An increasing (or nondecreasing) sequence \(\{ s_{n}\} \) that is bounded from above converges. A decreasing (or nonincreasing) sequence \(\{ s_{n}\} \) that is bounded from below converges.
To prove this theorem, we need the following property of real numbers. The set of real numbers is defined by a collection of axioms. One of these axioms is the Completeness Axiom.
If \(S\) is a nonempty set of real numbers that has an upper bound, then it has a least upper bound. Similarly, if \(S\) has a lower bound, then it has a greatest lower bound.
As an example, consider the set \(S\): \(\{x\,|\,x^{2} < 2,\,\,x>0\}\). The set of upper bounds to \(S\) is the set \(\{x\,|\,x^{2}\geq 2,\,\,x>0\}\).
We prove the theorem for a nondecreasing sequence \(\{ s_{n}\}\). The proofs for the other three cases are similar.
B-12
Suppose \(\{ s_{n}\} \) is a nondecreasing sequence that is bounded from above. Since \(\{ s_{n}\} \) is bounded from above, there is a positive number \(K\) (an upper bound), so that \(s_{n}\leq K\) for every \(n\). From the Completeness Axiom, the set \(\{ s_{n}\} \) has a least upper bound \(L.\) That is, \(s_{n}\leq L\) for every \(n.\)
Then for any \(\varepsilon >0,\) \(L-\varepsilon \) is not an upper bound of \( \{ s_{n}\} .\) That is, \(L- \varepsilon < s_{N}\) for some integer \( N\). Since \(\{ s_{n}\} \) is nondecreasing, \(s_{N}\leq s_{n}\) for all \(n > N.\) Then for all \(n > N,\) \begin{equation*} L - \varepsilon < s_{n}\leq L < L + \varepsilon \end{equation*}
That is, \(\vert s_{n}-L\vert < \varepsilon \) for all \(n > N,\) so the sequence \(\{ s_{n}\} \) converges to \(L\).