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Proof of the Monotone Convergence Theorem

See how completeness turns monotonicity and a one-sided bound into convergence, and follow the epsilon–N argument in both directions.

Intermediate 9 min read

What You'll Learn

  • Use the least-upper-bound property to locate a candidate limit for a bounded nondecreasing sequence
  • Prove the approximation property of a supremum and infimum
  • Convert monotonicity and a bound into the epsilon–N definition of convergence
  • Prove the nonincreasing case using the greatest lower bound
  • Apply the proof to bounded partial sums and finite products

Completeness Supplies the Candidate Limit

In “The Monotone Convergence Theorem,” we saw that a nondecreasing sequence bounded above converges, and that a nonincreasing sequence bounded below also converges. The key question is how to prove convergence when the limit has not yet been identified. The answer comes from completeness of the real numbers: a nonempty set of real numbers bounded above has a least upper bound, and a nonempty set bounded below has a greatest lower bound.

For a nondecreasing sequence, collect all its terms into a set. Its least upper bound is a natural candidate for the limit: every term is at most this number, and some term must come arbitrarily close to it. Monotonicity then ensures that once a term is close, every later term remains close. For a nonincreasing sequence, the same reasoning uses a greatest lower bound.

Lemma (Approximation Property of the Supremum and Infimum): Let \(S\) be a nonempty set of real numbers. If \(S\) is bounded above and \(L=\sup S\), then for every \(\varepsilon>0\) there is an \(s\in S\) such that \(L-\varepsilon<s\leq L\). If \(S\) is bounded below and \(L=\inf S\), then for every \(\varepsilon>0\) there is an \(s\in S\) such that \(L\leq s<L+\varepsilon\).

Proof. Suppose first that \(L=\sup S\). By definition, \(L\) is an upper bound for \(S\), so \(s\leq L\) for every \(s\in S\). Fix \(\varepsilon>0\). The number \(L-\varepsilon\) cannot be an upper bound for \(S\), because it is strictly less than the least upper bound \(L\). Therefore some \(s\in S\) satisfies \(s>L-\varepsilon\). Together these inequalities give \(L-\varepsilon<s\leq L\).

Now suppose \(L=\inf S\). Then \(L\) is a lower bound, so \(L\leq s\) for every \(s\in S\). The number \(L+\varepsilon\) cannot be a lower bound for \(S\), because it is strictly greater than the greatest lower bound \(L\). Thus some \(s\in S\) satisfies \(s<L+\varepsilon\). Consequently, \(L\leq s<L+\varepsilon\), as required. \(\square\)

Proof of the Monotone Convergence Theorem

Theorem (Monotone Convergence Theorem): Every nondecreasing real sequence that is bounded above converges to a finite real number. Every nonincreasing real sequence that is bounded below converges to a finite real number.

Proof. First suppose \((a_n)\) is nondecreasing and bounded above. Consider its range

$$ S=\{a_n:n\in\mathbb{N}_0\}. $$

This set is nonempty and bounded above. By the least-upper-bound property of the real numbers, \(L=\sup S\) exists as a real number. We prove that \(a_n\to L\) using the epsilon–N definition of convergence.

Let \(\varepsilon>0\). By the Approximation Property of the Supremum and Infimum, there is an \(a_N\in S\) such that \(L-\varepsilon<a_N\leq L\). Here \(N\in\mathbb{N}_0\) is an index for a term with this value. For every \(n\geq N\), nondecreasing behavior gives \(a_N\leq a_n\). Also, \(L\) is an upper bound for the range, so \(a_n\leq L\). Hence

$$ L-\varepsilon<a_N\leq a_n\leq L, \qquad\text{so}\qquad 0\leq L-a_n<\varepsilon. $$

It follows that \(|a_n-L|<\varepsilon\) for every \(n\geq N\). Since this holds for every positive \(\varepsilon\), the definition of convergence gives \(a_n\to L\). The limit is finite because \(L\) is a real number.

Now suppose \((a_n)\) is nonincreasing and bounded below. Its range \(S=\{a_n:n\in\mathbb{N}_0\}\) is nonempty and bounded below, so the greatest lower bound \(L=\inf S\) exists in \(\mathbb{R}\). Given \(\varepsilon>0\), the approximation lemma supplies an index \(N\) such that \(L\leq a_N<L+\varepsilon\). For \(n\geq N\), nonincreasing behavior gives \(a_n\leq a_N\), while \(L\leq a_n\) because \(L\) is a lower bound for the range. Therefore

$$ L\leq a_n\leq a_N<L+\varepsilon, \qquad\text{so}\qquad 0\leq a_n-L<\varepsilon. $$

Thus \(|a_n-L|<\varepsilon\) for every \(n\geq N\). The epsilon–N definition gives \(a_n\to L\), again to a finite real number. This proves both statements. \(\square\)

Worked Examples: Applying the Proof

Worked Example: Partial Sums of Reciprocal Squares

For \(n\in\mathbb{N}_0\), define \(s_n=\sum_{k=1}^{n+1}1/k^2\). The new term added when moving from \(s_n\) to \(s_{n+1}\) is positive, since

$$ s_{n+1}-s_n=\frac{1}{(n+2)^2}>0. $$

So \((s_n)\) is nondecreasing. To find an upper bound, for every integer \(k\geq2\) observe that \(k^2\geq k(k-1)>0\), and therefore

$$ \frac{1}{k^2}\leq\frac{1}{k(k-1)} =\frac{1}{k-1}-\frac{1}{k}. $$

For every \(n\geq0\), summing these inequalities gives

$$ s_n =1+\sum_{k=2}^{n+1}\frac{1}{k^2} \leq 1+\sum_{k=2}^{n+1}\left(\frac{1}{k-1}-\frac{1}{k}\right) =2-\frac{1}{n+1} <2. $$

When \(n=0\), the sum from \(k=2\) to \(n+1\) is empty and equals zero; the displayed bound reads \(s_0=1\leq1\), so this edge case is included. Thus the sequence is bounded above by \(2\). The Monotone Convergence Theorem proves that \((s_n)\) converges, even though this argument does not need to calculate its limit.

Worked Example: A Decreasing Finite Product

Set \(p_0=1\), and for \(n\geq1\) define

$$ p_n=\prod_{j=2}^{n+1}\left(1-\frac{1}{j^2}\right). $$

Every factor is positive and less than \(1\). In particular, for \(n\geq0\), \(p_{n+1}=p_n\left(1-1/(n+2)^2\right)\), so \(0<p_{n+1}<p_n\). The sequence is strictly decreasing. To verify a lower bound, rewrite each factor and telescope:

$$ p_n =\prod_{j=2}^{n+1}\frac{(j-1)(j+1)}{j^2} =\left(\prod_{j=2}^{n+1}\frac{j-1}{j}\right) \left(\prod_{j=2}^{n+1}\frac{j+1}{j}\right) =\frac{1}{n+1}\cdot\frac{n+2}{2} =\frac{n+2}{2(n+1)}. $$

For \(n=0\), the product is empty and equals \(1\); the final formula also gives \(2/2=1\). For every \(n\geq0\), the formula yields \(p_n=1/2+1/(2(n+1))\geq1/2\). Thus \((p_n)\) is nonincreasing and bounded below. The Monotone Convergence Theorem guarantees a finite limit. The formula and \(1/(n+1)\to0\) further identify that limit as \(1/2\).

Worked Example: A Telescoping Sum with a Supremum Limit

Define \(t_n=\sum_{k=1}^{n+1}1/[k(k+2)]\). Each added term is positive, so \(t_{n+1}-t_n=1/[(n+2)(n+4)]>0\), and \((t_n)\) is nondecreasing. The partial fractions satisfy

$$ \frac{1}{k(k+2)} =\frac12\left(\frac1k-\frac{1}{k+2}\right). $$

Indeed, the right-hand side is \(\frac12((k+2-k)/(k(k+2)))=1/[k(k+2)]\). Summing and cancelling the intermediate terms gives

$$ t_n =\frac34-\frac{1}{2(n+2)}-\frac{1}{2(n+3)} \leq\frac34. $$

The equality follows by expanding the finite sum: the uncancelled positive terms are \(1+1/2\), and the two remaining negative terms are \(1/(n+2)\) and \(1/(n+3)\), all multiplied by \(1/2\). Thus the sequence is bounded above by \(3/4\), and the theorem proves convergence. Since both reciprocal terms tend to zero, the formula also shows that the limit is \(3/4\). In the proof of the theorem, this number is precisely the supremum of the set of terms.

Why the Bound Must Be in the Direction of Motion

The proof depends on choosing a real least upper bound or greatest lower bound. For a nondecreasing sequence, an upper bound prevents the terms from increasing without limit; for a nonincreasing sequence, a lower bound prevents them from decreasing without limit. A bound in the other direction does not do this. For instance, \(a_n=n+1\) is nondecreasing and bounded below, but its range has no finite upper bound, so this proof cannot produce a real supremum.

A common gap is to say that the sequence “approaches its supremum” without showing why. The approximation lemma supplies the needed term within \(\varepsilon\) of the supremum. Monotonicity then carries that estimate forward to every later term. For a decreasing sequence, a term close to the infimum gives the corresponding estimate for every later term. These are the two precise steps that turn completeness into convergence.

The theorem proves that a limit exists; it does not always identify that limit in a convenient form. The third example identifies the limit by an explicit formula, while the first only needs monotonicity, a bound, and completeness to establish convergence. Keeping existence separate from calculation is useful whenever a sequence is defined recursively or by partial sums.

Check Your Understanding

Use the proof and examples to answer the following questions.

  1. Why can \(L-\varepsilon\) not be an upper bound when \(L\) is the supremum of a set?
  2. In the nondecreasing case of the proof, which inequality comes from monotonicity, and which comes from the supremum being an upper bound?
  3. How does the approximation property of the infimum enter the proof for a nonincreasing sequence?
  4. Why does the Monotone Convergence Theorem not apply to \(a_n=n+1\) as a sequence converging to a finite real number?
  5. For the product sequence \(p_n\), what lower bound allows the theorem to be applied?

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