Monotone Sequences

Monotonicity Orders the Terms

An upper or lower bound controls where all the terms of a sequence lie. Monotonicity describes a different feature: how the terms are ordered as the index increases. A sequence may rise, fall, or remain level from one term to the next, and the same direction of change must persist throughout the sequence for it to be monotone.

We begin with the non-strict forms of increase and decrease. Under these definitions, equal consecutive terms are permitted. Some texts use “increasing” and “decreasing” for strict inequalities instead, so the terms nondecreasing and nonincreasing make the intended meaning explicit.

Definition: A real sequence $(a_n)_{n=0}^{\infty}$ is nondecreasing if $$ a_n\leq a_{n+1} $$ for every $n\in\mathbb{N}_0$. It is nonincreasing if $$ a_n\geq a_{n+1} $$ for every $n\in\mathbb{N}_0$. A sequence is monotone if it is either nondecreasing or nonincreasing.

The direction must work at every index. A sequence can have many consecutive increases and still fail to be nondecreasing if even one step decreases. Likewise, a single increase prevents a sequence from being nonincreasing. A constant sequence satisfies both definitions, since equality is allowed in each.

Consecutive Terms Determine the Whole Order

The definition compares consecutive terms, but it also determines the order of any two terms whose indices are ordered. This is useful because a comparison between distant terms need not be checked by a separate calculation: it follows by chaining the consecutive inequalities.

Theorem: A sequence $(a_n)$ is nondecreasing if and only if $a_m\leq a_n$ whenever $m,n\in\mathbb{N}_0$ and $m\leq n$. It is nonincreasing if and only if $a_m\geq a_n$ whenever $m\leq n$.

Proof. Suppose first that $(a_n)$ is nondecreasing, so $a_k\leq a_{k+1}$ for every $k\in\mathbb{N}_0$. Fix $m\leq n$. If $m=n$, then $a_m=a_n$. If $m<n$, chaining the consecutive inequalities gives

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

Thus $a_m\leq a_n$. Conversely, suppose $a_m\leq a_n$ whenever $m\leq n$. Taking $m=k$ and $n=k+1$ gives $a_k\leq a_{k+1}$ for every $k$, so the sequence is nondecreasing.

For a nonincreasing sequence, the proof uses the reversed inequalities. If $a_k\geq a_{k+1}$ for every $k$, then for $m\leq n$ we have

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

Hence $a_m\geq a_n$. Conversely, the comparison for every pair of indices gives $a_k\geq a_{k+1}$ by taking $m=k$ and $n=k+1$. Both equivalences follow. $\square$

The pairwise version is particularly helpful when a formula makes $a_m\leq a_n$ easier to establish directly than a comparison of consecutive terms. The consecutive version is often shorter when the sequence has a simple recurrence or when subtracting adjacent terms gives a manageable expression.

Worked Example: A Rational Sequence That Increases

For $n\in\mathbb{N}_0$, define $$ a_n=\frac{3n+2}{n+1}. $$ Since $n+1>0$, we can rewrite the terms as

a_n=3-\frac{1}{n+1}.

Compare consecutive terms:

a_{n+1}-a_n =\left(3-\frac{1}{n+2}\right)-\left(3-\frac{1}{n+1}\right) =\frac{1}{n+1}-\frac{1}{n+2} =\frac{1}{(n+1)(n+2)}.

Both factors in the denominator are positive, so $a_{n+1}-a_n>0$ for every $n$. Thus $a_n<a_{n+1}$, which in particular shows that the sequence is nondecreasing. The first terms confirm the calculation: $a_0=2$, $a_1=5/2$, and $a_2=8/3$. The comparison theorem also shows that $a_m\leq a_n$ whenever $m\leq n$.

Testing Monotonicity with Differences

For any real sequence, define its consecutive difference at index $n$ to be $a_{n+1}-a_n$. The sign of this difference records the direction of that step: a nonnegative difference means the next term is at least as large, while a nonpositive difference means it is at most as large. The definition of monotonicity can therefore be converted into a sign test.

Theorem: A real sequence $(a_n)$ is nondecreasing if and only if $a_{n+1}-a_n\geq0$ for every $n\in\mathbb{N}_0$. It is nonincreasing if and only if $a_{n+1}-a_n\leq0$ for every $n\in\mathbb{N}_0$.

Proof. For each $n$, subtracting $a_n$ from both sides of $a_n\leq a_{n+1}$ gives $0\leq a_{n+1}-a_n$. Conversely, adding $a_n$ to both sides of $0\leq a_{n+1}-a_n$ gives $a_n\leq a_{n+1}$. Thus the consecutive inequalities for nondecrease hold exactly when all the differences are nonnegative.

Similarly, subtracting $a_n$ from $a_n\geq a_{n+1}$ gives $0\geq a_{n+1}-a_n$. Adding $a_n$ to both sides of that difference inequality recovers $a_n\geq a_{n+1}$. So the differences are all nonpositive exactly when the sequence is nonincreasing. $\square$

This test is a method for organizing an argument, not a requirement to use subtraction. A difference may simplify to an expression with a clear sign; alternatively, the consecutive terms may be easier to compare by factoring or by using an earlier inequality.

Worked Example: A Nondecreasing Sequence with Repeated Terms

Let $b_n=\lfloor n/2\rfloor$ for $n\in\mathbb{N}_0$, where $\lfloor x\rfloor$ denotes the greatest integer less than or equal to $x$. If $n=2k$ is even, then

b_{n+1}-b_n =\left\lfloor\frac{2k+1}{2}\right\rfloor-\left\lfloor\frac{2k}{2}\right\rfloor =k-k=0.

If $n=2k+1$ is odd, then

b_{n+1}-b_n =\left\lfloor\frac{2k+2}{2}\right\rfloor-\left\lfloor\frac{2k+1}{2}\right\rfloor =(k+1)-k=1.

Every consecutive difference is either $0$ or $1$, so the sequence is nondecreasing. It is not strictly increasing at every step, because $b_1=b_0=0$, and indeed each even-indexed step has difference $0$. This illustrates why nondecreasing allows equality.

Monotonicity and Bounds

Monotonicity and boundedness are different properties. A sequence can move consistently in one direction while continuing without any bound in that direction. Still, the first term of a monotone sequence automatically supplies a bound on one side: a nondecreasing sequence never falls below its first term, and a nonincreasing sequence never rises above its first term.

Theorem: If $(a_n)$ is nondecreasing, then $a_0\leq a_n$ for every $n\in\mathbb{N}_0$, so $a_0$ is a lower bound. If $(a_n)$ is nonincreasing, then $a_n\leq a_0$ for every $n$, so $a_0$ is an upper bound.

Proof. In the nondecreasing case, apply the pairwise comparison theorem with $m=0$. It gives $a_0\leq a_n$ for every $n$. This is exactly the definition of $a_0$ being a lower bound. In the nonincreasing case, the same theorem gives $a_0\geq a_n$ for every $n$, which says that $a_0$ is an upper bound. $\square$

This result supplies only one of the two bounds needed for boundedness. By the theorem from “Upper and Lower Bounds for Sequences,” boundedness requires both an upper and a lower bound. A nondecreasing sequence still needs an upper bound to be bounded; a nonincreasing sequence still needs a lower bound.

Worked Example: Monotone Does Not Mean Bounded

Consider $c_n=-2n+4$. Its consecutive difference is

c_{n+1}-c_n =\bigl(-2(n+1)+4\bigr)-(-2n+4) =-2.

Since $-2<0$, the sequence is nonincreasing. The theorem gives $c_n\leq c_0=4$ for every $n$, so it is bounded above. But it is not bounded below: given any proposed lower bound $L\in\mathbb{R}$, the Archimedean Property gives a nonnegative integer $n$ large enough that $2n>4-L$. Rearranging yields $4-2n<L$, so $c_n<L$. Therefore this sequence is monotone but not bounded.

Oscillation and a Common Pitfall

A sequence that changes direction is not monotone, even if its terms remain in a fixed interval or its values seem to follow a pattern. For example, $d_n=(-1)^n$ takes the values $1,-1,1,-1,\ldots$. It cannot be nondecreasing because $d_0=1$ and $d_1=-1$, so $d_0\leq d_1$ fails. It cannot be nonincreasing because $d_1=-1$ and $d_2=1$, so $d_1\geq d_2$ fails.

A related error is to infer monotonicity from a few terms. Checking several consecutive inequalities only verifies those particular steps. The definition requires the comparison for every index, so a proof must establish the sign of every consecutive difference, or otherwise verify the required comparisons for all pairs of indices. A formula involving $n$ may require a parity split or another case distinction before its sign is clear.

It is also important not to confuse “monotone” with “strictly increasing” or “strictly decreasing.” Under the definitions here, a monotone sequence may have repeated terms, and a constant sequence is monotone in both directions. The next tutorial considers increasing and decreasing sequences in further detail.

Check Your Understanding

Use the definitions and results in this tutorial to answer the following questions.

What inequalities define a nondecreasing sequence and a nonincreasing sequence?
Why does $a_{n+1}-a_n\geq0$ for every $n$ imply $a_m\leq a_n$ whenever $m\leq n$?
For $a_n=n^2+1$, compute $a_{n+1}-a_n$ and use its sign to classify the sequence as nondecreasing or nonincreasing.
What bound does the first term provide for a nonincreasing sequence? Why does this alone not establish boundedness?
Give a pair of consecutive terms that rules out nondecrease for $(-1)^n$, and a pair that rules out nonincrease.

What You'll Learn