Sequence Converges to Within Half Limit

Theorem

Sequence of Real Numbers

Let $\left \langle {x_n} \right \rangle$ be a sequence in $\R$ or $\Q$.

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

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

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

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

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

Then:

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

Proof

Proof for Sequence of Real Numbers

Suppose $l > 0$.

From the definition of convergence to a limit:

$\forall \epsilon > 0: \exists N: \forall n > N: \left\vert{x_n - l}\right\vert < \epsilon$

That is, $l - \epsilon < x_n < l + \epsilon$.

As this is true for all $\epsilon > 0$, it is also true for $\epsilon = \dfrac l 2$ for some value of $N$.

Thus:

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

as required.

Now suppose $l < 0$.

By a similar argument:

$\forall \epsilon > 0: \exists N: \forall n > N: l - \epsilon < x_n < l + \epsilon$

Thus it is also true for $\epsilon = - \dfrac l 2$ for some value of $N$.

Thus:

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

as required.

$\blacksquare$

Proof for Sequence of Complex Numbers

Suppose $l > 0$.

Let us choose $N$ such that $\displaystyle \forall n > N: \left\vert{z_n - l}\right\vert < \frac {\left\vert{l}\right\vert} 2$.

Then:

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

$\blacksquare$

Note

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.

This is used in the Quotient Rule in Combination Theorem for Sequences.

Sources

• K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 4.29 \ (2)$