Sequences · Definition

Monotone convergence theorem

real analysissequences

Before you start you need to know

It is the first tool for proving that a limit exists without knowing its value, and it rests directly on the completeness of \(\mathbb R\).

Definition

Every real sequence \((a_n)\), indexed by the natural numbers \(\mathbb N\), that is monotone and bounded converges to a real number.

\[a_n\ \text{increasing and bounded}\ \Rightarrow\ \lim a_n=\sup\{a_n:n\in\mathbb N\}\]

The idea of the proof for an increasing sequence that is bounded above is the following. The set \(\{a_n\}\) is nonempty and bounded above, so by the completeness axiom \(s=\sup\{a_n\}\) exists. Given \(\varepsilon>0\), the number \(s-\varepsilon\) is not an upper bound, so there is \(N\) with \(a_N>s-\varepsilon\). Because the sequence is increasing, for \(n\ge N\) we have \(s-\varepsilon<a_N\le a_n\le s\), that is, \(|a_n-s|<\varepsilon\).

The decreasing case reduces to the previous one by applying it to \(-a_n\): a decreasing sequence that is bounded below converges to its infimum. So the limit of a bounded monotone sequence is always identified with a supremum or an infimum, although computing it may require another technique.

A typical application is recursive sequences. If \(a_1=\sqrt2\) and \(a_{n+1}=\sqrt{2+a_n}\), induction shows \(a_n<2\) and \(a_n<a_{n+1}\); the theorem ensures convergence to some \(L\), and taking limits in the recurrence gives \(L^2=L+2\), hence \(L=2\). The order matters: first prove the limit exists, and only then solve the equation.

Both hypotheses are necessary. \(a_n=n\) is monotone but unbounded and diverges; \((-1)^n\) is bounded but not monotone and diverges. The theorem also fails in \(\mathbb Q\): an increasing sequence of rationals bounded by \(\sqrt2\) may fail to converge in \(\mathbb Q\), since its supremum is irrational. That difference is exactly completeness.

Think about it

Why is the theorem false if we work only with rational numbers?

Because \(\mathbb Q\) does not satisfy the completeness axiom: the decimal truncations of \(\sqrt2\) form an increasing bounded sequence of rationals whose supremum, \(\sqrt2\), is not rational.

If an increasing sequence converges to \(L\), can some term exceed \(L\)?

No: \(L\) is the supremum of the terms, so \(a_n\le L\) for all \(n\).

Exercises 10

  1. The sequence \(a_n=1-\frac1n\) is increasing and bounded above by 1. What does it converge to?
  2. The sequence \(a_n=\frac1{n^2}\) is decreasing and positive. What does it converge to?
  3. The sequence \(a_n=n\) is increasing but does not converge. Which hypothesis of the theorem fails?
  4. The sequence \(a_n=(-1)^n\) is bounded and does not converge. Why does this not contradict the theorem?
  5. Let \(a_1=\sqrt2\) and \(a_{n+1}=\sqrt{2+a_n}\), increasing and bounded by 2. If it converges to \(L\), what is \(L\)?
  6. Let \(a_1=0\) and \(a_{n+1}=\frac{a_n}{3}+2\), increasing and bounded by 3. What is its limit?
  7. An increasing sequence satisfies \(a_n\le7\) for all \(n\) and converges to \(L\). What follows?
  8. A decreasing sequence that is bounded below converges to:
  9. Which of these statements is FALSE?
  10. Let \(a_n=\sum_{k=0}^{n}\frac1{k!}\), increasing with \(a_n<3\) for all \(n\). What follows?

References

  • Stephen Abbott, Understanding Analysis, 2nd edition, Springer, 2015. §2.4, Teorema 2.4.2
  • Walter Rudin, Principles of Mathematical Analysis, 3rd edition, McGraw-Hill, 1976. Cap. 3, Teorema 3.14
  • Robert G. Bartle, Donald R. Sherbert, Introduction to Real Analysis, 4th edition, Wiley, 2011. §3.3, Teorema 3.3.2