Monotone Convergence Theorem¶

Theorem¶

Let $(a_n)_{n=1}^\infty$ be a sequence of real numbers which has some finite upper bound $M \in \mathbb{R}\text{,}$ and which is also monotonically increasing. Then $(a_n)_{n=1}^\infty$ is convergent, and $\displaystyle \lim_{n\to\infty} a_n=\sup\limits_{n\geq 1} a_n\leq M\text{,}$ where $\sup\limits_{n\geq 1} a_n$ is the supremum of the set $\{a_n\in\mathbb{R}\colon n\geq 1\}\text{.}$

Interpretation¶

The monotone convergence theorem gives convergence from two facts that are often easy to check: the sequence never goes down, and it never rises above a fixed number $M$ . It also names the limit, the supremum of the terms, so the limit need not be guessed in advance as in the definition of a limit.

An increasing sequence that stays below an upper bound M and approaches its supremum, drawn as a horizontal line below M.

The figure shows $a_n=2-\frac{3}{n+1}$ $\frac{3}{n + 1}$ , which starts at $a_1=\tfrac{1}{2}$ and rises with every step. The upper dashed line is one upper bound $M$ ; the lower one is the least upper bound, $\sup_n a_n=2$ $2$ , and the terms approach this line rather than $M$ .

Move the candidate bound $M$ : any upper bound works for the theorem, and the increasing sequence then converges to its supremum, which is the smallest of them.

The proof shows why the limit is the supremum. Given $\varepsilon>0$ $0$ , the number $\sup_n a_n-\varepsilon$ is not an upper bound, so some $a_N$ lies above it, and since the sequence is increasing, every later term lies between $a_N$ and $\sup_n a_n$ . Both hypotheses are needed: $a_n=n$ is increasing but unbounded, and the bounded sequence $(-1)^n$ is not monotone; neither converges. A decreasing sequence that is bounded below converges to its infimum.

The theorem relies on the completeness of the real numbers and fails in $\mathbb{Q}$ : the decimal truncations $1,\ 1.4,\ 1.41,\ 1.414,\dots$ of $\sqrt{2}$ form an increasing, bounded sequence of rationals with no rational limit. Euler's number is the limit of such an increasing bounded sequence, $\left(1+\frac{1}{n}\right)^n$ , as the Examples show.

A common slip is to take the bound $M$ for the limit. The theorem only gives $\lim_{n\to\infty}a_n\le M$ $M$ ; the limit is the least upper bound.

Key Results¶

Every bounded monotone sequence is convergent.
The limit is given by the supremum or infimum (given a decreasing sequence with lower bound) of the sequence.

Examples¶

$\displaystyle \lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n$ exists since $\displaystyle\left(1+\frac{1}{n}\right)^n$ is an increasing sequence with an upper bound $3\text{.}$
For $x\geq 0\text{,}$ the partial sums $\displaystyle s_n=\sum_{k=0}^{n}\frac{x^k}{k!}$ form an increasing sequence that is bounded above: for an integer $N\geq 2x\text{,}$ each term with $k>N$ is at most $\frac{x^N}{N!}\,2^{N-k}\text{,}$ so $s_n\leq s_N+\frac{x^N}{N!}$ for all $n\text{.}$ Hence $\displaystyle \sum_{k=0}^\infty \frac{x^k}{k!}$ exists by the monotone convergence theorem. (For $x<0$ the partial sums are not monotone, and the series converges for a different reason.)

Prerequisites¶

Euler's Number

Sources¶

Math I: Calculus · Lecture 3

Aliases: MCT Symbols: element of, epsilon, infinity, rational numbers, real numbers, square root, sum

Last updated: 20 October 2026 03:02