Monotone Convergence Theorem

Theorem

Every bounded monotone sequence is convergent.

Let $\left \langle {x_n} \right \rangle$ be a sequence in $\R$.

Increasing Sequence

Let $\left \langle {x_n} \right \rangle$ be increasing and bounded above.

Then $\left \langle {x_n} \right \rangle$ converges to its supremum.

Decreasing Sequence

Let $\left \langle {x_n} \right \rangle$ be decreasing and bounded below.

Then $\left \langle {x_n} \right \rangle$ converges to its infimum.

Proof

Proof for Increasing Sequence

Suppose $\left \langle {x_n} \right \rangle$ is increasing and bounded above.

Let its supremum be $B$.

We need to show that $x_n \to B$ as $n \to \infty$.

Let $\epsilon > 0$.

Since $B - \epsilon$ is not an upper bound, by the definition of supremum.

Thus $\exists x_N: x_N > B - \epsilon$.

But $\left \langle {x_n} \right \rangle$ is increasing.

Hence $\forall n > N: x_n \ge x_N > B - \epsilon$.

But $B$ is still an upper bound for $\left \langle {x_n} \right \rangle$.

Then:

	\(\displaystyle \)	\(\displaystyle \forall n > N: B - \epsilon\)	\(<\)	\(\displaystyle x_n \le B\)	\(\displaystyle \)
	\(\displaystyle \implies\)	\(\displaystyle \forall n > N: B - \epsilon\)	\(<\)	\(\displaystyle x_n < B + \epsilon\)	\(\displaystyle \)	Real Plus Epsilon
	\(\displaystyle \implies\)	\(\displaystyle \forall n > N: \left\vert{x_n - B}\right\vert\)	\(<\)	\(\displaystyle \epsilon\)	\(\displaystyle \)	Negative of Absolute Value: Corollary

Hence the result.

$\blacksquare$

Proof for Decreasing Sequence

If $\left \langle {x_n} \right \rangle$ is decreasing and bounded below then $\left \langle {-x_n} \right \rangle$ is increasing and bounded above.

Thus the above result applies and the proof follows.

$\blacksquare$

Sources

• W.A. Sutherland: Introduction to Metric and Topological Spaces (1975): Theorem $1.2.6$
• K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 4.17$