# Divergent Sequence may be Bounded/Proof 1

## Theorem

While every Convergent Sequence is Bounded, it does not follow that every bounded sequence is convergent.

That is, there exist bounded sequences which are divergent.

## Proof

Let $\sequence {x_n}$ be the sequence in $\R$ defined as:

- $x_n = \paren {-1}^n$

It is clear that $\sequence {x_n}$ is bounded: above by $1$ and below by $-1$.

Aiming for a contradiction, suppose $x_n \to l$ as $n \to \infty$.

Let $\epsilon > 0$.

Then $\exists N \in \R: \forall n > N: \size {\paren {-1}^n - l} < \epsilon$.

But there are values of $n > N$ for which $\paren {-1}^n = \pm 1$.

It follows that $\size {1 - l} < \epsilon$ and $\size {-1 - l} < \epsilon$.

From the Triangle Inequality for Real Numbers, we have:

\(\displaystyle 2\) | \(=\) | \(\displaystyle \size {1 - \paren {-1} }\) | |||||||||||

\(\displaystyle \) | \(\le\) | \(\displaystyle \size {1 - l} + \size {l - \paren {-1} }\) | |||||||||||

\(\displaystyle \) | \(<\) | \(\displaystyle 2 \epsilon\) |

This is a contradiction whenever $\epsilon < 1$.

Thus $\sequence {x_n}$ has no limit and, while definitely bounded, is unmistakably divergent.

$\blacksquare$

## Sources

- 1977: K.G. Binmore:
*Mathematical Analysis: A Straightforward Approach*... (previous) ... (next): $\S 4$: Convergent Sequences: $\S 4.27$: Example