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Section 9.2 Monotone Sequence

Definition 9.2.1.

\(\displaystyle \left\{ a_n \right\}_{n = 1}^{\infty} \) is called:

  • strictly increasing if \(a_1 \lt a_2 \lt \dots \lt a_n\)

  • increasing if \(a_1 \leq a_2 \leq \dots \leq a_n\)

  • strictly decreasing if \(a_1 \gt a_2 \gt \dots \gt a_n\)

  • decreasing if \(a_1 \geq a_2 \geq \dots \geq a_n\)

A sequence which is increasing or decreasing is monotone and a sequence which is strictly increasing or decreasing is strictly monotone.

How do we tell what a squence does?

  • strictly increasing if \(a_{n+1} - a_n \gt 0\) or if \(\dfrac{a_{n+1}}{a_n} \gt 1\)

  • increasing if \(a_{n+1} - a_n \geq 0\) or if \(\dfrac{a_{n+1}}{a_n} \geq 1\)

  • strictly decreasing if \(a_{n+1} - a_n \lt 0\) or if \(\dfrac{a_{n+1}}{a_n} \lt 1\)

  • decreasing if \(a_{n+1} - a_n \leq 0\) or if \(\dfrac{a_{n+1}}{a_n} \leq 1\)

Example 9.2.1.

Determine the type of sequence:

\begin{align*} a_n \amp = \dfrac{n}{n+1}\\ a_{n+1} \amp = \dfrac{n+1}{n+2} \end{align*}

Substitution method:

\begin{align*} a_{n+1} - a_n \amp = \dfrac{n+1}{n+2} - \dfrac{n}{n+1} \\ \amp = \dfrac{(n+1)(n+1)}{(n+2)(n+1)} - \dfrac{n(n_2)}{(n+2)(n+1)}\\ \amp = \dfrac{(n+1)(n+1) - n (n+2)}{(n+2)(n+1)}\\ \amp = \dfrac{n^2 + 2n + 1 - n^2 - 2n}{n^2+3n+2}\\ \amp = \dfrac{1}{n^2+3n+2} \gt 0\\ \amp \boxed{\text{strictly increasing}} \end{align*}

Division method:

\begin{align*} \dfrac{a_{n+1}}{a_n} \amp = \dfrac{\frac{n+1}{n+2}}{\frac{n}{n+1}} \\ \amp = \dfrac{n+1}{n+2} \cdot \dfrac{n+1}{n} \\ \amp = \dfrac{(n+1)(n+1)}{n(n+2)}\\ \amp = \dfrac{n^2+2n+1}{n^2+2n} \gt 1\\ \amp \boxed{\text{strictly increasing}} \end{align*}

Subsection 9.2.1 Eventually Monotone Sequences

Consider the sequence

\begin{equation*} 9, \quad -8, \quad -17, \quad 12, \quad 1, \quad 2, \quad 3, \quad 4 \end{equation*}

Definition 9.2.2.

If discarding finitely many terms from the beginning of a sequence produces a sequence with a certain proterty, we say that the original sequence has that property eventually.

Example 9.2.2.

Show \(a_n = \frac{10^n}{n!}\) is eventually strictly decreasing:

\begin{equation*} 10, \quad 50, \quad \frac{500}{3}, \quad \frac{10000}{24}, \quad \frac{100000}{5!}, \dots \end{equation*}
\begin{align*} a_n \amp = \dfrac{10^n}{n!}\\ a_{n+1} \amp = \dfrac{10^{n+1}}{(n+1)!} \end{align*}

Using the divisional method:

\begin{align*} \dfrac{a_{n+1}}{a_n} \amp = \dfrac{\frac{10^{n+1}}{(n+1)!}}{\frac{10^n}{n!}} \\ \amp = \dfrac{10^{n+1}}{(n+1)!} \cdot \dfrac{n!}{10^n}\\ \amp = \dfrac{10}{n+1}\\ \dfrac{a_{n+1}}{a_n} \amp = \dfrac{10}{n+1} \lt 1 \text{ for } n \geq 10 \\ \amp \boxed{\text{eventually strictly decreasing}} \end{align*}

Subsection 9.2.2 Convergence of Monotone Sequences

If a sequence is eventually increasing there are two possibilities:

  • there is a constant \(M\) called an upper bound for the sequence such that \(a_n \leq M\) for all \(n\text{,}\) in which case the sequence converges to a limit \(L\) with \(L \leq M\)

  • no upper bound exists so \(\displaystyle \lim_{n\to\infty} a_n = \infty\)

If a sequence is eventually decreasing, there are two possibilities:

  • there is a constant \(M\text{,}\) called a lower bound for the sequence that \(M \leq a_n\) for all \(n\) in which case the sequence converges to a limit \(L\) with \(M \leq L\)

  • no lower bound exists so \(displaystyle \lim_{n\to\infty} a_n = - \infty \)

Example 9.2.3.

Show \(a_n = \frac{10^n}{n!}\) converges. We know \(\left\{ \frac{10^n}{n!} \right\}\) is eventually strictly decreasing, so we need to find a lower bound.

Sequence: \(10, \quad 50, \quad \frac{500}{3}, \quad \frac{10000}{24}, \dots\)

Can we find a number smaller than any of these? Yes, \(0\) is a lower bound thus the sequence converges.

Now we know the sequence converges but can we find what it converges to?

\begin{align*} a_n \amp = \frac{10^n}{n!}\\ a_{n+1} \amp = \frac{10^{n+1}}{(n+1)!} \\ \amp = \frac{10 \cdot 10^n}{(n+1) \cdot n!}\\ \amp = \frac{10}{n+1} a_n \end{align*}
\begin{align*} \lim_{n\to\infty} a_n \amp = L \\ \lim_{n\to\infty} a_{n+1} \amp = L \end{align*}
\begin{align*} L \amp = \lim_{n\to\infty} a_{n+1}\\ \amp = \lim_{n\to\infty} \frac{10}{n+1}a_n\\ \amp = 0 \cdot L\\ \amp = 0 \\ \amp \boxed{\text{converges to } 0} \end{align*}