How to Prove That a Monotone Bounded Sequence Converges

A monotone sequence moves in only one direction: it either never decreases or never increases. The monotone convergence theorem says that if such a sequence is bounded in the direction it moves, it has a finite limit. The key idea is to identify that limit with a supremum or infimum and use its defining property to show the sequence gets arbitrarily close.

The theorem

Let $(a_n)$ be a sequence of real numbers.

• If $(a_n)$ is nondecreasing and bounded above, then it converges to $\sup\{a_n:n\in\mathbb{N}\}$.
• If $(a_n)$ is nonincreasing and bounded below, then it converges to $\inf\{a_n:n\in\mathbb{N}\}$.

Here, nondecreasing means $a_n\leq a_{n+1}$ for every $n$; nonincreasing means $a_n\geq a_{n+1}$ for every $n$. Some texts use “increasing” and “decreasing” to mean strictly increasing or strictly decreasing, so stating the inequalities avoids that convention issue.

Proof for a nondecreasing sequence

Suppose $(a_n)$ is nondecreasing and bounded above. Let

$$L=\sup\{a_n:n\in\mathbb{N}\}.$$

The set of terms is nonempty and bounded above, so the least-upper-bound property of $\mathbb{R}$ guarantees that this supremum exists as a real number. We show that $a_n\to L$.

Take any $\varepsilon>0$. Because $L$ is the least upper bound, $L-\varepsilon$ cannot also be an upper bound: it is strictly less than $L$. Therefore, some term satisfies

$$a_N>L-\varepsilon.$$

Since the sequence is nondecreasing, every later term is at least as large as $a_N$. Since $L$ is an upper bound, every term is at most $L$. Thus, for every $n\geq N$,

$$L-\varepsilon<a_N\leq a_n\leq L,$$

so $|a_n-L|<\varepsilon$. This is precisely the definition of convergence, and therefore $a_n\to L$.

The proof uses two distinct facts: monotonicity keeps later terms above the term chosen near $L$, and the supremum property ensures there is a term close to $L$ from below.

Worked example: $a_n=1-1/n$

For $n\geq1$, the terms are $0, 1/2, 2/3, 3/4,\ldots$. The sequence is nondecreasing because

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

It is bounded above by $1$. In fact, its supremum is $1$: all terms are less than $1$, and for every $\varepsilon>0$ there is an $n>1/\varepsilon$ for which $1-1/n>1-\varepsilon$. The theorem therefore gives $a_n\to1$.

You can also check the limit directly. Given $\varepsilon>0$, choose an integer $N>1/\varepsilon$. For $n\geq N$,

$$|a_n-1|=\frac{1}{n}\leq\frac{1}{N}<\varepsilon.$$

This verifies the limit using the epsilon definition as well as illustrating the theorem.

The nonincreasing case

If $(a_n)$ is nonincreasing and bounded below, set $L=\inf\{a_n:n\in\mathbb{N}\}$. For any $\varepsilon>0$, the number $L+\varepsilon$ is not a lower bound, so some $a_N<L+\varepsilon$. For $n\geq N$, monotonicity and the lower-bound property give

$$L\leq a_n\leq a_N<L+\varepsilon.$$

Hence $|a_n-L|<\varepsilon$, and the sequence converges to its infimum. Equivalently, apply the nondecreasing result to $(-a_n)$.

Common mistakes

• Using the wrong bound: A nondecreasing sequence needs an upper bound for this theorem; a lower bound alone does not prevent it from growing without limit. A nonincreasing sequence needs a lower bound.
• Assuming monotonicity alone is enough: The sequence $a_n=n$ is nondecreasing but does not converge to a finite real number because it is not bounded above.
• Assuming boundedness alone is enough: The sequence $a_n=(-1)^n$ is bounded but does not converge, since it keeps alternating between $-1$ and $1$.
• Treating a supremum as a maximum: The supremum need not be one of the sequence’s terms. For $1-1/n$, the supremum is $1$, but no term equals $1$. The proof needs terms arbitrarily close to the supremum, not a term that attains it.

The theorem is a statement about real sequences, whose completeness guarantees the existence of the supremum and infimum used in the proof. In a setting without that completeness property, the same argument may not provide a limit in the space.