TheoremProved

Monotone convergence theorem

Statement

If $(a_n)$ is increasing and bounded above, then it converges, and $a_1\le a_2\le\cdots\le M\ \Rightarrow\ \lim_{n\to\infty}a_n=\sup_n a_n$ (symmetrically for decreasing sequences bounded below).

Why is it true?

Intuitively, a sequence that only ever climbs but can never pass a ceiling has nowhere else to go except settle right below that ceiling.

Proof sketch

Let $(a_n)$ be increasing and bounded above. By the completeness axiom of $\mathbb{R}$ , the set $\{a_n : n\in\mathbb{N}^*\}$ has a least upper bound $L=\sup_n a_n$ .

Fix $\varepsilon>0$ . Since $L$ is the least upper bound, $L-\varepsilon$ is not an upper bound, so there exists an index $N$ with $a_N>L-\varepsilon$ .

Because $(a_n)$ is increasing, for every $n>N$ we have $a_n\ge a_N>L-\varepsilon$ . Also $a_n\le L$ for every $n$ , since $L$ is an upper bound of the whole sequence.

Combining both bounds, $L-\varepsilon<a_n\le L<L+\varepsilon$ for all $n>N$ , i.e. $|a_n-L|<\varepsilon$ . Since $\varepsilon>0$ was arbitrary, $\lim_{n\to\infty} a_n = L$ holds by the ε–N definition. $\blacksquare$

Topics that use this theorem

Limits of sequences

Step-by-step proofs

No step-by-step proof yet for this theorem.

References

James Stewart (2015). Calculus: Early Transcendentals
Judith V. Grabiner (1983). Who Gave You the Epsilon? Cauchy and the Origins of Rigorous Calculus