# Divergent Sequence may be Bounded/Proof 2

Jump to navigation
Jump to search

## 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

- 1977: K.G. Binmore:
*Mathematical Analysis: A Straightforward Approach*... (previous) ... (next): $\S 5$: Subsequences: $\S 5.3$ - 1977: K.G. Binmore:
*Mathematical Analysis: A Straightforward Approach*... (previous) ... (next): $\S 5$: Subsequences: Bolzano-Weierstrass Theorem: $\S 5.11$: Example $\text{(i)}$