# Boundedness of Sequences

${k_2} \in \mathbb{R}$ such that ${k_1} \leqslant {u_n} \leqslant {k_2}$, $\forall {\text{ }}n \in \mathbb{N}$. Equivalently, $u$ is bounded if $\exists {\text{ }}k \geqslant 0$ such that$\left| {{u_n}} \right| \leqslant k{\text{ }}\forall {\text{ }}n \in \mathbb{N}$.

Evidently, $R\left\{ u \right\}$ is bounded if and only if $u$ is bounded. Upper and lower bounds (supremum and infimum) of $R\left\{ u \right\}$, if exist, are called the upper and lower bounds of the sequence $u$. A constant sequence is obviously bounded.

In sequences, terms with equal values can occur. Therefore, a sequence may have more than one term with the smallest value. In such a case any of these is taken for the smallest value. In fact while talking about the smallest term we are substantially interested in the value of the term rather than the position of the term in the sequence. Similar explanation holds for the greatest term. Note that, like sets of real numbers, a sequence bounded below or above may or may not have a smallest or a greatest member accordingly. Clearly, an unbounded sequence cannot have a smallest or a greatest member.

Example: The sequence whose nth term is.

(i) $1 + {\left( { - 1} \right)^n}$is bounded and has smallest and greatest terms 0, 2. Every non-positive number is a lower bound and any member of $[2,\infty )$ is an upper bound of the sequence.

(ii) ${2^n}$is bounded below and has smallest term as 2. Every member of$( - \infty ,2]$, is a lower bound of the sequence and the sequence is unbounded above.

(iii) ${\left( { - 1} \right)^n}$ n is unbounded, both ways.

(iv) $1/n$ is bounded and has no smallest member whereas it has 1 as the greatest member$( - \infty ,0]$ and $[1,\infty )$ are its sets of lower and upper bounds.