Divergent Sequence may be Bounded

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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 $\R$ 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 $\R$ 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.

$\blacksquare$


Sources