Divergent Sequence may be Bounded/Proof 2

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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

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