Squeeze Theorem/Sequences/Complex Numbers

From ProofWiki
Jump to: navigation, search

Theorem

Let $\left \langle {a_n} \right \rangle$ be a sequence in $\R$ which is null, that is:

$a_n \to 0$ as $n \to \infty$

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


Suppose $\left \langle {a_n} \right \rangle$ dominates $\left \langle {z_n} \right \rangle$.

That is, suppose that:

$\forall n \in \N: \left|{z_n}\right| \le a_n$


Then $\left \langle {z_n} \right \rangle$ is a null sequence.


Proof

In order to show that $\left \langle {z_n} \right \rangle$ is a null sequence, we want to show that:

$\forall \epsilon > 0: \exists N: \forall n > N: \left|{z_n}\right| < \epsilon$

But since $\left \langle {a_n} \right \rangle$ is a null sequence:

$\exists N: \forall n > N: a_n < \epsilon$

So, using this value of $n$, we have: $\left \vert {z_n} \right \vert \le a_n < \epsilon$

Thus $\left\langle{ z_n }\right\rangle$ is a null sequence.

$\blacksquare$


Also known as

This result is also known, in the UK in particular, as the sandwich theorem.