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 $M$ .

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