Eventually Monotone Sequences

Monotonicity After a Finite Initial Segment

The previous tutorial considered sequences that increase or decrease at every step. Some sequences do not behave monotonically from their first term, but do settle into a consistent direction after a finite number of terms. The exceptional initial segment may contain rises, falls, or both; what matters is that every later step obeys the same non-strict inequality.

For example, a sequence might begin with a few irregular values and then increase forever. In that case, the initial terms prevent the whole sequence from being nondecreasing, but they do not prevent its tail from being nondecreasing. The phrase eventually monotone records this distinction.

The index \(N\) is a threshold: the required inequalities apply from \(N\) onward. They do not have to hold before \(N\). When \(N=0\), the inequalities apply to the whole sequence, so ordinary nondecreasing or nonincreasing sequences are particular cases of eventually monotone sequences.

This definition permits equality at some or all later steps. If the inequalities are required to be strict for every \(n\geq N\), the corresponding terms are called eventually strictly increasing or eventually strictly decreasing. Those are stronger conditions than eventual nondecrease or eventual nonincrease.

Comparisons Throughout the Monotone Tail

The consecutive-term definition gives comparisons between any two terms in the tail. This is the same chaining principle used for monotone sequences, with the additional requirement that both indices lie at or beyond the threshold.

Proof. Suppose first that the sequence is eventually nondecreasing. Choose \(N\) such that \(a_k\leq a_{k+1}\) for every \(k\geq N\). Fix \(m,n\) with \(N\leq m\leq n\). If \(m=n\), then \(a_m=a_n\). If \(m<n\), the consecutive inequalities give

By transitivity, \(a_m\leq a_n\). Conversely, if there is an \(N\) such that \(a_m\leq a_n\) whenever \(N\leq m\leq n\), take \(m=k\) and \(n=k+1\) for each \(k\geq N\). This gives \(a_k\leq a_{k+1}\) at every such index, which is eventual nondecrease.

For the nonincreasing case, suppose \(a_k\geq a_{k+1}\) for every \(k\geq N\). Chaining these inequalities gives \(a_m\geq a_n\) whenever \(N\leq m\leq n\); equality holds when \(m=n\). Conversely, the pairwise condition applied to \(m=k\) and \(n=k+1\) gives \(a_k\geq a_{k+1}\) for every \(k\geq N\). This proves both equivalences. \(\square\)

Finite Initial Segments and Eventual Behavior

Eventual monotonicity depends only on what happens sufficiently far along the sequence. Changing a finite number of terms can affect whether the entire sequence is monotone, but it cannot change the eventual behavior if the terms after some index stay unchanged.

Proof. Suppose \((a_n)\) is eventually nondecreasing, and choose a threshold \(N\) so that \(a_n\leq a_{n+1}\) whenever \(n\geq N\). Let \(K\) be the larger of \(M\) and \(N\). For each \(n\geq K\), both \(n\) and \(n+1\) are at least \(M\), so \(a_n=b_n\) and \(a_{n+1}=b_{n+1}\). Also \(n\geq N\), so

Thus \((b_n)\) is eventually nondecreasing. Interchanging the roles of the two sequences proves the converse. If the inequality for \((a_n)\) is \(a_n\geq a_{n+1}\), the same argument gives \(b_n\geq b_{n+1}\); interchanging the sequences again gives the converse for eventual nonincrease. A sequence is eventually monotone precisely when one of these two alternatives holds, so the equivalence follows. \(\square\)

This result explains why a finite disturbance does not destroy eventual monotonicity. It does not say that finite initial terms never matter: they can still affect whether the whole sequence is bounded or monotone. It says only that they cannot determine whether the sequence has a monotone tail.

One-Sided Bounds from Eventual Monotonicity

A monotone tail provides a bound in the direction opposite its movement: a nondecreasing tail has a lower bound, and a nonincreasing tail has an upper bound. The finite initial segment can be included by taking a minimum or maximum over its terms together with the threshold term.

Proof. First suppose \((a_n)\) is eventually nondecreasing, and choose \(N\in\mathbb{N}_0\) such that \(a_n\leq a_{n+1}\) for all \(n\geq N\). By the pairwise comparison theorem, \(a_n\geq a_N\) for every \(n\geq N\). The finite set

is nonempty, including when \(N=0\), and therefore has a minimum. Let \(L\) be the minimum of this set. For \(0\leq n\leq N\), the definition of \(L\) gives \(L\leq a_n\). For \(n\geq N\), we have \(L\leq a_N\leq a_n\). Hence \(L\leq a_n\) for every \(n\in\mathbb{N}_0\), so \(L\) is a lower bound for the sequence.

Now suppose \((a_n)\) is eventually nonincreasing, with threshold \(N\). The pairwise comparison theorem gives \(a_n\leq a_N\) for every \(n\geq N\). The finite nonempty set \(\{a_0,a_1,\ldots,a_N\}\) has a maximum; call it \(U\). For \(0\leq n\leq N\), we have \(a_n\leq U\). For \(n\geq N\), we have \(a_n\leq a_N\leq U\). Thus \(U\) is an upper bound for the entire sequence. \(\square\)

The finite initial segment must be included in the bound. It is not enough to use \(a_N\) as a lower bound for an eventually nondecreasing sequence, because an earlier term may be smaller than \(a_N\). Taking the minimum through \(N\) handles those terms and also covers the edge case \(N=0\).

What Eventual Monotonicity Does—and Does Not—Say

An eventually nondecreasing sequence is guaranteed to be bounded below, but the theorem does not guarantee an upper bound. For instance, the sequence \(d_0=0\) and \(d_n=n\) for \(n\geq1\) is nondecreasing from the start and has no upper bound. For every proposed real upper bound \(B\), there is a positive integer \(n>B\), and then \(d_n=n>B\).

Likewise, an eventually nonincreasing sequence is guaranteed to be bounded above, but need not be bounded below. The direction of the tail determines which one-sided bound follows. To conclude that a sequence is bounded, one needs both an upper and a lower bound, as established earlier in this course.

Eventual monotonicity also does not mean strict monotonicity. For example, \(e_n=\lfloor n/4\rfloor\) is nondecreasing: increasing \(n\) cannot decrease its integer part. But \(e_0=e_1=0\), so it is not strictly increasing. The non-strict inequalities in the definition are essential; equality is allowed at infinitely many steps.

Finally, eventual monotonicity is a statement about order on a tail, not a claim about a limit. The pairwise comparison theorem and one-sided boundedness theorem describe what follows directly from the definition. Any later conclusion about convergence requires its own hypotheses and result.