$$\def \C {\mathbb{C}}\def \R {\mathbb{R}}\def \N {\mathbb{N}}\def \Z {\mathbb{Z}}\def \Q {\mathbb{Q}}\def\defn#1{{\bf{#1}}}$$

Now we come to a very useful method to show convergence: A bounded and monotonic sequence converges. Let’s do bounded above and increasing first. The corresponding result for bounded below and decreasing follows as a simple corollary.

**Theorem**

If $(a_n)$ is increasing and bounded above, then $(a_n)$ is convergent.

**Proof**

Since $(a_n)$ is bounded above there exists at least one bound (and possibly many more). Let $L$ be the smallest possible bound. (That such a number exists is intuitively obvious but is a very technical and subtle point. In effect we have to take it as an axiom and this is referred to as the **completeness axiom**.)

Given $\varepsilon >0$, then there exists $N\in \N $ such that $L-\varepsilon \lt a_N$. To see this suppose for a contradiction that $a_n\leq L-\varepsilon$ for all $n\in \N$. Then $L-\varepsilon $ is a bound for $(a_n)$ that is smaller than $L$. This contradicts that $L$ is the smallest possible bound.

As $(a_n)$ is an increasing sequence we have, for all $n>N$,

\[

L-\varepsilon \lt a_N\leq a_n\leq L \lt L+\varepsilon .

\]

This means $|a_n-L| \lt \varepsilon $ for all $n>N$. That is, $a_n\to L$.

▢

There is a similar statement for bounded decreasing sequences.

**Theorem**

If $(a_n)$ is decreasing and bounded below, then $(a_n)$ is convergent.

**Proof**

This theorem is in fact a simple corollary of the previous theorem. We let $b_n=-a_n$ and apply the theorem to $b_n$ to deduce that $b_n$ has a limit, say $b$. From the Scalar Rule we know that $a_n$ has a limit and it is equal to $-b$.

▢

**Remark**

The proofs of the the preceding theorems tell us what the limit of the sequence is. For an increasing sequence the limit is the smallest upper bound for $(a_n)$.

For a decreasing sequence the limit is the largest lower bound for $(a_n)$.