Monotonic Sequences

Section2Monotonic Sequences

Another class of sequences whose behavior is well regulated is the class of sequences which "do not change direction." These are the monotonic sequences.

Definition2.1Monotonic sequence

A sequence \(s_n\) of real numbers is called monotonic if one of the following is true:

For all \(n\in\mathbb{N}\text{,}\) we have \(s_n \leq s_{n+1}.\)
For all \(n\in\mathbb{N}\text{,}\) we have \(s_n \geq s_{n+1}.\)

In the first case, we say the sequence is increasing. In the second case, we say the sequence is decreasing. If either inequality is a strict inequality (\(\lt\) or \(\gt\)), then we say the sequence is "strictly" increasing or decreasing respectively.

Monotonicity alone is not sufficient to guarantee convergence of a sequence. Indeed, many monotonic sequences diverge to infinity, such as the natural number sequence \(s_n = n\text{.}\) But, if we can force a monotonic sequence to remain trapped between a constant ceiling and floor, we can guarantee it will converge. This is the monotone convergence theorem.

Theorem2.2Monotone convergence theorem

Let \(s_n\) be an increasing (respectively, decreasing) sequence. If \(s_n\) is bounded above (respectively, bounded below), then \(s_n\) is convergent and

\begin{align*} \lim_{n\to\infty} s_n \amp = \sup \{ s_n \} \amp\amp \text{(respectively,}\quad \inf \{ s_n \}.) \end{align*}

In other words, an increasing sequence which is bounded above will converge to its supremum; a decreasing sequence which is bounded above will converge to its infimum.

Proof

We will show the proof for the increasing case. The decreasing case is analogous by reversing inequality symbols and trading supremum for infimum.

Let \(s_n\) be an increasing sequence that is bounded above. The completeness axiom of the real numbers guarantees that any bounded-above and nonempty set of real numbers has a least upper bound, so \(S = \sup \{s_n\}\) exists. We will show that the sequence \(s_n\) converges to \(S\text{.}\)

Let \(\epsilon \gt 0\) be arbitrarily chosen.

Since \(S = \sup\{s_n\}\) is the least upper bound of the set of all terms of the sequence, we know by a result from a previous quiz that there exists at least one term of the sequence which is greater than \(M-\epsilon\text{.}\)

Define \(N\in\mathbb{N}\) to be the index of this term, i.e. choose \(N\) such that

\begin{gather} s_N > M-\epsilon.\tag{2.1} \end{gather}

Now let \(n \geq N\) be arbitrarily chosen.

Because the sequence is increasing, we know that \(n \geq N\) implies \(s_n \geq s_N\text{.}\) But (2.1) then guarantees that

\begin{align} s_n \geq s_N \amp > S - \epsilon.\tag{2.2} \end{align}

Furthermore, since \(S\) is an upper bound for \(\{s_n\}\) we have \(s_n \leq S\text{.}\) This guarantees that \(s_n-S \leq 0\) and hence \(|s_n-S| = S-s_n.\) For this reason, (2.2) shows that

\begin{align*} |s_n-S| \amp = S-s_n \amp \lt S - (S-\epsilon) = \epsilon, \end{align*}

completing the proof.

The monotone convergence theorem provides a powerful one-two punch that is sufficient to prove that a sequence converges by proving two (probably) simpler properties: that it is monotonic, and that it is bounded. The following example illustrates how the monotone convergence theorem might be applied to a concrete example of a sequence.

Example2.3A recursively-defined sequence

Prove that the sequence which is defined recursively by

\begin{align*} s_1 = 1 \amp\amp s_{n+1}=\sqrt{1+s_n} \end{align*}

converges.

Proof: The sequence is monotonic: If this is indeed true, then all its terms will follow the pattern suggested by its first two, namely that

\begin{align*} s_1 = 1 \amp \amp \lt \amp \amp \sqrt{2} = s_2. \end{align*}

So we will try to prove that \(s_n\) is a stricly increasing sequence, i.e. we'll prove that

\begin{align} s_{n+1} \amp \gt s_n \amp\amp \forall n \geq 1\tag{2.3} \end{align}

Base case: \(n=1\text{.}\) We have

\begin{align*} s_{1+1}=s_n \amp = \sqrt{2} \amp \gt 1 = s_1 \end{align*}

establishing the base case.

Induction hypothesis: Assume that (2.3) is true.

Induction case: We must show that

\begin{align*} s_{(n+1)+1} \amp \gt s_{n+1} \amp\amp \text{i.e., that}\amp s_{n+2} \gt s_{n+1}. \end{align*}

But by definition of this sequence,

\begin{align*} s_{n+2} \amp = \sqrt{1+s_{n+1}} \amp\\ \amp \gt \sqrt{1+s_n} \amp \text{by the induction hypothesis } \knowl{./knowl/mdn-4.html}{\text{(2.3)}}\\ \amp = s_{n+1} \amp \text{by definition of this sequence.} \end{align*}

Thus we have proven the claim that this sequence is increasing.

The sequence is bounded above: Inspecting the first few terms of the sequence, we might conjecture that the whole sequence is bounded above by 2. This is what we'll prove, again by induction. We'll show that

\begin{align} |s_n| \amp \leq 2 \amp\amp \forall n \geq 1.\tag{2.4} \end{align}

Base case: \(n=1\text{.}\) We have

\begin{align*} |s_1| = 1 \amp \leq 2. \end{align*}

Induction hypothesis: Assume that (2.4) is true.

Induction case: We must show that

\begin{gather*} |s_{n+1}| \leq 2. \end{gather*}

By definition of this sequence,

\begin{align*} |s_{n+1}| \amp = s_{n+1} \amp \text{(since the sequence is nonnegative)}\\ \amp = \sqrt{1+s_n} \amp \\ \amp \leq \sqrt{1+2} \amp \text{by the induction hypothesis } \knowl{./knowl/mdn-5.html}{\text{(2.4)}}\\ \amp = \sqrt{3} \leq 2. \end{align*}

This proves the claim that this sequence is bounded above.

Since \(s_n\) is increasing and bounded above, by the monotone convergence theorem it is convergent.