monotone convergence theorem

Let \[\left\{ a_{n} \right\}\] be a set of values of a monotonic sequence. If \[\left\{ a_{n} \right\}\] is bounded then it will converge. In addition, if this is the case then:

• If it is monotonically increasing, then it converges to \[\sup \left\{ a_{n}\mid n\in\mathbb{N} \right\}\].
  
• If it is monotonically decreasing, then it converges to \[\inf \left\{ a_{n}\mid n\in\mathbb{N} \right\}\].
  

For the notations, see infimum and supremum. \[\sup \left\{ a_{n}\mid n\in\mathbb{N} \right\}\] here refers to the supremum of the set of all elements \[a_{n}\], or in other words, the supremum of set \[\left\{ a_{1},a_{2},a_{3},\dots,a_{n} \right\}\].

By assumption, \[\left\{ a_{n} \right\}\] is non-empty and bounded above. Define \[S\] to be the set of terms in \[\left\{ a_{n} \right\}\] and define \[L=\sup(S)\] which exists by the Completeness axiom since \[S\] is bounded. Our goal is to prove that \[\lim_{n\to\infty}a_{n}=L\].

According to the definition of limits at infinity:

\[\lim_{n\to\infty}a_{n}=L\] is true if for every number \[\epsilon>0\], there is a \[N>0\] such that \[\left| a_{n}-L \right|<\epsilon\] whenever \[n>N\].

Let \[\epsilon>0\], since \[L-\epsilon\] is not an upper bound for \[S\] (as \[L\] is already the least upper bound), we know there must be some \[N\] such that \[L\ge a_{N}>L-\epsilon\] for every \[\epsilon>0\]. This can simply be explained by listing the sequence out:

\begin{align*} a_{1}\le a_{2}\le a_{3}\le\dots\le L-\epsilon\le a_{N}\le \dots\le a_{n}\le L<\infty\\ \end{align*}

The concept is similar to the formal definition of limit, where you can always find a number that's closer to \[L\] no matter how small \[\epsilon\] is.

Since \[L\] is an upper bound (for every \[n\], \[a_{n}\le L\]) and is the least upper bound (for any number smaller than \[L\], say \[L-\epsilon\] there must be some term in the sequence \[\left\{ a_{n} \right\}\] that exceeds it), logically, \[a_{n}\le L<L+\epsilon\] and \[L\ge a_{n}\ge a_{N}>L-\epsilon\]. We can now conclude for all \[n>N\], \[L-\epsilon<a_{n}<L+\epsilon\], or \[\left| a_{n}-L \right|<\epsilon\].

Therefore, according to the definition of limit, \[\lim_{n\to\infty}a_{n}=L\].

The proof for a monotonically decreasing sequence is similar.