Increasing and Decreasing Sequences

When Every Step Must Be Strict

The previous tutorial defined nondecreasing and nonincreasing sequences, for which equality between consecutive terms is allowed. This tutorial focuses on the strict versions: a sequence must rise at every step to be strictly increasing, or fall at every step to be strictly decreasing. Keeping the distinction clear matters because a sequence may be monotone without being strictly monotone.

Some mathematical texts use “increasing” and “decreasing” to mean the strict versions, while others use those words for non-strict monotonicity. Here we will use the explicit terms strictly increasing and strictly decreasing, so the required inequalities are unambiguous.

The word “every” is essential: one equality or one step in the opposite direction is enough to rule out strict increase or strict decrease, respectively. A constant sequence is both nondecreasing and nonincreasing, as established earlier in this course, but it is neither strictly increasing nor strictly decreasing.

Strict Comparisons Between Distant Terms

Strict increase is defined using consecutive terms, but its effect extends to any two terms in their index order. If the index increases from \(m\) to \(n\), with \(m<n\), at least one strict step occurs—and in fact every step in between is strict.

Proof. Suppose first that \((a_n)\) is strictly increasing. Fix indices \(m<n\). The consecutive inequalities in the definition give

By transitivity of the order relation, \(a_m<a_n\). Conversely, suppose \(a_m<a_n\) whenever \(m<n\). Taking \(m=k\) and \(n=k+1\) gives \(a_k<a_{k+1}\) for every \(k\in\mathbb{N}_0\), so the sequence is strictly increasing.

For the decreasing case, if \(a_k>a_{k+1}\) at every step, then for \(m<n\),

and hence \(a_m>a_n\). Conversely, the pairwise comparison applied to \(m=k\) and \(n=k+1\) gives \(a_k>a_{k+1}\) for every \(k\). This proves both equivalences. \(\square\)

The condition \(m<n\), rather than \(m\leq n\), is important here. When \(m=n\), the two terms are equal, so a strict inequality cannot hold. For non-strict monotonicity, the earlier pairwise comparison theorem uses \(m\leq n\); the strict version uses \(m<n\).

Testing Strictness with Consecutive Differences

As in the previous tutorial, the consecutive difference \(a_{n+1}-a_n\) is a useful way to test how a sequence changes. For strict monotonicity, the difference must have a strictly positive or strictly negative sign at every index. A difference of zero indicates an equal step, which is compatible with non-strict monotonicity but not strict monotonicity.

Proof. For each \(n\), subtracting \(a_n\) from the inequality \(a_n<a_{n+1}\) gives \(0<a_{n+1}-a_n\). Conversely, adding \(a_n\) to both sides of \(0<a_{n+1}-a_n\) gives \(a_n<a_{n+1}\). Therefore all the consecutive inequalities for strict increase hold exactly when every difference is positive.

For strict decrease, subtracting \(a_n\) from \(a_n>a_{n+1}\) gives \(0>a_{n+1}-a_n\). Adding \(a_n\) back to that inequality recovers \(a_n>a_{n+1}\). Thus the consecutive inequalities for strict decrease hold exactly when every difference is negative. \(\square\)

The sign must be strict at every index. Finding that the differences are nonnegative proves only that the sequence is nondecreasing; it does not rule out repeated consecutive terms. Similarly, nonpositive differences establish nonincrease, not necessarily strict decrease.

Strict Monotonicity Forces Distinct Terms

A strictly monotone sequence cannot repeat a value. Indeed, any two different indices are ordered, and the theorem on distant terms then gives a strict inequality between the corresponding terms. This provides a necessary consequence of strict monotonicity, though it does not work in reverse: a sequence can have distinct terms while repeatedly changing direction.

Proof. Let \(m\) and \(n\) be distinct nonnegative integers. One of the two inequalities \(m<n\) or \(n<m\) must hold.

If the sequence is strictly increasing and \(m<n\), the pairwise comparison theorem gives \(a_m<a_n\), so \(a_m\neq a_n\). If instead \(n<m\), that theorem gives \(a_n<a_m\), again implying \(a_m\neq a_n\).

If the sequence is strictly decreasing and \(m<n\), then \(a_m>a_n\), so the terms are unequal. If \(n<m\), then \(a_n>a_m\), which also gives \(a_m\neq a_n\). Thus in either direction distinct indices have distinct terms, proving injectivity. \(\square\)

Adding and Scaling Preserve or Reverse Direction

Simple changes to the terms can preserve strict monotonicity or reverse it. Adding the same constant to every term preserves all comparisons. Multiplying by a positive number also preserves comparisons, while multiplying by a negative number reverses them. This can be useful when a sequence is a transformed version of one whose order is already known.

Proof. First suppose \((a_n)\) is strictly increasing, so \(a_n<a_{n+1}\) for every \(n\). If \(c>0\), multiplying by \(c\) preserves the inequality, and adding \(d\) to both sides gives

Thus the transformed sequence is strictly increasing. If \(c<0\), multiplication reverses the inequality, so \(c a_n>c a_{n+1}\); adding \(d\) preserves that comparison, and the transformed sequence is strictly decreasing. If \(c=0\), every term equals \(d\), so it is constant.

Now suppose \((a_n)\) is strictly decreasing, so \(a_n>a_{n+1}\). For \(c>0\), multiplication and addition preserve that inequality, giving a strictly decreasing transformed sequence. For \(c<0\), multiplication reverses it, giving a strictly increasing sequence. For \(c=0\), the transformed sequence is again constant. This proves all the claims. \(\square\)

Strict and Non-Strict: A Common Distinction

A strict inequality at every step implies the corresponding non-strict inequality at every step, so every strictly increasing sequence is also nondecreasing, and every strictly decreasing sequence is also nonincreasing. The converse fails: nondecreasing sequences may have equal consecutive terms, and nonincreasing sequences may also have equal consecutive terms.

For example, \(e_n=\lfloor n/3\rfloor\) is nondecreasing: as \(n\) advances, the integer part never falls. But \(e_0=e_1=0\), so it is not strictly increasing. The difference test captures the distinction precisely: nondecrease requires \(a_{n+1}-a_n\geq0\), while strict increase requires \(a_{n+1}-a_n>0\) at every index. The analogous distinction for decrease is between nonpositive and strictly negative differences.

Strict monotonicity should also not be confused with positivity or boundedness. A strictly increasing sequence can begin with negative terms, and a strictly decreasing sequence can begin with positive terms. The direction of change alone does not determine whether the sequence has an upper or lower bound. In particular, the results here concern order between terms; they do not assert that the sequence approaches a real number.