5.1 Sequences  (Page 7/25)

For example, the sequence $\left\{1\text{/}n\right\}$ is bounded above because $1\text{/}n\le 1$ for all positive integers $n.$ It is also bounded below because $1\text{/}n\ge 0$ for all positive integers n. Therefore, $\left\{1\text{/}n\right\}$ is a bounded sequence. On the other hand, consider the sequence $\left\{2^n\right\}.$ Because $2^n\ge 2$ for all $n\ge 1,$ the sequence is bounded below. However, the sequence is not bounded above. Therefore, $\left\{2^n\right\}$ is an unbounded sequence.

We now discuss the relationship between boundedness and convergence. Suppose a sequence $\left\{a_n\right\}$ is unbounded. Then it is not bounded above, or not bounded below, or both. In either case, there are terms $a_n$ that are arbitrarily large in magnitude as $n$ gets larger. As a result, the sequence $\left\{a_n\right\}$ cannot converge. Therefore, being bounded is a necessary condition for a sequence to converge.

Note that a sequence being bounded is not a sufficient condition for a sequence to converge. For example, the sequence $\left\{{\left(\mathrm{-1}\right)}^n\right\}$ is bounded, but the sequence diverges because the sequence oscillates between $1$ and $\mathrm{-1}$ and never approaches a finite number. We now discuss a sufficient (but not necessary) condition for a bounded sequence to converge.

Consider a bounded sequence $\left\{a_n\right\}.$ Suppose the sequence $\left\{a_n\right\}$ is increasing. That is, $a_1\le a_2\le a_3\text{…}.$ Since the sequence is increasing, the terms are not oscillating. Therefore, there are two possibilities. The sequence could diverge to infinity, or it could converge. However, since the sequence is bounded, it is bounded above and the sequence cannot diverge to infinity. We conclude that $\left\{a_n\right\}$ converges. For example, consider the sequence

Since this sequence is increasing and bounded above, it converges. Next, consider the sequence

Even though the sequence is not increasing for all values of $n,$ we see that $\mathrm{-1}\text{/}2<\text{−}1\text{/}3<\text{−}1\text{/}4<\text{⋯}.$ Therefore, starting with the eighth term, $a_8=\mathrm{-1}\text{/}2,$ the sequence is increasing. In this case, we say the sequence is eventually increasing. Since the sequence is bounded above, it converges. It is also true that if a sequence is decreasing (or eventually decreasing) and bounded below, it also converges.

We now have the necessary definitions to state the Monotone Convergence Theorem, which gives a sufficient condition for convergence of a sequence.

The proof of this theorem is beyond the scope of this text. Instead, we provide a graph to show intuitively why this theorem makes sense ( [link] ).