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.
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:
Theorem1.2Convergent implies bounded
Let \(s_n\) be a convergent sequence of real numbers. Then \(s_n\) is a bounded sequence.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:
- 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 \(|s_n|\leq m \leq M.\)
- 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.