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:

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

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

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

$\blacksquare$

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.

$\blacksquare$