Sequence Converges to Within Half Limit/Complex Numbers
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Theorem
Let $\sequence {z_n}$ be a sequence in $\C$.
Let $\sequence {z_n}$ be convergent to the limit $l$.
That is, let $\ds \lim_{n \mathop \to \infty} z_n = l$ where $l \ne 0$.
Then:
- $\exists N: \forall n > N: \cmod {z_n} > \dfrac {\cmod l} 2$
Proof
Suppose $l > 0$.
Let us choose $N$ such that:
- $\forall n > N: \cmod {z_n - l} < \dfrac {\cmod l} 2$
Then:
\(\ds \cmod {z_n - l}\) | \(<\) | \(\ds \frac {\cmod l} 2\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \cmod l - \cmod {z_n}\) | \(\le\) | \(\ds \cmod {z_n - l}\) | Reverse Triangle Inequality | ||||||||||
\(\ds \) | \(<\) | \(\ds \frac {\cmod l} 2\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \cmod {z_n}\) | \(>\) | \(\ds \cmod l - \frac {\cmod l} 2\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\cmod l} 2\) |
$\blacksquare$