Monotone Convergence Theorem (Real Analysis)

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 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 $$\left \langle {x_n} \right \rangle$$.

$$ $$ $$

Hence the result.

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.