Sequence Converges to Within Half Limit/Complex Numbers

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Theorem

Let $\left \langle {z_n} \right \rangle$ be a sequence in $\C$.

Let $\left \langle {z_n} \right \rangle$ be convergent to the limit $l$.

That is, let $\displaystyle \lim_{n \mathop \to \infty} z_n = l$ where $l \ne 0$.


Then:

$\exists N: \forall n > N: \left\vert{z_n}\right\vert > \dfrac {\left\vert{l}\right\vert} 2$


Proof

Suppose $l > 0$.

Let us choose $N$ such that:

$\forall n > N: \left\vert{z_n - l}\right\vert < \dfrac {\left\vert{l}\right\vert} 2$


Then:

\(\displaystyle \left\vert{z_n - l}\right\vert\) \(<\) \(\displaystyle \frac {\left\vert{l}\right\vert} 2\)
\(\displaystyle \implies \ \ \) \(\displaystyle \left\vert{l}\right\vert - \left\vert{z_n}\right\vert\) \(\le\) \(\displaystyle \left\vert{z_n - l}\right\vert\) Reverse Triangle Inequality
\(\displaystyle \) \(<\) \(\displaystyle \frac {\left\vert{l}\right\vert} 2\)
\(\displaystyle \implies \ \ \) \(\displaystyle \left\vert{z_n}\right\vert\) \(>\) \(\displaystyle \left\vert{l}\right\vert - \frac {\left\vert{l}\right\vert} 2\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {\left\vert{l}\right\vert} 2\)

$\blacksquare$