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Section1Bounded Sequences

Convergence is a very strong property for a sequence to have, since it requires the tails of the sequence to all grow arbitrarily close to a specified real number (its limit). Let's look at some simpler properties, each of which is weaker than convergence, and their relationships to convergence.

Definition1.1Bounded sequence
A sequence \(s_n\) of real numbers is called bounded if there exists a real number \(M \in\mathbb{R}\) such that, for all \(n \in\mathbb{N}\text{,}\) we have \begin{equation*} |s_n| \leq M. \end{equation*}

Intuitively, we might say that all terms of a bounded sequence lie between a constant "ceiling" and a constant "floor:" another way to write the inequality at the end of the definition is \(-M \leq |s_n| \leq M.\)

Boundedness is indeed a weaker condition than convergence; while it is not true that every bounded sequence is convergent, is is true that every convergent sequence is bounded:

Intuitively, convergence is a strong condition. Given any \(\epsilon>0\text{,}\) it produces an \(N\in\mathbb{N}\) which divides the sequence into a (finite) head and an (infinite) tail. We imagine that each will have a ceiling of its own:

Now, let us define \(\epsilon = 1\text{.}\) Since \(s_n\) is a convergent sequence, let us denote \(L = \lim_{n\to\infty} s_n.\)

Then by definition of convergence there exists an \(N\in\mathbb{N}\) such that, for all \(n \geq N\text{,}\) we have

\begin{equation*} |s_n - L| \lt \epsilon. \end{equation*}

This defines for us a head of the sequence, \(\{s_1,s_2,\ldots,s_N\}\text{,}\) and a tail of the sequence, \(\{s_N,s_{N+1},s_{N+2},\ldots\}\text{,}\) and all of the terms in the tail are within a distance of \(\epsilon\) of the limit \(L\text{.}\)

Using an add-subtract trick can shift the inequality \(|s_n-L|\lt\epsilon\) from a measurement of the sequence's distance from \(L\) into a measurement of its distance from zero (i.e., its absolute value):

\begin{align} |s_n| = |s_n - L + L| \leq |s_n-L| + |L| \lt \epsilon + |L| \amp\amp \text{for all } n \geq N. \tag{1.1} \end{align}

In other words, \(\epsilon+|L|\) is an upper bound for the tail of the sequence.

Meanwhile, since the head of the sequence is a finite set, it will in particular have a largest element that can be used as an upper bound for that set. So we define \(m = \max \{ |s_1|, |s_2|, \ldots, |s_N| \}.\)

Now define \(M = \max\{ m , \epsilon+|L|\}.\)

Let \(n\in\mathbb{N}\) be arbitrarily chosen. Then there are two cases, depending on whether the \(n\)th term belongs to the head of the sequence or the tail:

  1. If \(n \leq N\), then \(s_n\) belongs to the head of the sequence and \(|s_n|\) is one of the values in the finite list which was used to define the maximum of the head, \(m\text{.}\) Hence \(|s_n|\leq m \leq M.\)
  2. If \(n \gt N\), then \(s_n\) belongs to the tail of the sequence and \(|s_n|\) is governed by the tail inequality (1.1). Hence \(|s_n| \lt \epsilon+|L| \leq M.\)

This covers all cases, so we conclude that for all \(n\in\mathbb{N}\) we have \(|s_n|\leq M.\)

The contrapositive of Theorem 2 then says that "unbounded implies not convergent." But be careful not to confuse this with "unbounded implies divergent to infinity."

Find an example of an unbounded sequence \(s_n\) for which \(s_n\) does not diverge to infinity.