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Proof of the Week

Monotone Convergence Theorem

Theorem

Let \left \langle {x_n} \right \rangle be a sequence in \mathbb{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 ε > 0.

Since B − ε 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.


\forall n > N: B - \epsilon < x_n \le B                    
\Longrightarrow \forall n > N: B - \epsilon < xn < B + ε          Real Plus Epsilon          
\Longrightarrow \forall n > N: \left|{x_n - B}\right| < ε          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


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