# Divergent Sequence may be Bounded

## 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 1

Let $\sequence {x_n}$ be the sequence in $\R$ which forms the basis of Grandi's series, 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:

 $\ds 2$ $=$ $\ds \size {1 - \paren {-1} }$ $\ds$ $\le$ $\ds \size {1 - l} + \size {l - \paren {-1} }$ $\ds$ $<$ $\ds 2 \epsilon$

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

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

$\blacksquare$

## Proof 2

Let $\sequence {x_n}$ be the sequence in $\R$ which forms the basis of Grandi's series, defined as:

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

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

Note the following subsequences of $\sequence {x_n}$:

$(1): \quad \sequence {x_{n_r} }$ where $\sequence {n_r}$ is the integer sequence defined as $n_r = 2 r$
$(2): \quad \sequence {x_{n_s} }$ where $\sequence {n_s}$ is the integer sequence defined as $n_s = 2 s + 1$.

We have that:

$\sequence {x_{n_r} }$ is the sequence $1, 1, 1, 1, \ldots$
$\sequence {x_{n_s} }$ is the sequence $-1, -1, -1, -1, \ldots$

So $\sequence {x_n}$ has two subsequences with different limits.

From Limit of Subsequence equals Limit of Real Sequence, that means $\sequence {x_n}$ can not be convergent.

$\blacksquare$