Convergent Real Sequence is Bounded/Proof 2
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Theorem
Let $\sequence {x_n}$ be a sequence in $\R$.
Let $l \in A$ such that $\ds \lim_{n \mathop \to \infty} x_n = l$.
Then $\sequence {x_n}$ is bounded.
Proof
Let $\sequence {x_n}$ be a sequence in $\R$.
Let $x_n \to l$ as $n \to \infty$.
To show that $\sequence {x_n}$ is bounded sequence, we need to find $K$ such that:
- $\forall n \in \N: \size {x_n} \le K$
Because $\sequence {x_n}$ converges:
- $\forall \epsilon > 0: \exists N: n > N \implies \size {x_n - l} < \epsilon$
In particular, this is true when $\epsilon = 1$.
That is:
- $\exists N_1: \forall n > N_1: \size {x_n - l} < 1$
By Backwards Form of Triangle Inequality:
- $\forall n > N_1: \size {x_n} - \size l \le \size {x_n - l} < 1$
That is:
- $\size {x_n} < \size l + 1$
So we set:
- $K = \max \set {\size {x_1}, \size {x_2}, \ldots, \size {x_{N_1} }, \size l + 1}$
and the result follows.
$\blacksquare$
Sources
- 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 4$: Convergent Sequences: Some simple properties of convergent sequences: $\S 4.25$