Monotone Sequences and Their Order

What Does It Mean for a Sequence to Be Monotone?

A sequence can rise, fall, stay level, or change direction as its index increases. To describe its order precisely, we compare each term with the term immediately after it. A sequence whose terms never decrease is called increasing; one whose terms never increase is called decreasing. A sequence that is both is constant.

These descriptions concern order, not how quickly terms change or whether they approach a particular number. The same definitions apply whether the terms are integers, fractions, or other real numbers.

Definition: Let \((a_n)\) be a real sequence, indexed by \(n\in\mathbb{N}\). It is increasing (or nondecreasing) if \(a_n\le a_{n+1}\) for every \(n\in\mathbb{N}\). It is decreasing (or nonincreasing) if \(a_n\ge a_{n+1}\) for every \(n\in\mathbb{N}\). It is constant if \(a_n=a_{n+1}\) for every \(n\in\mathbb{N}\).

In these definitions, equality is allowed. To require every step to be a genuine rise or fall, use the terms strictly increasing and strictly decreasing: the requirements are \(a_n<a_{n+1}\) for every \(n\), and \(a_n>a_{n+1}\) for every \(n\), respectively. Some contexts use “increasing” to mean strictly increasing. Here, “increasing” means nondecreasing; the word “strictly” makes the stronger condition explicit.

From Consecutive Terms to the Whole Sequence

Although the definitions compare neighbors, they also determine the order of every pair of terms. If an increasing sequence is considered at indices \(m\) and \(n\) with \(m<n\), then its terms satisfy \(a_m\le a_n\). For a decreasing sequence, the order is reversed. This fact follows directly by chaining the consecutive comparisons.

Theorem: A sequence is increasing exactly when \(a_m\le a_n\) whenever \(m\le n\). It is decreasing exactly when \(a_m\ge a_n\) whenever \(m\le n\). Replacing the weak inequalities by strict inequalities for \(m<n\) gives the corresponding characterizations of strictly increasing and strictly decreasing sequences.

Justification. First suppose the sequence is increasing, so \(a_k\le a_{k+1}\) for every \(k\). For any \(m<n\), the consecutive comparisons from \(m\) to \(n\) form the chain

a_m\le a_{m+1}\le a_{m+2}\le\cdots\le a_n.

Transitivity of the order on the real numbers gives \(a_m\le a_n\). When \(m=n\), the conclusion is simply \(a_m=a_n\). Conversely, if \(a_m\le a_n\) whenever \(m\le n\), choose \(m=n\) and \(n=n+1\) in the pairwise statement (equivalently, apply it to any adjacent pair). It gives \(a_n\le a_{n+1}\), which is the definition of increasing. The decreasing case follows by reversing each inequality. In the strict case, the same chain consists of strict inequalities when \(m<n\), and the converse follows by considering adjacent indices.

This characterization is useful for understanding what monotonicity says: an increasing sequence cannot later fall below one of its earlier terms. It does not mean that every term must be different from the preceding one; equality is permitted unless strict monotonicity is specified.

Worked Classifications

Worked Example: A Strictly Increasing Rational Sequence

Classify the sequence \(a_n=\frac{2n+1}{n+2}\), where \(n\in\mathbb{N}\).

Compare consecutive terms. Substituting \(n+1\) gives

a_{n+1}=\frac{2(n+1)+1}{(n+1)+2}=\frac{2n+3}{n+3}.

Subtract \(a_n\) from \(a_{n+1}\), using a common denominator:

\begin{aligned} a_{n+1}-a_n &=\frac{2n+3}{n+3}-\frac{2n+1}{n+2}\\ &=\frac{(2n+3)(n+2)-(2n+1)(n+3)}{(n+2)(n+3)}\\ &=\frac{(2n^2+7n+6)-(2n^2+7n+3)}{(n+2)(n+3)}\\ &=\frac{3}{(n+2)(n+3)}. \end{aligned}

For every \(n\in\mathbb{N}\), both denominator factors are positive, so \(a_{n+1}-a_n>0\). Therefore \(a_n<a_{n+1}\) at every step: the sequence is strictly increasing. As a check, \(a_1=1\), \(a_2=\frac54\), and their difference is \(\frac14\), matching \(\frac{3}{(1+2)(1+3)}=\frac14\).

Worked Example: A Sequence That Changes Direction

Classify \(b_n=(-1)^n\). The first four terms are

b_1=-1,\qquad b_2=1,\qquad b_3=-1,\qquad b_4=1.

For an increasing sequence, every adjacent pair must satisfy \(b_n\le b_{n+1}\). The pair \(b_2,b_3\) fails this condition because \(1\not\le -1\). Thus the sequence is not increasing. For a decreasing sequence, every adjacent pair must satisfy \(b_n\ge b_{n+1}\). The pair \(b_1,b_2\) fails because \(-1\not\ge 1\). Thus the sequence is not decreasing, either.

In fact, no tail of this sequence is monotone. However far along the sequence one starts, there is a next adjacent pair that rises from \(-1\) to \(1\), followed by one that falls from \(1\) to \(-1\). Both directions therefore occur arbitrarily far along the sequence. The checks use exactly the required adjacent comparisons: \(-1\le1\) holds for the first pair, while \(1\le-1\) fails for the next; the reverse inequality fails for the first pair.

Worked Example: Monotonicity After a Finite Initial Segment

Define a sequence by \(c_1=0\) and \(c_n=\frac{1}{n-1}\) for \(n\ge2\). Its first terms are

c_1=0,\qquad c_2=1,\qquad c_3=\frac12,\qquad c_4=\frac13.

The sequence is not increasing, since \(c_2\le c_3\) would require \(1\le\frac12\), which is false. It is not decreasing, since \(c_1\ge c_2\) would require \(0\ge1\), also false.

Now compare consecutive terms from index \(n=2\) onward. For \(n\ge2\),

c_n=\frac{1}{n-1}, \qquad c_{n+1}=\frac{1}{n}.

Since \(n-1\) and \(n\) are positive and \(n-1<n\), their reciprocals satisfy \(\frac{1}{n-1}>\frac{1}{n}\). Hence \(c_n>c_{n+1}\) for every \(n\ge2\). The sequence is strictly decreasing from its second term onward, even though it is not decreasing from the start.

For example, the comparison at \(n=2\) is \(c_2=1>c_3=\frac12\), and at \(n=3\) it is \(c_3=\frac12>c_4=\frac13\), as the general calculation predicts.

Eventually Monotone Sequences

A sequence can fail to be monotone because of a finite initial segment, yet settle into one direction afterward. The appropriate definition states exactly where the consecutive comparisons must begin.

Definition: A sequence is eventually increasing if there is some \(N\in\mathbb{N}\) such that \(a_n\le a_{n+1}\) for every \(n\ge N\). It is eventually decreasing if there is some \(N\in\mathbb{N}\) such that \(a_n\ge a_{n+1}\) for every \(n\ge N\). Using strict inequalities defines eventually strictly increasing or eventually strictly decreasing. In each case, the sequence from index \(N\) onward is called its tail.

The index \(N\) is allowed to depend on the sequence; it need not be the first index from which the comparisons hold. An eventually increasing sequence is not necessarily increasing from its first term, as the example above illustrates. Conversely, if the comparisons hold from the first index, then the sequence is eventually monotone by taking \(N=1\).

Changing or removing finitely many initial terms can affect whether a sequence is monotone from the beginning, but it cannot affect a property that holds on a sufficiently late tail. This distinction is useful whenever a sequence has a special initial value or a finite starting irregularity.

Common Pitfall: Confusing the Order of Denominators

When comparing fractions, a larger positive denominator gives a smaller reciprocal, not a larger one. For instance, it would be incorrect to argue that \(1/(n+1)>1/n\) because \(n+1>n\). At \(n=1\), that claim says \(\frac12>1\), which is false.

For every \(n\in\mathbb{N}\), both \(n\) and \(n+1\) are positive, and the exact difference is

\frac{1}{n+1}-\frac{1}{n} = \frac{n-(n+1)}{n(n+1)} = -\frac{1}{n(n+1)} <0.

Therefore \(1/(n+1)<1/n\): the sequence \(1/n\) is strictly decreasing. The correction depends on the sign of the difference, not merely on noticing that the denominator changes.

Why the Order Classification Matters

Monotonicity gives a concise description of how terms are arranged across their indices. A sequence may be strictly increasing, weakly increasing with some repeated terms, decreasing, constant, or neither increasing nor decreasing. It may also have a tail with one of these patterns even when its initial terms do not.

These are distinct possibilities. For example, \(a_n=n\) is strictly increasing, while the constant sequence \(a_n=4\) is both increasing and decreasing under the non-strict definitions. The sequence \((-1)^n\) changes direction repeatedly, so it is neither. Recognizing the correct classification keeps later reasoning about sequences precise: one must check the required comparisons at every relevant index, not infer a pattern from just a few terms.

Check Your Understanding

Answer each question using comparisons of consecutive terms.

State the condition that makes a sequence increasing under the convention used here. Does the condition allow equal consecutive terms?
Can a sequence be both increasing and decreasing? If so, what must be true of its consecutive terms?
Classify \(a_n=3-n\) as increasing, decreasing, or neither. Is it strict?
For \(d_1=2\) and \(d_n=1/(n-1)\) for \(n\ge2\), is the sequence monotone from the beginning? From some later index onward?
What does it mean for a sequence to be eventually decreasing, and how does that differ from being decreasing?