Sequence Converges to Within Half Limit

Theorem

Sequence of Real Numbers

Let $\sequence {x_n}$ be a sequence in $\R$.

Let $\sequence {x_n}$ be convergent to the limit $l$.

That is, let $\displaystyle \lim_{n \mathop \to \infty} x_n = l$.

Suppose $l > 0$.

Then:

$\exists N: \forall n > N: x_n > \dfrac l 2$

Similarly, suppose $l < 0$.

Then:

$\exists N: \forall n > N: x_n < \dfrac l 2$

Sequence of Complex Numbers

Let $\sequence {z_n}$ be a sequence in $\C$.

Let $\sequence {z_n}$ be convergent to the limit $l$.

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

Then:

$\exists N: \forall n > N: \cmod {z_n} > \dfrac {\cmod l} 2$

Sequence in Normed Division Ring

Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring with zero $0$.

Let $\sequence {x_n}$ be a sequence in $R$.

Let $\sequence {x_n}$ be convergent in the norm $\norm {\, \cdot \,}$ to the following limit:

$\displaystyle \lim_{n \mathop \to \infty} x_n = l \ne 0$

Then:

$\exists N: \forall n > N: \norm {x_n} > \dfrac {\norm l} 2$

Also see

This is used in the Quotient Rule for Sequences.

Although this result seems a little trivial, it is often crucial to know that a sequence will be "eventually non-zero" so we know we can legitimately divide by it.