Eventual Monotonicity And Convergence

When Monotonicity Begins Later

A sequence need not be monotone from its first term in order to have a monotone tail. Its first few terms may behave differently, while all later terms move in one direction. This is enough to use the monotone convergence theorem, provided the tail also has the appropriate bound.

The reason is that convergence describes what happens when the index becomes large. Any fixed, finite collection of initial terms lies before that stage. It can affect the sequence’s first values, but not whether the terms eventually approach a limit.

Eventual Monotonicity

Definition: A sequence \((a_n)\) is eventually nondecreasing if there is an index \(N\in\mathbb{N}\) such that \(a_{n+1}\ge a_n\) for every \(n\ge N\). It is eventually nonincreasing if there is an index \(N\in\mathbb{N}\) such that \(a_{n+1}\le a_n\) for every \(n\ge N\). The terms from \(a_N\) onward form the tail beginning at \(N\).

The index \(N\) need not be the first index at which the sequence starts moving in one direction. Any index from which the required comparisons hold is suitable. Also, the inequalities allow equal consecutive terms: “nondecreasing” does not require strict increase, and “nonincreasing” does not require strict decrease.

For example, if \(a_{n+1}\ge a_n\) for all \(n\ge 4\), then the tail \(a_4,a_5,a_6,\ldots\) is nondecreasing. There is no requirement that \(a_2\ge a_1\) or \(a_3\ge a_2\).

The Tail Principle for Convergence

Deleting finitely many initial terms does not change whether a sequence converges, and it does not change its limit. More precisely, if \(N\) is fixed, the sequence \((a_n)\) converges to \(L\) if and only if its tail \((a_n)_{n\ge N}\) converges to \(L\).

This follows directly from the definition of convergence. A sequence converges to \(L\) when, for every \(\varepsilon>0\), there is an index \(K\) such that \(|a_n-L|<\varepsilon\) whenever \(n\ge K\). This condition only concerns terms after some index. Starting the sequence at a fixed later index \(N\) does not alter the condition: for the tail, choose an index at least as large as both \(N\) and the index required by \(\varepsilon\). Conversely, if the tail converges, then all sufficiently late terms of the original sequence belong to that tail and satisfy the same inequalities.

Theorem: If a sequence is eventually nondecreasing and its tail is bounded above, then the sequence converges. If it is eventually nonincreasing and its tail is bounded below, then the sequence converges. In the nondecreasing case, the limit is the supremum of the tail; in the nonincreasing case, it is the infimum of the tail.

To apply this result, use the tail starting at an index where monotonicity holds. The earlier material on the monotone convergence theorem establishes that a bounded monotone sequence converges: a nondecreasing sequence bounded above converges to its supremum, and a nonincreasing sequence bounded below converges to its infimum. Apply that theorem to the tail, then use the tail principle to conclude that the original sequence has the same limit.

It is enough to have the required bound on the tail. The original sequence is also bounded if its tail is bounded: there are only finitely many terms before the tail, and finitely many real numbers have both a largest and a smallest value. Those initial terms do not need to satisfy the tail’s monotonicity pattern.

Worked Examples

Worked Example: An Eventually Increasing Sequence

Define \(a_1=12\), and for \(n\ge 2\), let \(a_n=2-\frac{1}{n}\). Determine whether the sequence converges and find its limit.

The first terms are \(a_1=12\), \(a_2=2-\frac12=\frac32\), \(a_3=2-\frac13=\frac53\), and \(a_4=2-\frac14=\frac74\). The first step is a decrease, so the whole sequence is not nondecreasing. For \(n\ge 2\), however,

a_{n+1}-a_n =\left(2-\frac{1}{n+1}\right)-\left(2-\frac{1}{n}\right) =\frac{1}{n}-\frac{1}{n+1} =\frac{1}{n(n+1)}>0.

Thus the tail beginning at \(n=2\) is increasing. It is bounded above by \(2\), because \(a_n=2-\frac1n<2\) for each \(n\ge2\). The monotone convergence theorem shows that this tail converges, and so does the full sequence. Its limit is \(2\), since

|a_n-2|=\frac{1}{n}\longrightarrow 0 \qquad (n\ge2).

This also checks the proposed limit directly. The single initial value \(a_1=12\) does not affect it.

Worked Example: An Eventually Decreasing Sequence

Define \(b_1=-10\), and for \(n\ge2\), let \(b_n=5+\frac{1}{n}\). Show that the sequence converges and identify its limit.

The first terms are \(b_1=-10\), \(b_2=5+\frac12=\frac{11}{2}\), \(b_3=5+\frac13=\frac{16}{3}\), and \(b_4=5+\frac14=\frac{21}{4}\). For \(n\ge2\),

b_{n+1}-b_n =\left(5+\frac{1}{n+1}\right)-\left(5+\frac{1}{n}\right) =\frac{1}{n+1}-\frac{1}{n} =-\frac{1}{n(n+1)}<0.

The tail beginning at \(n=2\) is decreasing. Each of its terms is greater than \(5\), so it is bounded below by \(5\). By the monotone convergence theorem, the tail and hence the original sequence converge. For \(n\ge2\),

|b_n-5|=\frac{1}{n}\longrightarrow 0.

Therefore the limit is \(5\). The initial value \(-10\) does not violate the tail’s lower bound and does not change the limit.

Worked Example: Finding the Correct Tail Index

Let \(c_1=100\), \(c_2=-40\), \(c_3=7\), and for \(n\ge4\), let \(c_n=1-\frac{1}{n-3}\). Find a tail on which the sequence is monotone, and determine its limit.

The first terms of the tail are \(c_4=1-1=0\), \(c_5=1-\frac12=\frac12\), and \(c_6=1-\frac13=\frac23\). For \(n\ge4\), the consecutive difference is

c_{n+1}-c_n =\left(1-\frac{1}{n-2}\right)-\left(1-\frac{1}{n-3}\right) =\frac{1}{n-3}-\frac{1}{n-2} =\frac{1}{(n-3)(n-2)}>0.

Thus the sequence is increasing from \(n=4\) onward. Its tail is bounded above by \(1\), since \(c_n=1-\frac{1}{n-3}<1\) for \(n\ge4\). It converges by the monotone convergence theorem. Moreover,

|c_n-1|=\frac{1}{n-3}\longrightarrow 0 \qquad (n\ge4),

so its limit is \(1\). The values \(100,-40,7\) show why it matters to identify the correct starting index: they are not part of the increasing tail. They are nevertheless only finitely many terms and do not change the limit.

A Complete Justification of the Method

The argument used in each example has two parts. First, the monotone convergence theorem applies to a tail; second, convergence of that tail transfers to the original sequence. The transfer can be made explicit with the definition of convergence.

Choose a tail index.
Let \(N\) be an index after which the sequence is monotone. Consider the sequence beginning with \(a_N\).

Apply monotone convergence to the tail.
If the tail is nondecreasing and bounded above, it converges to its supremum. If it is nonincreasing and bounded below, it converges to its infimum. Call this limit \(L\).

Return to the original sequence.
For any \(\varepsilon>0\), tail convergence gives an index \(K\ge N\) such that \(|a_n-L|<\varepsilon\) for every \(n\ge K\). These are also terms of the original sequence, so the original sequence converges to \(L\).

This justification also explains why the limit is not determined by the initial segment. Once \(n\ge K\), the finitely many terms with index less than \(K\) are irrelevant to the inequalities that define convergence.

Common Pitfall: Requiring Monotonicity from the First Term

A common mistake is to test only whether every consecutive pair in the full sequence moves in the same direction. For the sequence in the first example, \(a_2-a_1=\frac32-12=-\frac{21}{2}<0\). It would be incorrect to say that the entire sequence is increasing. But it would also be incorrect to conclude that the monotone convergence theorem cannot help.

a_2-a_1=-\frac{21}{2}<0, \qquad a_{n+1}-a_n=\frac{1}{n(n+1)}>0\quad(n\ge2).

The correction is to state the index range precisely: the sequence is not nondecreasing from \(n=1\), but it is nondecreasing for \(n\ge2\). Apply the theorem to that bounded tail, then use the tail principle. In general, do not let a finite initial irregularity obscure what happens for all sufficiently large indices.

Why Eventual Monotonicity Is Useful

In applications, a sequence may become monotone only after an initial adjustment. The first few terms could be selected separately, or the inequalities needed to prove monotonicity might hold only beyond a particular index. Eventual monotonicity lets you use the monotone convergence theorem without imposing an unnecessary condition on those early terms.

The method still requires both parts of the theorem’s hypothesis: monotonicity on a tail and a bound in the correct direction. An eventually increasing tail needs an upper bound; an eventually decreasing tail needs a lower bound. Once those conditions are verified, the finite initial segment neither prevents convergence nor changes the limit.

Check Your Understanding

For each question, pay attention to the index from which the stated property holds.

A sequence is nonincreasing for every \(n\ge5\) and its tail from \(n=5\) is bounded below. What conclusion follows, and what additional fact lets you apply it to the full sequence?
If a sequence converges to \(L\), must the tail beginning at \(n=10\) converge to \(L\)? Explain using the definition of convergence.
For \(d_n=3+\frac{1}{n}\) when \(n\ge2\), calculate \(d_{n+1}-d_n\) and determine whether this tail is increasing or decreasing.
Why does a finite initial term that lies above every term of a monotone tail not automatically prevent convergence?
What bound is needed to apply the monotone convergence theorem to an eventually nondecreasing sequence?

Eventual Monotonicity And Convergence

What You'll Learn

When Monotonicity Begins Later