Divergent Sequence may be Bounded/Proof 1
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 $\R$ 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.
$\blacksquare$
Sources
- 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 4$: Convergent Sequences: $\S 4.27$: Example