end{equation*}

This shows that \(a(x)\) and hence \(a(n)=a_n\) is decreasing.

Example 5.5.

An arithmetic sequence (aka arithmetic progression) is a sequence of the form

\begin{equation*} a, a+d, a+2d, \ldots \end{equation*}

for some constants \(a\text{,}\) called the initial term, and \(d\text{,}\) called the common difference of the arithmetic sequence. For example, wehen \(a=1\) and \(d=2\text{,}\) we obtain the sequence:

\begin{equation*} 1,3,5,\ldots, 1+2n, \ldots \end{equation*}

Example 5.6.

An geometric sequence (aka geometric progression) is a sequence of the form

\begin{equation*} a, ar, ar^2, \ldots \end{equation*}

for some constants \(a\text{,}\) called the initial term, and \(r \neq 0\text{,}\) called the common ratio of the geometric sequence. For example, wehen \(a=1\) and \(r=1/2\text{,}\) we obtained the sequence:

\begin{equation*} 1,\frac{1}{2},\frac{1}{4}, \ldots, \frac{1}{2^n}, \ldots \end{equation*}

Subsection 5.1.2 Algebraic Operations on Sequences

We can extend an operation on the terms of sequences to the sequences themselves term-wise. For instance, we can add and multiply sequences of real numbers as real-valued functions. Suppose \((a_n)\) and \((b_n)\) are sequences of real numbers. Then, respectively, their sum and product are defined by

\(\displaystyle (a+b)_n = a_n+b_n\)
\(\displaystyle (ab)_n = a_nb_n\)

Identifying a constant \(c\) with the sequence whose terms are all \(c\text{.}\) then \((c+b)_n = c+b_n\) and \((cb)_n = cb_n\text{.}\)

Checkpoint 5.7.

Write out the \(n\)-th term of the sum and product of the sequences \(1,1/2,1/3,1/4,\ldots\) and \(1,4,9,16,\ldots\text{.}\)

Solution.

The \(n\)-th of their sum is \(n^2 + 1/n\) and that of their product is \((n^2)(1/n) = n\text{.}\)

Subsection 5.1.3 Convergence of Sequence

In Calculus, we are interested in the eventual behavior of an infinite sequence. This leads to the concept of a limit of a sequence. Roughly speaking, a number \(L\) is a limit of a sequence of numbers if its terms all congregate to \(L\) eventually. To make this more precise, we first define what means by \(0\) is a limit of a sequence. A sequence \((a_n)\) is null if for any \(\varepsilon >0\text{,}\) there is some \(N\) such that

\begin{equation*} |a_n| \lt \varepsilon \qquad \text{whenever}\ n \ge N. \end{equation*}

The \(\varepsilon\) in the definition should be regarded as a prescribed distance. In other words, \((a_n)\) is null if for each prescribed distance \(\varepsilon \gt 0\text{,}\) there is a number \(N\) (depending on the chosen \(\varepsilon\)), so that eventually every term, meaning every \(a_n\) with index \(n \ge N\text{,}\) falls within the prescribed distance from \(0\text{.}\) Another way of saying this is that for every \(\varepsilon \gt 0\text{,}\) \(|a_n| \lt \varepsilon\) for all but finitely many indices \(n\text{.}\)

We say that a real number \(L\) is a limit of \((a_n)\) if the sequence \((a_n-L)\) is null. So a sequence is null precisely means 0 is a limit of the sequence. A convergent sequence is a sequence that has a limit. A sequence is divergent if it is not convergent.

We say that a sequence \((a_n)\) diverges to \(+\infty\), written as \(a_n \to +\infty\) or \(\lim_{n\to \infty} a_n = +\infty\text{,}\) if any number \(M\text{,}\) there exists some integer \(N\) so that \(a_n \gt M\) whenever \(n \ge N\text{.}\) Likewise, we say that a sequence \((a_n)\) diverges to \(-\infty\), written as \(a_n \to -\infty\) or \(\lim_{n\to \infty} a_n = -\infty\text{,}\) if any number \(M\text{,}\) there exists some integer \(N\) so that \(a_n \lt M\) whenever \(n \ge N\text{.}\) Equivalently, \(a_n \to -\infty\) if \(-a_n \to +\infty\text{.}\)

It is clear from the definition of convergence that altering and/or dropping finitely many terms of a sequence has no bearing on its convergence.

Example 5.8.

The sequence \((1/n)\) is null. Plot out the first few terms and convince yourself that is the case. To verify this using the definition, take an integer \(N\) larger that \(1/\varepsilon\) for the given \(\varepsilon \gt 0\text{.}\) Then whenever \(n \ge N\text{,}\) \(|1/n| = 1/n \le 1/N \lt \varepsilon\text{.}\)

For example, if \(\varepsilon = 0.1\text{,}\) then we can choose any integer \(N \gt 1/\varepsilon = 10\text{,}\) say \(N=20\text{.}\) Then whenever \(n \ge N\text{,}\) we will have \(1/n \le 1/20 \lt 0.1\text{.}\)

Since \(|a+b| \le |a|+|b|\) and \(|ca| = |c||a|\text{,}\) it follows that the sum of two null sequences is null and a constant multiple of a null sequence is also null. Consequently, if \(L_1,L_2\) are both limits of a sequence \((a_n)\text{,}\) then the constant sequence \((L_1-L_2) = (a_n-L_2)+(-1)(a_n-L_1)\) must also be null and hence \(L_1 = L_2\text{.}\) This shows

Proposition 5.9.

A convergent sequence has a unique limit.

Because of this we can speak of the limit of a convergent sequence. We write \(\lim_{n\to \infty} a_n = L\) or \(\lim_n a_n =L\) or simply \(a_n \to L\) if \(L\) is the limit of \((a_n)\text{.}\)

Suppose \(a_n \to a\text{.}\) So there is some \(N\) so that \(a_n\) are within \(1\) from \(a\) for all \(n \ge N\text{.}\) That means the terms are bounded between \(\max\{a_1,\ldots, a_{N-1}, a+1\}\) and \(\min\{a_1,\ldots, a_{N-1},a-1\}\text{.}\) We have just argued that

Proposition 5.10.

Every convergent sequence is bounded.

Proposition 5.11. Squeeze Lemma.

Suppose \(a_n \le b_n \le c_n\) for all \(n\) and that \((a_n)\) and \((c_n)\) both converge to the same limit \(L\text{.}\) Then \(b_n \to L\text{.}\)

Proof.

Since \((a_n - L)\) and \((c_n-L)\) are both null sequences. For any \(\varepsilon > 0\text{,}\) the conditions \(|a_n| \lt \varepsilon\) and \(|c_n| \lt \varepsilon\) each fails for only finitely many \(n\)’s. Thus, these conditions hold simultaneously for all but finitely many \(n\)’s. Consequently,

\begin{equation*} -\varepsilon \lt a_n - L \le b_n - L \le c_n-L \lt \varepsilon \end{equation*}

for all but finitely many \(n\)’s. This shows that \((b_n)\) converges to \(L\) as well.

Algebraic operations on sequences respect convergence.

Proposition 5.12.

Suppose \(a_n \to a\) and \(b_n \to b\text{.}\) Then

\(\displaystyle a_n+b_n \to a+b\)
\(\displaystyle a_nb_n \to ab.\)
\(\dfrac{a_n}{b_n} \to \dfrac{a}{b}\) provided that \(b_n \neq 0\) and \(b\neq 0\)

Proof.

Both \((a_n-a)\) and \((b_n-b)\) are null and hence so is

\begin{equation*} (a_n-a+b_n-b) = (a_n+b_n-(a+b))\text{.} \end{equation*}

That means \(a_n+b_n \to a+b\text{.}\) Also,

\begin{align*} (a_nb_n) & = (a_n-a)(b_n-b) + (ab_n) + (ba_n) - (ab)\\ & \to 0 +ab + ba -ab = ab \end{align*}

Since \(b \neq 0\text{,}\) there are \(0\lt m \lt M\text{,}\) such that for all \(n\) sufficiently large, \(m \lt |b_n| \lt M\text{.}\) And so,

\begin{equation*} \frac{|b-b_n|}{bM} \lt \frac{|b-b_n|}{|bb_n|} \lt \frac{|b-b_n|}{bm} \end{equation*}

Since \(|b-b_n| \to 0\text{,}\) the above inequalities show that (Proposition 5.11)

\begin{equation*} \left| \frac{1}{b_n} - \frac{1}{b} \right| = \frac{|b-b_n|}{bb_n} \to 0. \end{equation*}

We should say a few words about the quotient operation on sequences. If \(b_n \to b\) and \(b\neq 0\text{,}\) then eventually all \(b_n\)’s are non-zero hence \(a_n/b_n\) is defined for all sufficiently large \(n\text{.}\) Also, if \((b_n)\) is null, then the convergence of \((a_n/b_n)\) is not certain. For example, take \(a_n=1\) and \(b_n = (-1)^n/n\text{,}\) then \(a_n/b_n = (-1)^n n\) which form a divergent sequence. On the other hand, take \(a_n = A/n\) and \(b_n=1/n\text{,}\) then \((a_n/b_n)\) is simply the constant sequence \((A)\) which converges to \(A\text{.}\)

Certainly, if \((a_n)\) converges to a non-zero number but \((b_n)\) is null, then \((|a_n/b_n|)\) diverges to \(+\infty\text{.}\)

Example 5.13.

Let us argue that the flip-flop sequence \(((-1)^n)\) is divergent. Suppose on the contrary that it converges to some number \(L\text{.}\) Then the sequence \(((-1)^{n+1})\text{,}\) viewed as the sequence obtained by dropping the first term of \(((-1)^n)\text{,}\) must also converge to \(L\text{.}\) And so, their sum, which is constantly 0, must converge to \(2L\text{.}\) That means \(L=0\text{.}\) However, the sequence \(((-1)^n)\) is clearly not null since the absolute values of its terms are all \(1\text{.}\)

Note also that the flip-flop sequence is bounded, so neither does it does diverges to \(+\infty\) nor to \(-\infty\text{.}\)

Checkpoint 5.14.

Give an unbounded sequence (and hence must be divergent) that neither diverges to \(+\infty\) nor diverges to \(-\infty\text{.}\)

Example 5.15.

For any \(M > 0\text{,}\) and any \(n > e^M, \ln n> M\text{.}\) Therefore, \(\ln(n) \to +\infty\) and \(\ln(1/n) = - \ln n \to -\infty\text{.}\)

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